Physics Chapt 19

1867

Chapter 19 The Second Law of Thermodynamics Conceptual Problems 1 • Modern automobile gasoline engines have efficiencies of about 25%. About what percentage of the heat of combustion is not used for work but released as heat? (a) 25%, (b) 50%, (c) 75%, (d) 100%, (e) You cannot tell from the data given. Determine the Concept The efficiency of a heat engine is the ratio of the work done per cycle W to the heat absorbed from the high-temperature reservoir Qh. The percentage of the heat of combustion (heat absorbed from the high-temperature reservoir) is the ratio of Qc to Qh. We can use the relationship between W, Qh, and Qc ( ch QQW !" ) to find Qc/ Qh. Use the definition of efficiency and the relationship between W, Qh, and Qc to obtain:

h

c

h

ch

h

1QQ

QQQ

QW

!"!

""#

Solving for Qc/ Qh yields: #!"1

h

c

QQ

Substitute for # to obtain: 75.025.01

h

c "!"QQ

and $ %c is correct. 2 • If a heat engine does 100 kJ of work per cycle while releasing 400 kJ of heat, what is its efficiency? (a) 20%, (b) 25%, (c) 80%, (d) 400%, (e) You cannot tell from the data given. Determine the Concept The efficiency of a heat engine is the ratio of the work done per cycle W to the heat absorbed from the high-temperature reservoir Qh. We can use the relationship between W, Qh, and Qc ( ch QQW !" ) to express the efficiency of the heat engine in terms of Qc and W. Use the definition of efficiency and the relationship between W, Qh, and Qc to obtain:

WQQW

WQW

cch 1

1

&"

&""#

Chapter 19

1868

Substitute for Qc and W to obtain:

2.0

kJ 100kJ 4001

1"

&"#

and $ %a is correct. 3 • If the heat absorbed by a heat engine is 600 kJ per cycle, and it releases 480 kJ of heat each cycle, what is its efficiency? (a) 20%, (b) 80%, (c) 100%, (d) You cannot tell from the data given. Determine the Concept The efficiency of a heat engine is the ratio of the work done per cycle W to the heat absorbed from the high-temperature reservoir Qh. We can use the relationship between W, Qh, and Qc ( ch QQW !" ) to express the efficiency of the heat engine in terms of Qc and Qh. Use the definition of efficiency and the relationship between W, Qh, and Qc to obtain:

h

c

h

ch

h

1QQ

QQQ

QW

!"!

""#

Substitute for Qc and Qh to obtain: 2.0

kJ 600kJ 4801 "!"#

and $ %a is correct. 4 • Explain what distinguishes a refrigerator from a !heat pump.! Determine the Concept The job of a refrigerator is to move heat from its cold interior to the warmer kitchen environment. This process moves heat in a direction that is opposite its 'natural' direction of flow, analogous to the use of a water pump to pump water out of a boat. The term heat pump is used to describe devices, such as air conditioners, that are used to cool living and working spaces in the summer and warm them in the winter. 5 • [SSM] An air conditioner’s COP is mathematically identical to that

of a refrigerator, that is, cAC refCOP COP" "

QW

. However a heat pump’s COP is

defined differently, as hhpCOP "

QW

. Explain clearly why the two COPs are

defined differently. Hint: Think of the end use of the three different devices. Determine the Concept The COP is defined so as to be a measure of the effectiveness of the device. For a refrigerator or air conditioner, the important quantity is the heat drawn from the already colder interior, Qc. For a heat pump, the ideas is to focus on the heat drawn into the warm interior of the house, Qh.

The Second Law of Thermodynamics

1869

6 • Explain why you cannot cool your kitchen by leaving your refrigerator door open on a hot day. (Why does turning on a room air conditioner cool down the room, but opening a refrigerator door does not?) Determine the Concept As described by the second law of thermodynamics, more heat must be transmitted to the outside world than is removed by a refrigerator or air conditioner. The heating coils on a refrigerator are inside the room and so the refrigerator actually heats the room in which it is located. The heating coils on an air conditioner are outside one’s living space, so the waste heat is vented to the outside. 7 • Why do steam-power-plant designers try to increase the temperature of the steam as much as possible? Determine the Concept Increasing the temperature of the steam increases the Carnot efficiency, and generally increases the efficiency of any heat engine. 8 • To increase the efficiency of a Carnot engine, you should (a) decrease the temperature of the hot reservoir, (b) increase the temperature of the cold reservoir, (c) increase the temperature of the hot reservoir, (d) change the ratio of maximum volume to minimum volume. Determine the Concept Because the efficiency of a Carnot cycle engine is given

byh

cC 1

TT

!"# , you should increase the temperature of the hot reservoir. $ %c is

correct. 9 •• [SSM] Explain why the following statement is true: To increase the efficiency of a Carnot engine, you should make the difference between the two operating temperatures as large as possible; but to increase the efficiency of a Carnot cycle refrigerator, you should make the difference between the two operating temperatures as small as possible. Determine the Concept A Carnot-cycle refrigerator is more efficient when the temperatures are close together because it is easier to extract heat from an already cold interior if the temperature of the exterior is close to the temperature of the interior of the refrigerator. A Carnot-cycle heat engine is more efficient when the temperature difference is large because then more work is done by the engine for each unit of heat absorbed from the hot reservoir. 10 •• A Carnot engine operates between a cold temperature reservoir of 27oC and a high temperature reservoir of 127(C. Its efficiency is (a) 21%, (b) 25%, (c) 75%, (d) 79%.

Chapter 19

1870

Determine the Concept The efficiency of a Carnot cycle engine is given by

h

cC 1

TT

!"# where Tc and Th (in kelvins) are the temperatures of the cold and hot

reservoirs, respectively. Substituting numerical values for Tc and Th yields: 25.0

K 400K 3001C "!"#

$ %b is correct. 11 •• The Carnot engine in Problem 10 is run in reverse as a refrigerator. Its COP is (a) 0.33, (b) 1.3, (c) 3.0 (d) 4.7. Determine the Concept The coefficient of performance of a Carnot cycle engine

run in reverse as refrigerator is given byWQc

refCOP " . We can use the relationship

between W, Qc, and Qh to eliminate W from this expression and then use the

relationship, applicable only to a device operating in a Carnot cycle, h

c

h

c

TT

QQ

" to

express the refrigerator’s COP in terms of Tc and Th. The coefficient of performance of a refrigerator is given by: W

QcrefCOP "

or, because ch QQW !" ,

ch

crefCOP

QQQ!

"

Dividing the numerator and denominator of this fraction by Qc yields: 1

1COP

c

href

!"

QQ

For a device operating in a Carnot cycle: h

c

h

c

TT

QQ

"

Substitute in the expression for COPref to obtain:

1

1COP

c

hC ref,

!"

TT


1871

Substitute numerical values and evaluate COPref, C: 0.3

1K 300K 4001COP C ref, "!

"

$ %c is correct. 12 •• On a humid day, water vapor condenses on a cold surface. During condensation, the entropy of the water (a) increases, (b) remains constant, (c) decreases, (d) may decrease or remain unchanged. Explain your answer. Determine the Concept When water vapor condenses, its entropy decreases (the liquid state is a more ordered state than is the vapor state) and the entropy of the universe increases. )(a is correct. 13 •• An ideal gas is taken reversibly from an initial state Pi, Vi, Ti to the final state Pf, Vf, Tf. Two possible paths are (A) an isothermal expansion followed by an adiabatic compression and (B) an adiabatic compression followed by an isothermal expansion. For these two paths, (a) )Eint A > )Eint B, (b) )SA > )SB, (c) )SA < )SB, (d) None of the above. Determine the Concept The two paths are shown on the PV diagram to the right. We can use the concept of a state function to choose from among the alternatives given as possible answers to the problem.

P

V iV fV

fP

iP i

f

BB

A

A

iT

fT

(a) Because Eint is a state function and the initial and final states are the same for the two paths, B int,A int, EE )") .

(b) and (c) S, like Eint, is a state function and its change when the system moves from one state to another depends only on the system’s initial and final states. It is not dependent on the process by which the change occurs. Thus BA SS )") . (d) )(d is correct. 14 •• Figure 19-12 shows a thermodynamic cycle for an ideal gas on an ST diagram. Identify this cycle and sketch it on a PV diagram.

Chapter 19

1872

Determine the Concept The processes A*B and C*D are adiabatic and the processes B*C and D*A are isothermal. Therefore, the cycle is the Carnot cycle shown in the adjacent PV diagram.

P

V

A

B

C

D

15 •• Figure 19-13 shows a thermodynamic cycle for an ideal gas on an SV diagram. Identify the type of engine represented by this diagram. Determine the Concept Note that A*B is an adiabatic expansion, B*C is a constant-volume process in which the entropy decreases, C*D is an adiabatic compression and D*A is a constant-volume process that returns the gas to its original state. The cycle is that of the Otto engine (see Figure 19-3). The points A, B, C, and D in Figure 19-13 correspond to points c, d, a, and b, respectively, in Figure 19-3. 16 •• Sketch an ST diagram of the Otto cycle. (The Otto cycle is discussed in Section 19-1.) Determine the Concept The Otto cycle consists of four quasi-static steps. Refer to Figure 19-3. There a*b is an adiabatic compression, b*c is a constant volume heating, c*d is an adiabatic expansion and d*a is a constant-volume cooling. So, from a to b, S is constant and T increases, from b to c, heat is added to the system and both S and T increase, from c*d S is constant while T decreases, and from d to a both S and T decrease. To determine how S depends on T along b*c and d*a, consider the entropy change of the gas from point b to an arbitrary point on the path b*c where the entropy and temperature of the gas are S and T, respectively:

TQS ")

where, because heat is entering the system, Q is positive.

Because Won = 0 for this constant-volume process:

$ %bTTCTCQQE !")""") VVinint

Substituting for Q yields: $ %+,-

./0 !"

!")

TTC

TTTCS bb 1V

V


1873

On path b*c the entropy is given by: +

,-

./0 !&")&"

TTCSSSS b

bb 1V

The first and second derivatives,

dTdS and 22 dTSd , give the slope and concavity of the path. Calculate these derivatives assuming CV is constant. (For an ideal gas CV is a positive constant.):

2V TTC

dTdS b"

3V2

2

2TTC

dTSd b!"

These results tell us that, along path b*c, the slope of the path is positive and the slope decreases as T increases. The concavity of the path is negative for all T. Following the same procedure on path d*a gives:

+,-

./0 !&"

TTCSS d

d 1V

2V TTC

dTdS d"

3V2

2

2TTC

dTSd d!"

These results tell us that, along path d*a, the slope of the path is positive and the slope decreases as T increases. The concavity of the path is negative for all T. An ST diagram for the Otto cycle is shown to the right.

S

T

a b

cd

17 •• [SSM] Sketch an SV diagram of the Carnot cycle for an ideal gas. Determine the Concept Referring to Figure 19-8, process 1*2 is an isothermal expansion. In this process heat is added to the system and the entropy and volume increase. Process 2*3 is adiabatic, so S is constant as V increases. Process 3*4 is an isothermal compression in which S decreases and V also decreases. Finally, process 4*1 is adiabatic, that is, isentropic, and S is constant while V decreases.

Chapter 19

1874

During the isothermal expansion (from point 1 to point 2) the work done by the gas equals the heat added to the gas. The change in entropy of the gas from point 1 (where the temperature is T1) to an arbitrary point on the curve is given by:

1TQS ")

For an isothermal expansion, the work done by the gas, and thus the heat added to the gas, are given by:

++,

-../

0""

11 ln

VVnRTWQ

Substituting for Q yields: ++

,

-../

0")

1

lnVVnRS

Since SSS )&" 1 , we have: ++

,

-../

0&"

11 ln

VVnRSS

The graph of S as a function of V for an isothermal expansion shown to the right was plotted using a spreadsheet program. This graph establishes the curvature of the 1*2 and 3*4 paths for the SV graph.

V

S

An SV graph for the Carnot cycle (see Figure 19-8) is shown to the right.

V

S

1

2 3

4

18 •• Sketch an SV diagram of the Otto cycle. (The Otto cycle is discussed in Section 19-1.)


1875

Determine the Concept The Otto cycle is shown in Figure 19-3. Process a*b takes place adiabatically and so both Q = 0 and )S = 0 along this path. Process b*c takes place at constant volume. Qin, however, is positive and so, while )V = 0 along this path, Q > 0 and, therefore )S > 0. Process c*d also takes place adiabatically and so, again, both Q = 0 and )S = 0 along this path. Finally, process d*a is a constant-volume process, this time with heat leaving the system and )S < 0. A sketch of the SV diagram for the Otto cycle follows:

ab

c d

S

V 19 •• Figure 19-14 shows a thermodynamic cycle for an ideal gas on an SP diagram. Make a sketch of this cycle on a PV diagram. Determine the Concept Process A*B is at constant entropy; that is, it is an adiabatic process in which the pressure increases. Process B*C is one in which P is constant and S decreases; heat is exhausted from the system and the volume decreases. Process C*D is an adiabatic compression. Process D*A returns the system to its original state at constant pressure. The cycle is shown in the adjacent PV diagram.

P

V

A

BC

D

20 •• One afternoon, the mother of one of your friends walks into his room and finds a mess. She asks your friend how the room came to be in such a state, and your friend replies, !Well, it is the natural destiny of any closed system to degenerate toward greater and greater levels of entropy. That’s all, Mom.! Her reply is a sharp !Nevertheless, you’d better clean your room!! Your friend retorts, !But that can’t happen. It would violate the second law of thermodynamics.! Critique your friend’s response. Is his mother correct to ground him for not cleaning his room, or is cleaning the room really impossible? Determine the Concept The son is out of line, here, but besides that, he’s also wrong. While it is true that systems tend to degenerate to greater levels of

Chapter 19

1876

disorder, it is not true that order cannot be brought forth from disorder. What is required is an agent doing work – for example, your friend – on the system in order to reduce the level of chaos and bring about order. His cleanup efforts will be rewarded with an orderly system after a sufficient time for him to complete the task. It is true that order will not come about from the disordered chaos of his room – unless he applies some elbow grease. Estimation and Approximation 21 • Estimate the change in COP of your electric food freezer when it is removed from your kitchen to its new location in your basement, which is 8°C cooler than your kitchen. Picture the Problem We can use the definition of the coefficient of performance to express the ratio of the coefficient of performance in your basement to the coefficient of performance in the kitchen. If we further assume that the freezer operates in a Carnot cycle, then we can use the proportion chch TTQQ " to express the ratio of the coefficients of performance in terms of the temperatures in the kitchen, basement, and freezer. The ratio of the coefficients of performance in the basement and kitchen is given by:

kitc,

kitc,

basementc,

basementc,

kit

basement

COPCOP

WQ

WQ

"

Because ch QQW !" for a heat engine or refrigerator:

kitc,kith,

kitc,

basementc,basementh,

basementc,

kit

basement

COPCOP

QQQ

QQQ

!

!"

Divide the numerators and denominators by Qc,basement and Qc,kit and simplify to obtain:

1

1

1

1

1

1

COPCOP

basementc,

basementh,

kitc,

kith,

kitc,

kith,

basementc,

basementh,

kit

basement

!

!"

!

!"

QQ

QQQQ

QQ


1877

If we assume that the freezer unit operates in a Carnot cycle, then

c

h

c

h

TT

QQ

" and our expression for the

ratio of the COPs becomes:

1

1

COPCOP

basementc,

basementh,

kitc,

kith,

kit

basement

!

!"

TT

TT

Assuming that the temperature in your kitchen is 20(C and that the temperature of the interior of your freezer is !5(C, substitute numerical values and evaluate the ratio of the coefficients of performance:

47.11

K 268K 285

1K 268K 932

COPCOP

kit

basement "!

!"

or an increase of %47 in the

performance of the freezer!

22 •• Estimate the probability that all the molecules in your bedroom are located in the (open) closet which accounts for about 10% of the total volume of the room. Picture the Problem The probability that all the molecules in your bedroom are

located in the (open) closet is given by N

VVp ++,

-../

0"

1

2 where N is the number of air

molecules in your bedroom and V1 and V2 are the volumes of your bedroom and closet, respectively. We can use the ideal-gas law to find the number of molecules N. We’ll assume that the volume of your room is about 50 m3 and that the temperature of the air is 20(C. If the original volume of the air in your bedroom is V1, the probability p of finding the N molecules, normally in your bedroom, confined to your closet whose volume is V2 is given by:

N

VVp ++,

-../

0"

1

2

or, because 1101

2 VV " , N

p +,-

./0"101 (1)

Use the ideal-gas law to express N: kT

PVN "

Substitute numerical values and evaluate N:

$ %$ %$ %$ %

molecules 10252.1K 293J/K 10381.1

m 50kPa 325.101

27

23

3

1"

1" !N

Chapter 19

1878

Substitute for N in equation (1) and evaluate p:

27

27

27

27

10

10252.110252.1

10252.1

10

1010

1101

!

1!1

1

2

""+,-

./0"p

23 •• [SSM] Estimate the maximum efficiency of an automobile engine that has a compression ratio of 8.0:1.0. Assume the engine operates according to the Otto cycle and assume 3 = 1.4. (The Otto cycle is discussed in Section 19-1. ) Picture the Problem The maximum efficiency of an automobile engine is given by the efficiency of a Carnot engine operating between the same two temperatures. We can use the expression for the Carnot efficiency and the equation relating V and T for a quasi-static adiabatic expansion to express the Carnot efficiency of the engine in terms of its compression ratio. Express the Carnot efficiency of an engine operating between the temperatures Tc and Th:

h

cC 1

TT

!"#

Relate the temperatures Tc and Th to the volumes Vc and Vh for a quasi-static adiabatic compression from Vc to Vh:

1hh

1cc

!! " 33 VTVT 41

c

h1

c

1h

h

c

!

!

!

++,

-../

0""

3

3

3

VV

VV

TT

Substitute for h

c

TT to obtain:

1

c

hC 1

!

++,

-../

0!"

3

#VV

Express the compression ratio r:

h

c

VVr "

Substituting for r yields:

1C11 !!" 3#

r

Substitute numerical values for r and 3 (1.4 for diatomic gases) and evaluate #C:

$ %%56

0.811 14.1C 2!" !#

24 •• You are working as an appliance salesperson during the summer. One day, your physics professor comes into your store to buy a new refrigerator. Wanting to buy the most efficient refrigerator possible, she asks you about the efficiencies of the available models. She decides to return the next day to buy the most efficient refrigerator. To make the sale, you need to provide her with the following estimates: (a) the highest COP possible for a household refrigerator, and (b) and the highest rate possible for the heat to be released by the refrigerator if the refrigerator uses 600 W of electrical power.


1879

Picture the Problem If we assume that the temperature on the inside of the refrigerator is 0(C (273 K) and the room temperature to be about 30(C (303 K), then the refrigerator must be able to maintain a temperature difference of about 30 K. We can use the definition of the COP of a refrigerator and the relationship between the temperatures of the hot and cold reservoir and Qh and Qc to find an upper limit on the COP of a household refrigerator. In (b) we can solve the definition of COP for Qc and differentiate the resulting equation with respect to time to estimate the rate at which heat is being drawn from the refrigerator compartment. (a) Using its definition, express the COP of a household refrigerator:

WQcCOP " (1)

Apply conservation of energy to the refrigerator to obtain:

hc QQW "& 4 ch QQW !"

Substitute for W and simplify to obtain:

1

1COP

c

hch

c

!"

!"

QQQQ

Q

Assume, for the sake of finding the upper limit on the COP, that the refrigerator is a Carnot refrigerator and relate the temperatures of the hot and cold reservoirs to hQ and Qc:

c

h

c

h

TT

QQ

"

Substitute for c

h

QQ to obtain:

1

1COP

c

hmax

!"

TT

Substitute numerical values and evaluate COPmax: 1.9

1K273K3031COPmax "!

"

(b) Solve equation (1) for Qc: $ %COPc WQ " (2)

Differentiate equation (2) with respect to time to obtain:

$ %dt

dWdt

dQ COPc "

Substitute numerical values and

evaluate dt

dQc : $ %$ % kW5.5J/s6009.1c ""

dtdQ

Chapter 19

1880

25 •• [SSM] The average temperature of the surface of the Sun is about 5400 K, the average temperature of the surface of Earth is about 290 K. The solar constant (the intensity of sunlight reaching Earth’s atmosphere) is about 1.37 kW/m2. (a) Estimate the total power of the sunlight hitting Earth. (b) Estimate the net rate at which Earth’s entropy is increasing due to this solar radiation. Picture the Problem We can use the definition of intensity to find the total power of sunlight hitting the earth and the definition of the change in entropy to find the changes in the entropy of Earth and the Sun resulting from the radiation from the Sun. (a) Using its definition, express the intensity of the Sun’s radiation on Earth in terms of the power P delivered to Earth and Earth’s cross sectional area A:

API "

Solve for P and substitute for A to obtain:

2RIIAP 5"" where R is the radius of Earth.

Substitute numerical values and evaluate P:

$ %$ %W1075.1W10746.1

m1037.6kW/m37.11717

262

1"1"

1" 5P

(b) Express the rate at which Earth’s entropy SEarth changes due to the flow of solar radiation: Earth

Earth

TP

dtdS

"

Substitute numerical values and

evaluate dt

dSEarth : sJ/K1002.6

K290W10746.1

14

17Earth

61"

1"

dtdS

26 •• A 1.0-L box contains N molecules of an ideal gas, and the positions of the molecules are observed 100 times per second. Calculate the average time it should take before we observe all N molecules in the left half of the box if N is equal to (a) 10, (b) 100, (c) 1000, and (d) 1.0 mole. (e) The best vacuums that have been created to date have pressures of about 10–12 torr. If a vacuum chamber has the same volume as the box, how long will a physicist have to wait before all of the gas molecules in the vacuum chamber occupy only the left half of it? Compare that to the expected lifetime of the universe, which is about 1010 years. Picture the Problem If you had one molecule in a box, it would have a 50% chance of being on one side or the other. We don’t care which side the molecules are on as long as they all are on one side, so with one molecule you have a 100% chance of it being on one side or the other. With two molecules, there are four


1881

possible combinations (both on one side, both on the other, one on one side and one on the other, and the reverse), so there is a 25% (1 in 4) chance of them both being on a particular side, or a 50% chance of them both being on either side. Extending this logic, the probability of N molecules all being on one side of the box is P = 2/2N, which means that, if the molecules shuffle 100 times a second, the time it would take them to cover all the combinations and all get on one side

or the other is $ %10022N

t " . In (e) we can apply the ideal gas law to find the number

of molecules in 1.0 L of air at a pressure of 10!12 torr and an assumed temperature of 300 K. (a) Evaluate t for N = 10 molecules: $ % s5s12.5

s 10022

1

10

2"" !t

(b) Evaluate t for N = 100 molecules: $ %

y102s 103.156

y 1s1034.6

s 10022

20

727

1

100

12

111"

" !t

(c) Evaluate t for N = 1000 molecules: $ %1

1000

s 10022

!"t

To evaluate 10002 let 1000210 "x and take the logarithm of both sides of the equation to obtain:

$ % 10ln2ln1000 x" 4 301"x

Substitute to obtain: $ %

y102s 103.156

y 1s105.0

s 100210

291

7299

1

301

12

111"

" !t

(d) Evaluate t for N = 1.0 mol =6.022 11023 molecules:

$ %1

10022.6

s 10022

23

!

1

"t

To evaluate

2310022.62 1 let 2310022.6210 1"x and take the logarithm

of both sides of the equation to obtain:

$ % 10ln2ln10022.6 23 x"1 4 23102x

Chapter 19

1882

Substituting for x yields:

$ %y10

s 103.156y 1

s 100210

23

23

10

71

10

2

+,-

./0

12 !t

(e) Solve the ideal gas law for the number of molecules N in the gas:

kTPVN "

Assuming the gas to be at room temperature (300 K), substitute numerical values and evaluate N:

$ %$ %$ %$ %$ %

molecules1022.3K300J/K10381.1

L 0.1Pa/torr32.133torr10

7

23

12

1"

1" !

!

N

Evaluate t for N = 3.221107 molecules:

$ %1

1022.3

s 10022

7

!

1

"t

To evaluate

71022.32 1 let 71022.3210 1"x and take the logarithm

of both sides of the equation to obtain:

$ % 10ln2ln1022.3 7 x"1 4 7102x

Substituting for x yields: $ %

y10

s 103.156y 1

s 100210

7

7

10

71

10

2

11" !t

Express the ratio of this waiting time to the lifetime of the universe tuniverse:

77

1010

10

universe

10y10y102"

tt

or

universe107

10 tt 2 Heat Engines and Refrigerators 27 • [SSM] A heat engine with 20.0% efficiency does 0.100 kJ of work during each cycle. (a) How much heat is absorbed from the hot reservoir during each cycle? (b) How much heat is released to the cold reservoir during each cycle? Picture the Problem (a) The efficiency of the engine is defined to be

hQW"# where W is the work done per cycle and Qh is the heat absorbed from the hot reservoir during each cycle. (b) Because, from conservation of energy,

ch QWQ &" , we can express the efficiency of the engine in terms of the heat Qc


1883

released to the cold reservoir during each cycle.

(a) Qh absorbed from the hot reservoir during each cycle is given by:

J5000.200

J100h """

#WQ

(b) Use ch QWQ &" to obtain: J 400J 100J 500hc "!"!" WQQ

28 • A heat engine absorbs 0.400 kJ of heat from the hot reservoir and does 0.120 kJ of work during each cycle. (a) What is its efficiency? (b) How much heat is released to the cold reservoir during each cycle? Picture the Problem (a) The efficiency of the engine is defined to be

hQW"# where W is the work done per cycle and Qh is the heat absorbed from the hot reservoir during each cycle. (b) We can apply conservation of energy to the engine to obtain ch QWQ &" and solve this equation for the heat Qc released to the cold reservoir during each cycle.

(a) The efficiency of the heat engine is given by:

%30J400J120

h

"""QW#

(b) Apply conservation of energy to the engine to obtain:

ch QWQ &" 4 WQQ !" hc

Substitute numerical values and evaluate Qc:

J280J 120J 400c "!"Q

29 • A heat engine absorbs 100 J of heat from the hot reservoir and releases 60 J of heat to the cold reservoir during each cycle. (a) What is its efficiency? (b) If each cycle takes 0.50 s, find the power output of this engine. Picture the Problem We can use its definition to find the efficiency of the engine and the definition of power to find its power output.

(a) The efficiency of the heat engine is given by:

h

c

h

ch

h Q1 Q

QQQ

QW

!"!

""#

Substitute numerical values and evaluate #:

%40J100J601 "!"#

Chapter 19

1884

(b) The power output P of this engine is the rate at which it does work:

dtdQQ

dtd

dtdWP h

h ## """

Substitute numerical values and evaluate P: $ % W80

s0.500J1000.40 "++,

-../

0"P

30 • A refrigerator absorbs 5.0 kJ of heat from a cold reservoir and releases 8.0 kJ to a hot reservoir. (a) Find the coefficient of performance of the refrigerator. (b) The refrigerator is reversible. If it is run backward as a heat engine between the same two reservoirs, what is its efficiency? Picture the Problem We can apply their definitions to find the COP of the refrigerator and the efficiency of the heat engine.

(a) The COP of a refrigerator is defined to be:

WQcCOP "

Apply conservation of energy to relate the work done per cycle to Qh and Qc:

ch QQW !"

Substitute for W to obtain: ch

cCOPQQ

Q!

"

Substitute numerical values and evaluate COP:

7.1kJ5.0kJ 0.8

kJ5.0COP "!

"

(b) The efficiency of a heat pump is defined to be: hQ

W"#

Apply conservation of energy to the heat pump to obtain:

h

c

h

ch 1QQ

QQQ

!"!

"#

Substitute numerical values and evaluate # :

%38kJ8.0kJ 0.51 "!"#

31 •• [SSM] The working substance of an engine is 1.00 mol of a monatomic ideal gas. The cycle begins at P1 = 1.00 atm and V1 = 24.6 L. The gas is heated at constant volume to P2 = 2.00 atm. It then expands at constant pressure until its volume is 49.2 L. The gas is then cooled at constant volume until its


1885

pressure is again 1.00 atm. It is then compressed at constant pressure to its original state. All the steps are quasi-static and reversible. (a) Show this cycle on a PV diagram. For each step of the cycle, find the work done by the gas, the heat absorbed by the gas, and the change in the internal energy of the gas. (b) Find the efficiency of the cycle. Picture the Problem To find the heat added during each step we need to find the temperatures in states 1, 2, 3, and 4. We can then find the work done on the gas during each process from the area under each straight-line segment and the heat that enters the system from TCQ )" V and .P TCQ )" We can use the 1st law of thermodynamics to find the change in internal energy for each step of the cycle. Finally, we can find the efficiency of the cycle from the work done each cycle and the heat that enters the system each cycle.

(a) The cycle is shown to the right:

Apply the ideal-gas law to state 1 to find T1:

$ %$ %$ %

K300

KmolatmL108.206mol1.00

L24.6atm1.002

111 "

+,-

./0

66

1""

!nRVPT

The pressure doubles while the volume remains constant between states 1 and 2. Hence:

KTT 6002 12 ""

The volume doubles while the pressure remains constant between states 2 and 3. Hence:

KTT 12002 23 ""

The pressure is halved while the volume remains constant between states 3 and 4. Hence:

KTT 600321

4 ""

Chapter 19

1886

For path 1*2:

0" 1212 "" VPW

and

$ % kJ74.3K300K600Kmol

J8.314"" 23

1223

12V12 "!+,-

./0

6""" TRTCQ

The change in the internal energy of the system as it goes from state 1 to state 2 is given by the 1st law of thermodynamics:

oninint" WQE &"

Because 012 "W : kJ 74.3" 1212int, "" QE

For path 2*3:

$ %$ % kJ99.4atmL

J101.325L24.6L49.2atm2.00" 2323on !"+,-

./0

6!!"!"!" VPWW

$ % kJ5.12K600K1200Kmol

J8.314"" 25

2325

23P23 "!+,-

./0

6""" TRTCQ

Apply oninint" WQE &" to obtain:

kJ 5.7kJ 99.4kJ 5.12" 23 int, "!"E

For path 3*4:

03434 ")" VPW

and

$ % kJ48.7K0021K600Kmol

J8.314""" 23

3423

34V34int,34 !"!+,-

./0

6"""" TRTCEQ


kJ 48.70kJ 48.7" 34 int, !"&!"E

For path 4*1:

$ %$ % kJ49.2atmL

J101.325L2.94L24.6atm1.00" 4141on "+,-

./0

6!!"!"!" VPWW

and

$ % kJ24.6K600K003Kmol

J8.314"" 25

4125

41P41 !"!+,-

./0

6""" TRTCQ


1887


kJ 75.3kJ 49.2kJ 24.6" 41 int, !"&!"E

For easy reference, the results of the preceding calculations are summarized in the following table:

Process onW , kJ inQ , kJ $ %oninint" WQE &" , kJ 1*2 0 3.74 3.74 2*3 !4.99 12.5 7.5 3*4 0 !7.48 !7.48 4*1 2.49 !6.24 !3.75

(b) The efficiency of the cycle is given by:

$ %2312

4123

in

by

QQWW

QW

&!&!

""#


%15kJ5.12kJ3.74kJ2.49kJ4.99

2&!

"#

Remarks: Note that the work done per cycle is the area bounded by the rectangular path. Note also that, as expected because the system returns to its initial state, the sum of the changes in the internal energy for the cycle is zero. 32 •• The working substance of an engine is 1.00 mol of a diatomic ideal gas. The engine operates in a cycle consisting of three steps: (1) an adiabatic expansion from an initial volume of 10.0 L to a pressure of 1.00 atm and a volume of 20.0 L, (2) a compression at constant pressure to its original volume of 10.0 L, and (3) heating at constant volume to its original pressure. Find the efficiency of this cycle. Picture the Problem The three steps in the process are shown on the PV diagram. We can find the efficiency of the cycle by finding the work done by the gas and the heat that enters the system per cycle.

V(L)

23

1P (atm)

2.639

2

1

00 10.0 20.0

The pressures and volumes at the end points of the adiabatic expansion are related according to:

332211 VPVP " 4 2

1

21 P

VVP

3

++,

-../

0"

Chapter 19

1888

Substitute numerical values and evaluate P1: $ % atm 639.2atm 00.1

L 0.10L 0.20 4.1

1 "+,-

./0"P

Express the efficiency of the cycle: hQ

W"# (1)

No heat enters or leaves the system during the adiabatic expansion:

012 "Q

Find the heat entering or leaving the system during the isobaric compression:

$ %$ %Latm35.0

L20.0L10.0atm1.00"""

27

2327

2327

23V23

6!"

!"

""" VPTRTCQ

Find the heat entering or leaving the system during the constant-volume process:

$ %$ %Latm0.41

L10.0atm1.00atm2.639"""

25

3125

3125

31V31

6"

!"

""" PVTRTCQ

Apply the 1st law of thermodynamics to the cycle ( 0cycle int, ")E ) to obtain:

Latm6.0Latm 41.0Latm 35.00

"

312312

inininton

6"6&6!"

&&"!"!"

QQQQQEW

Substitute numerical values in equation (1) and evaluate # :

%15Latm41Latm6.0"

66

"#

33 •• An engine using 1.00 mol of an ideal gas initially at a volume of 24.6 L and a temperature of 400 K performs a cycle consisting of four steps: (1) an isothermal expansion at 400 K to twice its initial volume, (2) cooling at constant volume to a temperature of 300 K (3) an isothermal compression to its original volume, and (4) heating at constant volume to its original temperature of 400 K. Assume that Cv = 21.0 J/K. Sketch the cycle on a PV diagram and find its efficiency. Picture the Problem We can find the efficiency of the cycle by finding the work done by the gas and the heat that enters the system per cycle.


1889

The PV diagram of the cycle is shown to the right. A, B, C, and D identify the four states of the gas and the numerals 1, 2, 3, and 4 represent the four steps through which the gas is taken.

0 10 20 30 40 50 60

2

1

1.5

0

0.5 400 K300 K

P (a

tm)

V (L)

1

23

4

A

B

C

D

Express the efficiency of the cycle: 4h,3h,2h,1h,

4321

h QQQQWWWW

QW

&&&&&&

""#

Because steps 2 and 4 are constant-volume processes, W2 = W4 = 0:

4h,3h,2h,1h,

31

h

00QQQQ

WWQW

&&&&&&

""#

Because the internal energy of the gas increases in step 4 while no work is done, and because the internal energy does not change during step 1 while work is done by the gas, heat enters the system only during these processes:

4h,1h,

31

h QQWW

QW

&&

""# (1)

The work done during the isothermal expansion (1) is given by:

++,

-../

0"

A

B1 ln

VVnRTW

The work done during the isothermal compression (3) is given by:

++,

-../

0"

C

Dc3 ln

VVnRTW

Because there is no change in the internal energy of the system during step 1, the heat that enters the system during this isothermal expansion is given by:

++,

-../

0""

A

Bh11 ln

VVnRTWQ

The heat that enters the system during the constant-volume step 4 is given by:

$ %chVV4 " TTCTCQ !""

Chapter 19

1890

Substituting in equation (1) yields:

$ %chVA

Bh

C

Dc

A

Bh

ln

lnln

TTCVVnRT

VVnRT

VVnRT

!&++,

-../

0

++,

-../

0&++,

-../

0

"#

Noting the 2A

B "VV and

21

C

D "VV

, substitute and simplify to obtain:

$ %

$ % $ %$ % $ %

$ % $ % $ % $ %chV

h

ch

chV

h

ch

chV

h

ch

2ln2ln

2ln2ln

2ln

21ln2ln

TTnR

CT

TT

TTnRCT

TT

TTnRCT

TT

!&

!"

!&

!"

!&

+,-

./0&

"#


$ % $ %$ %

%1.13

K 300K 4002ln

KmolJ 314.8mol 00.1

KJ 0.21

K 400

K 300K 400"

!+,-

./0

6

&

!"#

34 •• Figure 19-15 shows the cycle followed by 1.00 mol of an ideal monatomic gas initially at a volume of 25.0 L. All the processes are quasi-static. Determine (a) the temperature of each numbered state of the cycle, (b) the heat transfer for each part of the cycle, and (c) the efficiency of the cycle. Picture the Problem We can use the ideal-gas law to find the temperatures of each state of the gas and the heat capacities at constant volume and constant pressure to find the heat flow for the constant-volume and isobaric processes. Because the change in internal energy is zero for the isothermal process, we can use the expression for the work done on or by a gas during an isothermal process to find the heat flow during such a process. Finally, we can find the efficiency of the cycle from its definition. (a) Use the ideal-gas law to find the temperature at point 1:

$ %$ %$ %

K301

KmolJ8.314mol1.00

L25.0kPa100111

"

+,-

./0

6

""nRVPT


1891

Use the ideal-gas law to find the temperatures at points 2 and 3: $ %$ %

$ %

K601

KmolJ8.314mol1.00

L25.0kPa200

2232

"

+,-

./0

6

"

""nRVPTT

(b) Find the heat entering the system for the constant-volume process from 1 * 2:

$ % kJ3.74K301K601Kmol

J8.314"" 23

1223

12V12 "!+,-

./0

6""" TRTCQ

Find the heat entering or leaving the system for the isothermal process from 2 * 3:

$ % $ % kJ46.3L0.52L0.05lnK016

KmolJ8.314mol1.00ln

2

3223 "++

,

-../

0+,-

./0

6"++

,

-../

0"

VVnRTQ

Find the heat leaving the system during the isobaric compression from 3 * 1:

$ % kJ24.6K601K301Kmol

J8.314"" 25

3125

31P31 !"!+,-

./0

6""" TRTCQ

(c) Express the efficiency of the cycle:

2312in QQW

QW

&""# (1)

Apply the 1st law of thermodynamics to the cycle: kJ0.96

kJ6.24kJ3.46kJ.743312312

"!&"

&&""7 QQQQW

because, for the cycle, 0" int "E .


%13kJ3.46kJ3.74

kJ0.96"

&"#

35 •• An ideal diatomic gas follows the cycle shown in Figure 19-16. The temperature of state 1 is 200 K. Determine (a) the temperatures of the other three numbered states of the cycle and (b) the efficiency of the cycle. Picture the Problem We can use the ideal-gas law to find the temperatures of each state of the gas. We can find the efficiency of the cycle from its definition;

Chapter 19

1892

using the area enclosed by the cycle to find the work done per cycle and the heat entering the system between states 1 and 2 and 2 and 3 to determine Qin. (a) Use the ideal-gas law for a fixed amount of gas to find the temperature in state 2 to the temperature in state 1:

2

22

1

11

TVP

TVP" 4

1

21

11

2212 P

PTVPVPTT ""

Substitute numerical values and evaluate T2:

$ % $ %$ % K600

atm1.0atm3.0K2002 ""T

Apply the ideal-gas law for a fixed amount of gas to states 2 and 3 to obtain:

2

32

22

3323 V

VTVPVPTT ""


$ % $ %$ % K1800

L100L300K6003 ""T

Apply the ideal-gas law for a fixed amount of gas to states 3 and 4 to obtain:

3

43

33

4434 P

PTVPVPTT ""


$ % $ %$ % K600

atm3.0atm1.0K18004 ""T

(b) The efficiency of the cycle is: inQ

W"# (1)

Use the area of the rectangle to find the work done each cycle:

$ %$ %Latm400

atm1.0atm3.0L100L300""

6"!!"

" VPW

Apply the ideal-gas law to state 1 to find the product of n and R:

$ %$ %

atm/KL0.50K200

L100atm1.0

1

11

6"

""TVPnR

Noting that heat enters the system between states 1 and 2 and states 2 and 3, express Qin: $ %nRTT

TnRTnRTCTCQQQ

2327

1225

2327

1225

23P12V2312in

)&)"

)&)"

)&)"&"


1893

Substitute numerical values and evaluate Qin:

$ %8 $ %9 Latm2600KatmL50.0K600K1800K200K600 2

725

in 6"+,-

./0 6

!&!"Q


%15Latm2600Latm400"

66

"#

36 ••• Recently, an old design for a heat engine, known as the Stirling engine has been promoted as a means of producing power from solar energy. The cycle of a Stirling engine is as follows: (1) isothermal compression of the working gas (2) heating of the gas at constant volume, (3) an isothermal expansion of the gas, and (4) cooling of the gas at constant volume. (a) Sketch PV and ST diagrams for the Stirling cycle. (b) Find the entropy change of the gas for each step of the cycle and show that the sum of these entropy changes is equal to zero. Picture the Problem (a) The PV and ST cycles are shown below. (b) We can show that the entropy change during one Stirling cycle is zero by adding up the entropy changes for the four processes. P

V

T

Tc

h

(1)

(2)

(3)

(4)

1

2

3

4

S

T

Th

cT

V = 0

V = 0

)

)

(b) The change in entropy for one Stirling cycle is the sum of the entropy changes during the cycle:

41342312cycle """"" SSSSS &&&" (1)

Express the entropy change for the isothermal process from state 1 to state 2:

++,

-../

0"

1

212 ln"

VVnRS

Chapter 19

1894

Similarly, the entropy change for the isothermal process from state 3 to state 4 is:

++,

-../

0"

3

434 ln"

VVnRS

or, because V2 = V3 and V1 = V4,

++,

-../

0!"++

,

-../

0"

1

2

2

134 lnln"

VVnR

VVnRS

The change in entropy for a constant-volume process is given by:

++,

-../

0"

"" ::

i

fV

Visochoric

ln

"f

i

TTnC

TdTnC

TdQS

T

T

For the constant-volume process from state 2 to state 3:

++,

-../

0"

h

cV23 ln"

TTCS

For the constant-volume process from state 4 to state 1: ++

,

-../

0!"++

,

-../

0"

h

cV

c

hV41 lnln"

TTC

TTCS

Substituting in equation (1) yields:

0lnlnlnln"h

cV

1

2

h

cV

1

2cycle "++

,

-../

0!++,

-../

0!++,

-../

0&++,

-../

0!"

TTC

VVnR

TTC

VVnRS

37 •• !As far as we know, Nature has never evolved a heat engine!—Steven Vogel, Life’s Devices, Princeton University Press (1988). (a) Calculate the efficiency of a heat engine operating between body temperature (98.6ºF) and a typical outdoor temperature (70ºF), and compare this to the human body’s efficiency for converting chemical energy into work (approximately 20%). Does this efficiency comparison contradict the second law of thermodynamics? (b) From the result of Part (a), and a general knowledge of the conditions under which most warm-blooded organisms exist, give a reason why no warm-blooded organisms have evolved heat engines to increase their internal energies. Picture the Problem We can use the efficiency of a Carnot engine operating between reservoirs at body temperature and typical outdoor temperatures to find an upper limit on the efficiency of an engine operating between these temperatures. (a) Express the maximum efficiency of an engine operating between body temperature and 70°F:

h

cC 1

TT

!"#


1895

Use $ % 27332F95 &!" tT to obtain: K 310body "T and K 294room "T

Substitute numerical values and evaluate C# :

%16.5K310K2941C "!"#

The fact that this efficiency is considerably less than the actual efficiency of a human body does not contradict the second law of thermodynamics. The application of the second law to chemical reactions such as the ones that supply the body with energy have not been discussed in the text but we can note that we don’t get our energy from heat swapping between our body and the environment. Rather, we eat food to get the energy that we need. (b) Most warm-blooded animals survive under roughly the same conditions as humans. To make a heat engine work with appreciable efficiency, internal body temperatures would have to be maintained at an unreasonably high level. 38 ••• The diesel cycle shown in Figure 19-17 approximates the behavior of a diesel engine. Process ab is an adiabatic compression, process bc is an expansion at constant pressure, process cd is an adiabatic expansion, and process da is cooling at constant volume. Find the efficiency of this cycle in terms of the volumes Va, Vb and Vc. Picture the Problem The working fluid will be modeled as an ideal gas and the process will be modeled as quasistatic. To find the efficiency of the diesel cycle we can find the heat that enters the system and the heat that leaves the system and use the expression that gives the efficiency in terms of these quantities. Note that no heat enters or leaves the system during the adiabatic processes ab and cd. Heat enters the system during the isobaric process bc and leaves the system during the isovolumetric process da. Express the efficiency of the cycle in terms of Qc and Qh: h

c

h

ch

h

1QQ

QQQ

QW

!"!

""#

Express Q for the isobaric warming process bc:

$ %bcbc TTCQQ !"" Ph

Because CV is independent of T, Qda (the constant-volume cooling process) is given by:

$ %adda TTCQQ !"" Vc

Substitute for Qh and Qc and simplify using VP CC"3 to obtain:

$ %$ %

$ %$ %bc

ad

bc

ad

TTTT

TTCTTC

!!

!"!!

!"3

# 11P

V

Chapter 19

1896

Using an equation for a quasistatic adiabatic process, relate the temperatures Ta and Tb to the volumes Va and Vb:

11 !! " 33bbaa VTVT 4 1

1

!

!

" 3

3

a

bba V

VTT (1)

Proceeding similarly, relate the temperatures Tc and Td to the volumes Vc and Vd:

11 !! " 33ddcc VTVT 4 1

1

!

!

" 3

3

d

ccd V

VTT (2)

Use equations (1) and (2) to eliminate Ta and Td: $ %bc

a

bb

d

cc

TTVVT

VVT

!

++,

-../

0!

!"!

!

!

!

3#

3

3

3

3

1

1

1

1

1

Because Va = Vd:

++,

-../

0!

++

,

-

.

.

/

0++,

-../

0!++

,

-../

0

!"

!!

c

b

a

b

c

b

a

c

TT

VV

TT

VV

11

11

3#

33

Noting that Pb = Pc, apply the ideal-gas law to relate Tb and Tc:

c

b

c

b

VV

TT

"

Substitute for the ratio of Tb to Tc and simplify to obtain:

$ %bca

bc

a

b

a

c

a

b

a

c

a

b

a

c

a

b

c

b

a

c

a

c

a

c

c

b

a

b

c

b

a

c

VVVVV

VV

VV

VV

VV

VV

VV

VV

VV

VV

VVVV

VV

VV

VV

VV

!!

!"

++,

-../

0!

++,

-../

0!++

,

-../

0

!"

++,

-../

0!

++,

-../

0!++

,

-../

0

!"6

++,

-../

0!

++,

-../

0!++

,

-../

0

!"

!

!!!!

1

1111

11

11

1

3

33

33

3333

33

33#

Second Law of Thermodynamics 39 •• [SSM] A refrigerator absorbs 500 J of heat from a cold reservoir and releases 800 J to a hot reservoir. Assume that the heat-engine statement of the second law of thermodynamics is false, and show how a perfect engine working


1897

with this refrigerator can violate the refrigerator statement of the second law of thermodynamics. Determine the Concept The following diagram shows an ordinary refrigerator that uses 300 J of work to remove 500 J of heat from a cold reservoir and releases 800 J of heat to a hot reservoir (see (a) in the diagram). Suppose the heat-engine statement of the second law is false. Then a 'perfect' heat engine could remove energy from the hot reservoir and convert it completely into work with 100 percent efficiency. We could use this perfect heat engine to remove 300 J of energy from the hot reservoir and do 300 J of work on the ordinary refrigerator (see (b) in the diagram). Then, the combination of the perfect heat engine and the ordinary refrigerator would be a perfect refrigerator; transferring 500 J of heat from the cold reservoir to the hot reservoir without requiring any work (see (c) in the diagram).This violates the refrigerator statement of the second law.

;

;

;

;

;;;500 J

Cold reservoir at temperature Tc

Hot reservoir at temperature Th

800 J

300 J 300 J

300 J

Ordinaryrefrigerator

Perfectrefrigerator

500 J

500 J

a b c( ) ( ) ( )

Perfectheatengine

40 •• If two curves that represent quasi-static adiabatic processes could intersect on a PV diagram, a cycle could be completed by an isothermal path between the two adiabatic curves shown in Figure 19-18. Show that such a cycle violates the second law of thermodynamics. Determine the Concept The work done by the system is the area enclosed by the cycle, where we assume that we start with the isothermal expansion. It is only in this expansion that heat is extracted from a reservoir. There is no heat transfer in the adiabatic expansion or compression. Thus, we would completely convert heat to mechanical energy, without exhausting any heat to a cold reservoir, in violation of the second law of thermodynamics.

Carnot Cycles 41 • [SSM] A Carnot engine works between two heat reservoirs at temperatures Th = 300 K and Tc = 200 K. (a) What is its efficiency? (b) If it

Chapter 19

1898

absorbs 100 J of heat from the hot reservoir during each cycle, how much work does it do each cycle? (c) How much heat does it release during each cycle? (d) What is the COP of this engine when it works as a refrigerator between the same two reservoirs? Picture the Problem We can find the efficiency of the Carnot engine using

hc /1 TT!"# and the work done per cycle from ./ hQW"# We can apply conservation of energy to find the heat rejected each cycle from the heat absorbed and the work done each cycle. We can find the COP of the engine working as a refrigerator from its definition.

(a) The efficiency of the Carnot engine depends on the temperatures of the hot and cold reservoirs:

%3.33K300K20011

h

cC "!"!"

TT

#

(b) Using the definition of efficiency, relate the work done each cycle to the heat absorbed from the hot reservoir:

$ %$ % J33.3J1000.333hC """ QW #

(c) Apply conservation of energy to relate the heat given off each cycle to the heat absorbed and the work done:

J67

J 66.7J33.3J100hc

"

"!"!" WQQ

(d) Using its definition, express and evaluate the refrigerator’s coefficient of performance:

0.2J33.3J66.7COP c """

WQ

42 • An engine absorbs 250 J of heat per cycle from a reservoir at 300 K and releases 200 J of heat per cycle to a reservoir at 200 K. (a) What is its efficiency? (b) How much additional work per cycle could be done if the engine were reversible? Picture the Problem We can find the efficiency of the engine from its definition and the additional work done if the engine were reversible from ,hCQW #" where #C is the Carnot efficiency.

(a) Express the efficiency of the engine in terms of the heat absorbed from the high-temperature reservoir and the heat exhausted to the low-temperature reservoir:

%0.20J250J2001

1h

c

h

ch

h

"!"

!"!

""QQ

QQQ

QW#


1899

(b) Express the additional work done if the engine is reversible:

$ %aWWW PartCarnot !") (1)

Relate the work done by a reversible engine to its Carnot efficiency:

hh

chC 1 Q

TTQW ++,

-../

0!"" #

Substitute numerical values and evaluate W: $ % J3.83J250

K300K2001 "++,

-../

0!"W

Substitute numerical values in equation (1) and evaluate )W:

J33J50J3.38" "!"W

43 •• A reversible engine working between two reservoirs at temperatures Th and Tc has an efficiency of 30%. Working as a heat engine, it releases 140 J per cycle of heat to the cold reservoir. A second engine working between the same two reservoirs also releases 140 J per cycle to the cold reservoir. Show that if the second engine has an efficiency greater than 30%, the two engines working together would violate the heat-engine statement of the second law. Determine the Concept Let the first engine be run as a refrigerator. Then it will remove 140 J from the cold reservoir, deliver 200 J to the hot reservoir, and require 60 J of energy to operate. Now take the second engine and run it between the same reservoirs, and let it eject 140 J into the cold reservoir, thus replacing the heat removed by the refrigerator. If #2, the efficiency of this engine, is greater than 30%, then Qh2, the heat removed from the hot reservoir by this engine, is 140 J/(1 ! #2) > 200 J, and the work done by this engine is W = #2Qh2 > 60 J. The end result of all this is that the second engine can run the refrigerator, replacing the heat taken from the cold reservoir, and do additional mechanical work. The two systems working together then convert heat into mechanical energy without rejecting any heat to a cold reservoir, in violation of the second law.

44 •• A reversible engine working between two reservoirs at temperatures Th and Tc has an efficiency of 20%. Working as a heat engine, it does 100 J of work per cycle. A second engine working between the same two reservoirs also does 100 J of work per cycle. Show that if the efficiency of the second engine is greater than 20%, the two engines working together would violate the refrigerator statement of the second law. Determine the Concept If the reversible engine is run as a refrigerator, it will require 100 J of mechanical energy to take 400 J of heat from the cold reservoir and deliver 500 J to the hot reservoir. Now let the second engine, with #2 > 0.2, operate between the same two heat reservoirs and use it to drive the refrigerator.

Chapter 19

1900

Because #2 > 0.2, this engine will remove less than 500 J from the hot reservoir in the process of doing 100 J of work. The net result is then that no net work is done by the two systems working together, but a finite amount of heat is transferred from the cold reservoir to the hot reservoir, in violation of the refrigerator statement of the second law.

45 •• A Carnot engine works between two heat reservoirs as a refrigerator. During each cycle, 100 J of heat are absorbed and 150 J are released to the hot reservoir. (a) What is the efficiency of the Carnot engine when it works as a heat engine between the same two reservoirs? (b) Show that no other engine working as a refrigerator between the same two reservoirs can have a COP greater than 2.00. Picture the Problem We can use the definition of efficiency to find the efficiency of the Carnot engine operating between the two reservoirs. (a) The efficiency of the Carnot engine is given by:

%33J150J50

hC """

QW#

(b) If the COP > 2, then 50 J of work will remove more than 100 J of heat from the cold reservoir and put more than 150 J of heat into the hot reservoir. So running engine described in Part (a) to operate the refrigerator with a COP > 2 will result in the transfer of heat from the cold to the hot reservoir without doing any net mechanical work in violation of the second law. 46 •• A Carnot engine works between two heat reservoirs at temperatures Th = 300 K and Tc = 77.0 K. (a) What is its efficiency? (b) If it absorbs 100 J of heat from the hot reservoir during each cycle, how much work does it do? (c) How much heat does it release to the low-temperature reservoir during each cycle? (d) What is the coefficient of performance of this engine when it works as a refrigerator between these two reservoirs? Picture the Problem We can use the definitions of the efficiency of a Carnot engine and the coefficient of performance of a refrigerator to find these quantities. The work done each cycle by the Carnot engine is given by hCQW #" and we can use the conservation of energy to find the heat rejected to the low-temperature reservoir. (a) The efficiency of a Carnot engine depends on the temperatures of the hot and cold reservoirs:

74.3%K300K77.011

h

cC "!"!"

TT#


1901

(b) Express the work done each cycle in terms of the efficiency of the engine and the heat absorbed from the high-temperature reservoir:

$ %$ % J3.74J100743.0hC """ QW #

(c) Apply conservation of energy to obtain:

J26J74.3J001hc "!"!" WQQ

(d) Using its definition, express and evaluate the refrigerator’s coefficient of performance:

35.0J74.3

J26COP c """WQ

47 •• [SSM] In the cycle shown 1 in Figure 19-19, 1.00 mol of an ideal diatomic gas is initially at a pressure of 1.00 atm and a temperature of 0.0ºC. The gas is heated at constant volume to T2 = 150ºC and is then expanded adiabatically until its pressure is again 1.00 atm. It is then compressed at constant pressure back to its original state. Find (a) the temperature after the adiabatic expansion, (b) the heat absorbed or released by the system during each step, (c) the efficiency of this cycle, and (d) the efficiency of a Carnot cycle operating between the temperature extremes of this cycle. Picture the Problem We can use the ideal-gas law for a fixed amount of gas and the equations of state for an adiabatic process to find the temperatures, volumes, and pressures at the end points of each process in the given cycle. We can use

TQ )" VC and TQ )" PC to find the heat entering and leaving during the constant-volume and isobaric processes and the first law of thermodynamics to find the work done each cycle. Once we’ve calculated these quantities, we can use its definition to find the efficiency of the cycle and the definition of the Carnot efficiency to find the efficiency of a Carnot engine operating between the extreme temperatures.

(a) Apply the ideal-gas law for a fixed amount of gas to relate the temperature at point 3 to the temperature at point 1:

3

33

1

11

TVP

TVP"

or, because P1 = P3,

1

313 VVTT " (1)

Apply the ideal-gas law for a fixed amount of gas to relate the pressure at point 2 to the temperatures at points 1 and 2 and the pressure at 1:

2

22

1

11

TVP

TVP" 4

12

2112 TV

TVPP "

Chapter 19

1902

Because V1 = V2:

$ % atm1.55K273K423atm1.00

1

212 """

TTPP

Apply an equation for an adiabatic process to relate the pressures and volumes at points 2 and 3:

333311 VPVP " 4

31

3

113 ++

,

-../

0"

PPVV

Noting that V1 = 22.4 L, evaluate V3: $ % L30.6

atm1atm1.55L22.4

1.41

3 "++,

-../

0"V

Substitute numerical values in equation (1) and evaluate T3 and t3:

$ % K373L22.4L30.6K2733 ""T

and C10027333 ("!" Tt

(b) Process 1*2 takes place at constant volume (note that 3 = 1.4 corresponds to a diatomic gas and that CP – CV = R):

$ %

kJ3.12

K273K423Kmol

J8.314

""C

25

1225

12V12

"

!+,-

./0

6"

"" TRTQ

Process 2*3 takes place adiabatically:

023 "Q

Process 3*1 is isobaric (note that CP = CV + R): $ %

kJ2.91

K373K732Kmol

J8.314

""C

27

1227

31P31

!"

!+,-

./0

6"

"" TRTQ

(c) The efficiency of the cycle is given by:

inQW

"# (2)

Apply the first law of thermodynamics to the cycle:

oninint" WQE &" or, because 0cycle int, ")E (the system

begins and ends in the same state) and ingas by theon QWW "!" .

Evaluating W yields: kJ0.21kJ2.910kJ3.12

312312

"!&"

&&""7 QQQQW


1903


%7.6kJ3.12kJ0.21

""#

(d) Express and evaluate the efficiency of a Carnot cycle operating between 423 K and 273 K:

35.5%K234K73211

h

cC "!"!"

TT#

48 •• You are part of a team that is completing a mechanical-engineering project. Your team built a steam engine that takes in superheated steam at 270ºC and discharges condensed steam from its cylinder at 50.0ºC. Your team has measured its efficiency to be 30.0%. (a) How does this efficiency compare with the maximum possible efficiency for your engine? (b) If the useful power output of the engine is known to be 200 kW, how much heat does the engine release to its surroundings in 1.00 h? Picture the Problem We can find the maximum efficiency of the steam engine by calculating the Carnot efficiency of an engine operating between the given temperatures. We can apply the definition of efficiency to find the heat discharged to the engine’s surroundings in 1.00 h.

(a) The efficiency of the steam engine as a percentage of the maximum possible efficiency is given by:

maxmax

enginesteam 300.0##

#"

The efficiency of a Carnot engine operating between temperatures Fc and Th is:

%52.40K543K32311

h

cmax "!"!"

TT

#

Substituting for #max yields: %05.74

4052.0300.0

max

enginesteam ""#

#

or

maxenginesteam 740.0 ## "

(b) Relate the heat Qc discharged to the engine’s surroundings to Qh and the efficiency of the engine:

h

ch

h QQQ

QW !

""# 4 $ % hc 1 QQ #!"

Chapter 19

1904

Using its definition, relate the efficiency of the engine to the heat intake of the engine and the work it does each cycle:

##tPWQ )

""h

Substitute for Qh in the expression for cQ and simplify to obtain: $ % tPtPQ "11"1c +

,-

./0 !"!"##

#

Substitute numerical values and evaluate $ %h 00.1cQ :

$ % $ % GJ68.1s3600skJ2001

300.01h 00.1c "+

,-

./0+,-

./0 !"Q

*Heat Pumps 49 • [SSM] As an engineer, you are designing a heat pump that is capable of delivering heat at the rate of 20 kW to a house. The house is located where, in January, the average outside temperature is –10ºC. The temperature of the air in the air handler inside the house is to be 40ºC. (a) What is maximum possible COP for a heat pump operating between these temperatures? (b) What must be the minimum power of the electric motor driving the heat pump? (c) In reality, the COP of the heat pump will be only 60 percent of the ideal value. What is the minimum power of the engine when the COP is 60 percent of the ideal value? Picture the Problem We can use the definition of the COPHP and the Carnot efficiency of an engine to express the maximum efficiency of the refrigerator in terms of the reservoir temperatures. We can apply the definition of power to find the minimum power needed to run the heat pump.

(a) Express the COPHP in terms of Th and Tc:

ch

h

h

c

h

c

h

hhHP

1

1

1

1

COP

TTT

TT

QQ

QQQ

WQ

c

!"

!"

!"

!""

Substitute numerical values and evaluate COPHP:

3.6

26.6K263K313

K133COPHP

"

"!

"


1905

(b) The COPHP is also given by: motor

outHPCOP

PP

" 4 HP

outmotor COP

PP "

Substitute numerical values and evaluate Pmotor:

kW2.36.26

kW20motor ""P

(c) The minimum power of the engine is given by:

$ %maxHP,

c

HP

c

min COP##dt

dQdt

dQ

P ""

where #HP is the efficiency of the heat pump.

Substitute numerical values and evaluate Pmin: $ %$ % kW3.5

6.2660.0kW20

min ""P

50 • A refrigerator is rated at 370 W. (a) What is the maximum amount of heat it can absorb from the food compartment in 1.00 min if the food-compartment temperature of the refrigerator is 0.0ºC and it releases heat into a room at 20.0ºC? (b) If the COP of the refrigerator is 70% of that of a reversible refrigerator, how much heat can it absorb from the food compartment in 1.00 min under these conditions? Picture the Problem We can use the definition of the COP to relate the heat removed from the refrigerator to its power rating and operating time. By expressing the COP in terms of Tc and Th we can write the amount of heat removed from the refrigerator as a function of Tc, Th, P, and )t.

(a) Express the amount of heat the refrigerator can remove in a given period of time as a function of its COP:

$ %$ % tP

WQ)"

"COPCOPc

Express the COP in terms of Th and Tc and simplify to obtain:

ch

c

h

c

h

h

h

cc

11

1111

COP

TTT

TT

QWQ

QQ

WQ

!"

!!

"!"!

"

!"""

###

##

Chapter 19

1906

Substituting for COP yields: tP

TTTQ "

ch

cc ++

,

-../

0!

"


$ % MJ 30.0kJ 303min

s 60min00.1W370K 273K 293

K 273c ""+

,-

./0 1+

,-

./0

!"Q

(b) If the COP is 70% of the efficiency of an ideal pump:

$ %$ % MJ10.2kJ 30370.0c ""'Q

51 • A refrigerator is rated at 370 W. (a) What is the maximum amount of heat it can absorb for the food compartment in 1.00 min if the temperature in the compartment is 0.0ºC and it releases heat into a room at 35ºC? (b) If the COP of the refrigerator is 70% of that of a reversible pump, how much heat can it absorb from the food compartment in 1.00 min? Is the COP for the refrigerator greater when the temperature of the room is 35ºC or 20ºC? Explain. Picture the Problem We can use the definition of the COP to relate the heat removed from the refrigerator to its power rating and operating time. By expressing the COP in terms of Tc and Th we can write the amount of heat removed from the refrigerator as a function of Tc, Th, P, and )t.

(a) Express the amount of heat the refrigerator can remove in a given period of time as a function of its COP:

$ %$ % tP

WQ)"

"COPCOPc

Express the COP in terms of Th and Tc and simplify to obtain:

ch

c

h

c

h

h

h

cc

11

1

111

COP

TTT

TT

QWQ

QQ

WQ

!"!

!"

!"!

"

!"""

###

##

Substituting for COP yields:

tPTT

TQ "ch

cc ++

,

-../

0!

"


1907


$ % MJ 71.0kJ 731min

s 60min00.1W370K 273K 083

K 732c ""+

,-

./0 1+

,-

./0

!"Q

(b) If the COP is 70% of the efficiency of an ideal pump:

$ %$ % MJ12.0kJ 73170.0c ""'Q

Because the temperature difference increases when the room is warmer, the COP decreases. 52 ••• You are installing a heat pump, whose COP is half the COP of a reversible heat pump. You will use the pump on chilly winter nights to increase the air temperature in your bedroom. Your bedroom’s dimensions are 5.00 m × 3.50 m × 2.50 m. The air temperature should increase from 63°F to 68°F. The outside temperature is 35°F , and the temperature at the air handler in the room is 112°F. If the pump’s electric power consumption is 750 W, how long will you have to wait in order for the room’s air to warm (take the specific heat of air to be 1.005 kJ/(kg·°C)? Assume you have good window draperies and good wall insulation so that you can neglect the release of heat through windows, walls, ceilings and floors. Also assume that the heat capacity of the floor, ceiling, walls and furniture are negligible. Picture the Problem We can use the definition of the coefficient of performance of a heat pump and the relationship between the work done per cycle and the pump’s power consumption to find your waiting time. The coefficient of performance of the heat pump is defined as:

tPQ

WQ

"COP hh

HP "" 4 $ %PQt

HP

h

COP" "

where Qh is the heat required to raise the temperature of your bedroom, P is the power consumption of the heat pump, and )t is the time required to warm the bedroom.

We’re given that the coefficient of performance of the heat pump is half the coefficient of performance of an ideal heat pump: ++

,

-../

0!

"

""

ch

h21

max21h

HP COPCOP

TTT

WQ

Substituting for HPCOP yields:

PTT

TQt

++,

-../

0!

"

ch

h

h2"

Chapter 19

1908

The heat required to warm the room is related to the volume of the room, the density of air, and the desired increase in temperature:

TVcTmcQ ""h <"" where < is the density of air and c is its specific heat capacity.

Substitute for Qh to obtain:

PTT

TTVct

++,

-../

0!

"

ch

h

"2" <

Substitute numerical values and evaluate )t:

$ %

$ %s 56

W750K 275K 317

K 317F 9C 5F 5

CkgJ1005m 2.50m 3.50m 00.5

mkg293.12

"3

"+,-

./0

!

+,-

./0

((

1(++,

-../

0(6

11+,-

./0

"t

Entropy Changes 53 • [SSM] You inadvertently leave a pan of water boiling away on the hot stove. You return just in time to see the last drop converted into steam. The pan originally held 1.00 L of boiling water. What is the change in entropy of the water associated with its change of state from liquid to gas? Picture the Problem Because the water absorbed heat in the vaporization process

its change in entropy is positive and given byT

QS OHby

absorbed

OH2

2" " . See Table 18-2

for the latent heat of vaporization of water. The change in entropy of the water is given by:

T

QS OHby

absorbed

OH2

2" "

The heat absorbed by the water as it vaporizes is the product of its mass and latent heat of vaporization:

vvOHby

absorbed2

VLmLQ <""

Substituting for OHby

absorbed2

Q yields: TVLS v

OH2" <

"


1909

Substitute numerical values and evaluate OH2

"S : $ %

KkJ05.6

K 373kgkJ2257L 00.1

Lkg 00.1

" OH2

"

++,

-../

0+,-

./0

"S

54 • What is the change in entropy of 1.00 mol of liquid water at 0.0ºC that freezes to ice at 0.0°C? Picture the Problem We can use the definition of entropy change to find the change in entropy of the liquid water as it freezes. Because heat is removed from liquid water when it freezes, the change in entropy of the liquid water is negative. See Appendix C for the molar mass of water and Table 18-2 for the latent heat of fusion of water.

The change in entropy of the water is given by:: T

QS OH from

removed

OH2

2" "

The heat removed from the water as it freezes is the product of its mass and latent heat of fusion:

fOH from removed

2

mLQ !"

or, because OH2nMm " ,

fOHOH from removed 2

2

LnMQ !"

Substitute numerical values and evaluate OH2

"S :

$ %

KJ22.0

K273gJ333.5

molg18.015mol 00.1

" OH2!"

++,

-../

0+,-

./0!

"S

55 • Consider the freezing of 50.0 g of water once it is placed in the freezer compartment of a refrigerator. Assume the walls of the freezer are maintained at –10ºC. The water, initially liquid at 0.0ºC, is frozen into ice and cooled to –10ºC. Show that even though the entropy of the water decreases, the net entropy of the universe increases. Picture the Problem The change in the entropy of the universe resulting from the freezing of this water and the cooling of the ice formed is the sum of the entropy changes of the water-ice and the freezer. Note that, while the entropy of the water decreases, the entropy of the freezer increases.

Chapter 19

1910

The change in entropy of the universe resulting from this freezing and cooling process is given by:

freezerwateru SSS )&)") (1)

Express waterS) :

coolingfreezingwater SSS )&)") (2)

Express freezingS) :

freezing

freezingfreezing T

QS

!") (3)

where the minus sign is a consequence of the fact that heat is leaving the water as it freezes.

Relate freezingQ to the latent heat of

fusion and the mass of the water:

ffreezing mLQ "

Substitute in equation (3) to obtain: freezing

ffreezing T

mLS !")

Express coolingS) :

++,

-../

0")

i

fpcooling ln

TTmCS

Substitute in equation (2) to obtain: ++

,

-../

0&

!")

i

fp

freezing

fwater ln

TTmC

TmLS

Noting that the freezer gains heat (at 263 K) from the freezing water and cooling ice, express

freezerS) : freezer

p

freezer

f

freezer

ice cooling

freezer

icefreezer

TTmC

TmL

TQ

TQS

)&"

)&

)")

Substitute for waterS) and freezerS) in equation (1):

==>

?

@@A

B )&&++,

-../

0&

!"

)&&++

,

-../

0&

!")

freezer

pf

i

fp

freezing

f

freezer

p

freezer

f

i

fp

freezing

fu

ln

ln

TTCL

TTC

TLm

TTmC

TmL

TTmC

TmLS


1911

Substitute numerical values and evaluate )Su:

$ %

$ %J/K40.2

K263

K263K273Kkg

J2100kgJ105.333

K273K263ln

KkgJ2100

K273kgJ105.333

kg0500.0"

3

3

u

"

====

>

?!++

,

-../

06

&1&

@@@@

A

B

++,

-../

0++,

-../

06

&1

!"S

and, because )Su > 0, the entropy of the universe increases. 56 • In this problem, 2.00 mol of an ideal gas at 400 K expand quasi-statically and isothermally from an initial volume of 40.0 L to a final volume of 80.0 L. (a) What is the entropy change of the gas? (b) What is the entropy change of the universe for this process? Picture the Problem We can use the definition of entropy change and the 1st law of thermodynamics to express )S for the ideal gas as a function of its initial and final volumes.

(a) The entropy change of the gas is given by:

TQS "gas" (1)

Apply the first law of thermodynamics to the isothermal process to express Q in terms of Won:

onint" WEQ !" or, because )Eint = 0 for an isothermal expansion of a gas,

onWQ !"

The work done on the gas is given by: ++

,

-../

0"

f

ion ln

VVnRTW 4 ++

,

-../

0!"

f

ilnVVnRTQ

Substitute for Q in equation (1) to obtain: ++

,

-../

0!"

f

igas ln"

VVnRS

Substitute numerical values and evaluate )S:

$ %KJ11.5

L80.0L40.0ln

KmolJ8.314mol2.00" gas "++

,

-../

0+,-

./0

6!"S

Chapter 19

1912

(b) Because the process is reversible: 0u ")S

Remarks: The entropy change of the environment of the gas is !11.5 J/K. 57 •• [SSM] A system completes a cycle consisting of six quasi-static steps, during which the total work done by the system is 100 J. During step 1 the system absorbs 300 J of heat from a reservoir at 300 K, during step 3 the system absorbs 200 J of heat from a reservoir at 400 K, and during step 5 it absorbs heat from a reservoir at temperature T3. (During steps 2, 4 and 6 the system undergoes adiabatic processes in which the temperature of the system changes from one reservoir’s temperature to that of the next.) (a) What is the entropy change of the system for the complete cycle? (b) If the cycle is reversible, what is the temperature T3? Picture the Problem We can use the fact that the system returns to its original state to find the entropy change for the complete cycle. Because the entropy change for the complete cycle is the sum of the entropy changes for each process, we can find the temperature T3 from the entropy changes during the 1st two processes and the heat released during the third. (a) Because S is a state function of the system, and because the system’s final state is identical to its initial state:

0"cyclecomplete 1

system "S

(b) Relate the entropy changes for each of the three heat reservoirs and the system for one complete cycle of the system:

0"""" system321 "&&& SSSS

or

003

3

2

2

1

1 "&&&TQ

TQ

TQ

Substitute numerical values. Heat is rejected by the two high-temperature reservoirs and absorbed by the cold reservoir:

0J400K400

J200K300

J300

3

"&!

&!

T

Solving for T3 yields: K2673 "T

58 •• In this problem, 2.00 mol of an ideal gas initially has a temperature of 400 K and a volume of 40.0 L. The gas undergoes a free adiabatic expansion to twice its initial volume. What is (a) the entropy change of the gas and (b) the entropy change of the universe?


1913

Picture the Problem The initial and final temperatures are the same for a free expansion of an ideal gas. Thus, the entropy change )S for a free expansion from Vi to Vf is the same as )S for an isothermal process from Vi to Vf. We can use the definition of entropy change and the 1st law of thermodynamics to express )S for the ideal gas as a function of its initial and final volumes. (a) The entropy change of the gas is given by:

TQS "gas" (1)

Apply the first law of thermodynamics to the isothermal process to express Q:

onint" WEQ !" or, because )Eint = 0 for a free expansion of a gas,

onWQ !"

The work done on the gas is given by:

++,

-../

0"

f

ion ln

VVnRTW 4 ++

,

-../

0!"

f

ilnVVnRTQ

Substitute for Q in equation (1) to obtain: ++

,

-../

0!"

f

igas ln"

VVnRS

Substitute numerical values and evaluate )S:

$ %KJ11.5

L80.0L40.0ln

KmolJ8.314mol2.00" gas "++

,

-../

0+,-

./0

6!"S

(b) The change in entropy of the universe is the sum of the entropy changes of the gas and the surroundings:

gssurroundingasu """ SSS &"

For the change in entropy of the surroundings we use the fact that, during the free expansion, the surroundings are unaffected:

00" revgssurroundin """

TTQS

The change in entropy of the universe is the change in entropy of the gas: K

J5.11" u "S

59 •• A 200-kg block of ice at 0.0ºC is placed in a large lake. The temperature of the lake is just slightly higher than 0.0ºC, and the ice melts very

Chapter 19

1914

slowly. (a) What is the entropy change of the ice? (b) What is the entropy change of the lake? (c) What is the entropy change of the universe (the ice plus the lake)? Picture the Problem Because the ice gains heat as it melts, its entropy change is positive and can be calculated from its definition. Because the temperature of the lake is just slightly greater than 0(C and the mass of water is so much greater than that of the block of ice, the absolute value of the entropy change of the lake will be approximately equal to the entropy change of the ice as it melts.

(a) The entropy change of the ice is given by:

TmLS f

ice" "

Substitute numerical values and evaluate ice"S : $ %

KkJ244

K273kgkJ333.5kg200

" ice "++,

-../

0

"S

(b) Relate the entropy change of the lake to the entropy change of the ice:

KkJ244"" icelake !"!2 SS

(c) The entropy change of the universe due to this melting process is the sum of the entropy changes of the ice and the lake:

lakeiceu """ SSS &"

Because the temperature of the lake is slightly greater than that of the ice, the magnitude of the entropy change of the lake is less than 244 kJ/K and the entropy change of the universe is greater than zero. The melting of the ice is an irreversible process and 0u C)S . 60 •• A 100-g piece of ice at 0.0ºC is placed in an insulated calorimeter with negligible heat capacity containing 100 g of water at 100ºC. (a) What is the final temperature of the water once thermal equilibrium is established? (b) Find the entropy change of the universe for this process. Picture the Problem We can use conservation of energy to find the equilibrium temperature of the water and apply the equations for the entropy change during a melting process and for constant-pressure processes to find the entropy change of the universe (the entropy change of the piece of ice plus the entropy change of the water in the insulated container).


1915

(a) Apply conservation of energy to obtain:

0i

i "7Q

or 0

watercooling

waterwarming

icemelting "!& QQQ

Substitute to relate the masses of the ice and water to their temperatures, specific heats, and the final temperature of the water:

$ % $ %

$ % $ % 0C100Ckg

kJ18.4g100

CkgkJ18.4g100

kgkJ5.333g100

"!(++,

-../

0(6

!

++,

-../

0(6

&++,

-../

0

t

t

Solving for t yields: C1.10 ("t

(b) The entropy change of the universe is the sum of the entropy changes of the ice and the water:

watericeu SSS )&)")

Using the expression for the entropy change for a constant-pressure process, express the entropy change of the melting ice and warming ice-water:

++,

-../

0&"

)&)")

i

fP

f

f

waterwarmingicemeltingice

lnTTmc

TmL

SSS

Substitute numerical values to obtain:

$ %$ %

KJ137

K273K283ln

KkgkJ4.18kg0.100

K273kgkJ333.5kg0.100

" ice "++,

-../

0++,

-../

06

&++,

-../

0

"S

Find the entropy change of the cooling water:

$ %KJ 115

K373K283ln

KkgkJ4.18kg0.100" water !"++

,

-../

0++,

-../

06

"S

Substitute for )Sice and )Swater and evaluate the entropy change of the universe:

KJ22

KJ115

KJ371" u "!"S

Chapter 19

1916

Remarks: The result that )Su > 0 tells us that this process is irreversible. 61 •• [SSM] A 1.00-kg block of copper at 100ºC is placed in an insulated calorimeter of negligible heat capacity containing 4.00 L of liquid water at 0.0ºC. Find the entropy change of (a) the copper block, (b) the water, and (c) the universe. Picture the Problem We can use conservation of energy to find the equilibrium temperature of the water and apply the equations for the entropy change during a constant pressure process to find the entropy changes of the copper block, the water, and the universe.

(a) Use the equation for the entropy change during a constant-pressure process to express the entropy change of the copper block:

++,

-../

0")

i

fCuCuCu ln

TTcmS (1)

Apply conservation of energy to obtain:

0i

i "7Q

or 0waterwarmingblockcopper "&QQ

Substitute to relate the masses of the block and water to their temperatures, specific heats, and the final temperature Tf of the water:

$ % $ %

$ % $ % 0K273Kkg

kJ 4.18Lkg1.00L4.00

K373Kkg

kJ0.386kg1.00

f

f

"!++,

-../

06

+,-

./0&

!++,

-../

06

T

T

Solve for Tf to obtain: K275.26 f "T

Substitute numerical values in equation (1) and evaluate Cu"S :

$ %KJ117

K373K275.26ln

KkgkJ0.386kg1.00" Cu !"++

,

-../

0++,

-../

06

"S

(b) The entropy change of the water is given by:

++,

-../

0")

i

fwaterwaterwater ln

TTcmS


1917

Substitute numerical values and evaluate water"S :

$ %KJ138

K273K26.275ln

KkgkJ18.4kg00.4" water "++

,

-../

0++,

-../

06

"S

(c) Substitute for Cu"S and water"S

and evaluate the entropy change of the universe: K

J02

KJ138

KJ117""" waterCuu

"

&!"&" SSS

Remarks: The result that )Su > 0 tells us that this process is irreversible. 62 •• If a 2.00-kg piece of lead at 100ºC is dropped into a lake at 10ºC, find the entropy change of the universe. Picture the Problem Because the mass of the water in the lake is so much greater than the mass of the piece of lead, the temperature of the lake will increase only slightly and we can reasonably assume that its final temperature is 10(C. We can apply the equation for the entropy change during a constant pressure process to find the entropy changes of the piece of lead, the water in the lake, and the universe. Express the entropy change of the universe in terms of the entropy changes of the lead and the water in the lake:

wPbu SSS )&)") (1)

Using the equation for the entropy change during a constant-pressure process, express and evaluate the entropy change of the lead:

$ %KJ69.70

K373K283ln

KkgkJ0.128kg2.00ln"

i

fPbPbPb !"++

,

-../

0++,

-../

06

"++,

-../

0"

TTcmS

The entropy change of the water in the lake is given by: w

PbPbPb

w

Pb

w

ww

""T

TcmTQ

TQS """

Chapter 19

1918

Substitute numerical values and evaluate )Sw: $ % $ %

J/K41.81K283

K90Kkg

kJ0.128kg2.00" w

"

++,

-../

06

"S

Substitute numerical values in equation (1) and evaluate )Su: K

J11KJ81.41

KJ70.69" u "&!"S

Entropy and 'Lost' Work 63 •• [SSM] A a reservoir at 300 K absorbs 500 J of heat from a second reservoir at 400 K. (a) What is the change in entropy of the universe, and (b) how much work is lost during the process? Picture the Problem We can find the entropy change of the universe from the entropy changes of the high- and low-temperature reservoirs. The maximum amount of the 500 J of heat that could be converted into work can be found from the maximum efficiency of an engine operating between the two reservoirs. (a) The entropy change of the universe is the sum of the entropy changes of the two reservoirs: ++

,

-../

0!!"

&!")&)")

ch

chchu

11TT

Q

TQ

TQSSS


$ %

J/K0.42

K3001

K4001J500" u

"

++,

-../

0!!"S

(b) Relate the heat that could have been converted into work to the maximum efficiency of an engine operating between the two reservoirs:

hmaxQW #"

The maximum efficiency of an engine operating between the two reservoir temperatures is the efficiency of a Carnot device operating between the reservoir temperatures:

h

cCmax 1

TT

!"" ##


1919

Substitute for #max to obtain: h

h

c1 QTTW ++,

-../

0!"

Substitute numerical values and evaluate W: $ % J125J500

K400K3001 "++,

-../

0!"W

64 •• In this problem, 1.00 mol of an ideal gas at 300 K undergoes a free adiabatic expansion from V1 = 12.3 L to V2 = 24.6 L. It is then compressed isothermally and reversibly back to its original state. (a) What is the entropy change of the universe for the complete cycle? (b) How much work is lost in this cycle? (c) Show that the work lost is T)Su. Picture the Problem Although no heat is lost by the gas in the adiabatic free expansion, the process is irreversible and the entropy of the gas increases. In the isothermal reversible process that returns the gas to its original state, the gas releases heat to the surroundings. However, because the process is reversible, the entropy change of the universe is zero. Consequently, the net entropy change is the negative of that of the gas in the isothermal compression. (a) Relate the entropy change of the universe to the entropy changes of the gas during 1 complete cycle:

compresion isothermalduring gas

expansion freeduring gasu """ SSS &"

or, because 0"expansion freeduring gas "S ,

TQSS !

""compresion isothermal

during gasu ""

The work done by the gas during its isothermal compression is given by:

++,

-../

0!"!"!"

i

fonby ln

VVnRTQWW

Substituting for gas"S in the expression for u"S yields:

++,

-../

0!"

i

fu ln"

VVnRS (1)


$ %KJ5.76

KJ763.5

L6.24L3.12ln

KmolJ8.314mol1.00" u ""++

,

-../

0+,-

./0

6!"S

(b) Use Equation 19-22 to find the amount of energy that becomes unavailable for doing work during this process:

$ %

kJ1.73

KJ5.763K300" ulost

"

+,-

./0"" STW

Chapter 19

1920

(c) Use equation (1) to express the product of T and )Su:

ncompressio isothermalits during gasby

i

f

i

fu lnln"

W

VVnRT

VVnRTST

"

++,

-../

0!"+

+,

-../

0++,

-../

0!"

General Problems 65 • A heat engine with an output of 200 W has an efficiency of 30%. It operates at 10.0 cycles/s. (a) How much work is done by the engine during each cycle? (b) How much heat is absorbed from the hot reservoir and how much is released to the cold reservoir during each cycle? Picture the Problem We can use the definition of power to find the work done each cycle and the definition of efficiency to find the heat that is absorbed each cycle. Application of the first law of thermodynamics will yield the heat given off each cycle.

(a) Use the definition of power to relate the work done in each cycle to the frequency of each cycle:

fPtPW "" "cycle

where f is the frequency of the engine.

Substitute numerical values and evaluate Wcycle:

J 20.0s 10.0W200

1cycle "" !W

(b) Express the heat absorbed in each cycle in terms of the work done and the efficiency of the engine:

J670.30

J20.0cyclecycleh, """

#W

Q

Apply the 1st law of thermodynamics to find the heat given off in each cycle:

J47

J20J67cycleh,cyclec,

"

!"!" WQQ

66 • During each cycle, a heat engine operating between two heat reservoirs absorbs 150 J from the reservoir at 100ºC and releases 125 J to the reservoir at 20ºC. (a) What is the efficiency of this engine? (b) What is the ratio of its efficiency to that of a Carnot engine working between the same reservoirs? (This ratio is called the second law efficiency.) Picture the Problem We can use their definitions to find the efficiency of the engine and that of a Carnot engine operating between the same reservoirs.


1921

(a) The efficiency of the engine is given by: h

c

h

ch

h

1Q Q

QQQQW

!"!

""#


%7.16%67.16J150J1251 ""!"#

(b) Find the efficiency of a Carnot engine operating between the same reservoirs:

%45.21K 373K 29311

h

cC "!"!"

TT#

Express the ratio of the two efficiencies:

777.0%45.21%67.16

C

""##

67 • [SSM] An engine absorbs 200 kJ of heat per cycle from a reservoir at 500 K and releases heat to a reservoir at 200 K. Its efficiency is 85 percent of that of a Carnot engine working between the same reservoirs. (a) What is the efficiency of this engine? (b) How much work is done in each cycle? (c) How much heat is released to the low-temperature reservoir during each cycle? Picture the Problem We can use the definition of efficiency to find the work done by the engine during each cycle and the first law of thermodynamics to find the heat released to the low-temperature reservoir during each cycle.

(a) Express the efficiency of the engine in terms of the efficiency of a Carnot engine working between the same reservoirs:

++,

-../

0!""

h

cC 185.085.0

TT##

Substitute numerical values and evaluate # : %51510.0

K500K200185.0 ""++,

-../

0!"#

(b) Use the definition of efficiency to find the work done in each cycle:

$ %$ %MJ0.10

kJ 102kJ200.5100h

"

""" QW #

(c) Apply the first law of thermodynamics to the cycle to obtain: kJ89

kJ021kJ002cycleh,cyclec,

"

!"!" WQQ

68 • Estimate the change in entropy of the universe associated with an Olympic diver diving into the water from the 10-m platform.

Chapter 19

1922

Picture the Problem Assume that the mass of the diver is 75 kg and that the temperature of the water in the pool is 25(C. The energy added to the water in the pool is the change in the gravitational potential energy of the diver during the dive. The change in entropy of the universe associated with a dive is given by:

water

water toaddedwateru ""

TQSS ""

where water toaddedQ is the energy entering the water as a result of the kinetic energy of the diver as he enters the water.

The energy added to the water is the change in the gravitational potential energy of the diver:

wateru"

TmghS "

Substitute numerical values and evaluate u"S :

$ %$ %$ %$ %

KJ25

K27325m 10m/s 81.9kg 75"

2

u

2

&"S

69 • To maintain the temperature inside a house at 20ºC, the electric power consumption of the electric baseboard heaters is 30.0 kW on a day when the outside temperature is –7ºC. At what rate does this house contribute to the increase in the entropy of the universe? Picture the Problem The change in entropy of the universe is the change in entropy of the house plus the change in entropy of the environment. We can find the change in entropy of the house by exploiting the given information that the temperature inside the house is maintained at a constant temperature. We can find the change in entropy of the surrounding by dividing the heat added by the temperature.

Entropy is a state function, and the state of the house does not change. Therefore the entropy of the house does not change:

gssurroundinhouseu """ SSS &" or, because 0" house "S ,

gssurroundinu "" SS "

Heat is absorbed by the surroundings at the same rate R that energy is delivered to the house:

gssurroundingssurroundingssurroundin

""T

tRT

QS ""


1923

Substitute for )Ssurroundings yields: gssurroundin

u""

TtRS " 4

gssurroundin

u

""

TR

tS

"

Substitute numerical values and evaluate )Su/)t: K

W131K266kW30.0

"" u ""

tS

70 •• Calvin Cliffs Nuclear Power Plant, located on the Hobbes River, generates 1.00 GW of power. In this plant, liquid sodium circulates between the reactor core and a heat exchanger located in the superheated steam that drives the turbine. Heat is absorbed by the liquid sodium in the core, and released by the liquid sodium (and into the superheated steam) in the heat exchanger. The temperature of the superheated steam is 500 K. Heat is released into the river, and the water in the river flows by at a temperature of 25ºC. (a) What is the highest efficiency that this plant can have? (b) How much heat is released into the river every second? (c) How much heat must be released by the core to supply 1.00 GW of electrical power? (d) Assume that new environmental laws have been passed to preserve the unique wildlife of the river. Because of these laws, the plant is not allowed to heat the river by more than 0.50ºC. What is the minimum flow rate that the water in the Hobbes River must have? Picture the Problem We can use the expression for the Carnot efficiency of the plant to find the highest efficiency this plant can have. We can then use this efficiency to find the power that must be supplied to the plant to generate 1.00 GW of power and, from this value, the power that is wasted. The rate at which heat is being released to the river is related to the requisite flow rate of the river by .dtdVTcdtdQ <)"

(a) The Carnot efficiency of a plant operating between temperatures Tc and Th is given by:

h

cCmax 1

TT

!"" ##

Substitute numerical values and evaluate #C:

404.0K500K2981max "!"#

(c) The power that must be supplied, at 40.4% efficiency, to produce an output of 1.00 GW is given by: GW48.2

0.404GW00.1

max

outputsupplied

"

""#P

P

(b) Relate the wasted power to the power generated and the power supplied:

generatedsuppliedwasted PPP !"

Chapter 19

1924

Substitute numerical values and evaluate wastedP :

GW48.1

GW00.1GW48.2wasted

"

!"P

(d) Express the rate at which heat is being dumped into the river:

$ %

dtdVTc

VdtdTc

dtdmTc

dtdQ

<

<

)"

)")"

Solve for the flow rate dV/dt of the river:

<TcdtdQ

dtdV

)"

Substitute numerical values (see Table 19-1 for the specific heat of water) and evaluate dV/dt:

$ %

L/s101.7

mkg10K50.0

kgJ4180

sJ1048.1

5

33

9

1"

+,-

./0

++,

-../

0

1"

dtdV

71 •• An inventor comes to you to explain his new invention. It is a novel heat engine using water vapor as the working substance. He claims that the water vapor absorbs heat at 100°C, does work at the rate of 125 W, and releases heat to the air at the rate of only 25.0 W, when the air temperature is 25°C. (a) Explain to him why he cannot be correct. (b) After careful analysis of the data in his prospectus folder, you decide he has made an error in the measurement of his exhausted-heat value. What is the minimum rate of exhausting heat that would make you consider believing him? Picture the Problem We can use the inventor’s data to calculate the thermal efficiency of his steam engine and then compare this value to the efficiency of a Carnot engine operating between the same temperatures. (a) The Carnot efficiency of an engine operating between these temperatures is:

%1.20K 373K 29811

h

cC "!"!"

TT

#

The thermal efficiency of the inventor’s device, in terms of the rate at which it expels heat to the air and does work is:

%3.83 W0.25 W125

W125

ch

"&

"

&""

dtdQ

dtdW

dtdW

dtdQdt

dW

#


1925

You should explain to him that, because the efficiency he claims for his invention is greater than the efficiency of a Carnot engine operating between the same two temperatures, his data is not consistent with what is known about the thermodynamics of engines. He must have made a mistake in his analysis of his data!or he is a con man looking for people to swindle. (b) The maximum efficiency of a steam engine that has ever been achieved is about 50% of the Carnot efficiency of an engine operating between the same temperatures. Setting the efficiency of his steam engine equal to half the Carnot efficiency of the engine yields: dt

dQdt

dWdt

dW

dtdQdt

dW

chC2

1

&""#

Solve for dQc/dt to obtain: dt

dWdt

dQ++,

-../

0!" 12

C

c

#

Assuming that the inventor has measured the work done per cycle by his invention correctly:

$ % W1100 W1251201.02c 2+

,-

./0 !"

dtdQ

a value totally inconsistent with the inventor’s claims for his engine.

Ignoring his claim that 25.0 W of work are done per cycle, let’s assume that his device does take in 150 W of energy each cycle and find how much work would do with an efficiency half that of a Carnot engine:

dtdQdt

dW

hC2

1 "# 4dt

dQdt

dW hC2

1 #"

Substituting numerical values yields: $ %$ % W15 W150201.021 2"

dtdW

Because dt

dWdt

dQdt

dQ!" hc , a

reasonable value for dQc/dt is:

W135 W15 W150c "!"dt

dQ

72 •• The cycle represented in Figure 19-12 (next to Problem 19-14) is for 1.00 mol of an ideal monatomic gas. The temperatures at points A and B are 300 and 750 K, respectively. What is the efficiency of the cyclic process ABCDA?

Chapter 19

1926

Picture the Problem Because the cycle represented in Figure 19-12 is a Carnot cycle, its efficiency is that of a Carnot engine operating between the temperatures of its isotherms.

The Carnot efficiency of the cycle is given by:

h

cC 1

TT

!"#

Substitute numerical values and evaluate #C:

%0.60K750K3001C "!"#

73 •• [SSM] (a) Which of these two processes is more wasteful? (1) A block moving with 500 J of kinetic energy being slowed to rest by sliding (kinetic) friction when the temperature of the environment is 300 K, or (2) A reservoir at 400 K releasing 1.00 kJ of heat to a reservoir at 300 K? Explain your choice. Hint: How much of the 1.00 kJ of heat could be converted into work by an ideal cyclic process? (b) What is the change in entropy of the universe for each process? Picture the Problem All 500 J of mechanical energy are lost, i.e., transformed into heat in process (1). For process (2), we can find the heat that would be converted to work by a Carnot engine operating between the given temperatures and subtract that amount of work from 1.00 kJ to find the energy that is lost. In Part (b) we can use its definition to find the change in entropy for each process. (a) For process (2): inCrecoveredmax,2 QWW #""

The efficiency of a Carnot engine operating between temperatures Th and Tc is given by:

h

cC 1

TT

!"#

and hence

inh

crecovered 1 Q

TTW ++,

-../

0!"

Substitute for C# to obtain:

$ % J250kJ1.00K 400K 3001recovered "+,-

./0 !"W

or 750 J are lost.

Process (2) is more wasteful of available work.

(b) Find the change in entropy of the universe for process (1):

J/K1.67K300J500"" 1 """

TQS


1927

Express the change in entropy of the universe for process (2):

++,

-../

0!)"

)&

)!")&)")

hc

chch2

11TT

Q

TQ

TQSSS

Substitute numerical values and evaluate )S2:

$ %

J/K833.0

K4001

K3001kJ1.00" 2

"

++,

-../

0!"S

74 •• Helium, a monatomic gas, is initially at a pressure of 16 atm, a volume of 1.0 L, and a temperature of 600 K. It is quasi-statically expanded at constant temperature until its volume is 4.0 L and is then quasi-statically compressed at constant pressure until its volume and temperature are such that a quasi-static adiabatic compression will return the gas to its original state. (a) Sketch this cycle on a PV diagram. (b) Find the volume and temperature after the compression at constant pressure. (c) Find the work done during each step of the cycle. (d) Find the efficiency of the cycle. Picture the Problem Denote the three states of the gas as 1, 2, and 3 with 1 being the initial state. We can use the ideal-gas law and the equation of state for an adiabatic process to find the temperatures, volumes, and pressures at points 1, 2, and 3. To find the work done during each cycle, we can use the equations for the work done during isothermal, isobaric, and adiabatic processes. Finally, we find the efficiency of the cycle from the work done each cycle and the heat that enters the system during the isothermal expansion. (a) The PV diagram of the cycle is shown to the right.

(b) Apply the ideal-gas law to the isothermal expansion 1*2 to find P2:

$ % atm4.0L4.0L1.0atm16

2

112 "++

,

-../

0""

VVPP

Chapter 19

1928

Apply an equation for an adiabatic process to relate the pressures and volumes at 1 and 3:

333311 VPVP " 4

31

3

113 ++

,

-../

0"

PPVV

Substitute numerical values and evaluate V3: $ %

L2.3

L2.294atm4.0atm16L1.0

1.671

3

"

"++,

-../

0"V

Apply an equation for an adiabatic process (3 =1.67) to relate the temperatures and volumes at 1 and 3:

111

133

!! " 33 VTVT 41

3

113

!

++,

-../

0"

3

VVTT

Substitute numerical values and evaluate T3: $ %

K103.4

K 344L2.294

L1.0K600

2

11.67

3

1"

"++,

-../

0"

!

T

(c) Express the work done each cycle:

312312 WWWW &&" (1)

For the process 1*2:

$ %$ %

Latm22.18L1.0L4.0lnL1.0atm16

lnln1

211

1

2112

6"

++,

-../

0"

++,

-../

0"++

,

-../

0"

VVVP

VVnRTW


$ %$ %Latm824.6

L4.00L2.294atm4.0" 23223

6!"!"

" VPW


$ % $ %$ %$ % $ %$ %8 9

Latm0.241L2.294atm4.0L1.0atm16

"

23

331123

3123

31V31

6!"

!!"

!!"!!"!" VPVPTTnRTCW

Substitute numerical values in equation (1) and evaluate W: Latm5Latm5.116

Latm10.24Latm6.824Latm18.22

6"6"

6!6!6"W


1929

(d) Use its definition to express the efficiency of the cycle:

1212in WW

QW

QW

"""#


%20Latm22.18Latm5.1162

66

"#

75 •• [SSM] A heat engine that does the work of blowing up a balloon at a pressure of 1.00 atm absorbs 4.00 kJ from a reservoir at 120ºC. The volume of the balloon increases by 4.00 L, and heat is released to a reservoir at a temperature Tc, where Tc < 120ºC. If the efficiency of the heat engine is 50% of the efficiency of a Carnot engine working between the same two reservoirs, find the temperature Tc. Picture the Problem We can express the temperature of the cold reservoir as a function of the Carnot efficiency of an ideal engine and, given that the efficiency of the heat engine is half that of a Carnot engine, relate Tc to the work done by and the heat input to the real heat engine. Using its definition, relate the efficiency of a Carnot engine working between the same reservoirs to the temperature of the cold reservoir:

h

cC 1

TT

!"# 4 $ %Chc 1 #!" TT

Relate the efficiency of the heat engine to that of a Carnot engine working between the same temperatures:

C21

in

## ""QW

4 in

C2QW

"#

Substitute for C# to obtain: ++

,

-../

0!"

inhc

21QWTT

The work done by the gas in expanding the balloon is:

$ %$ %Latm4.00

L4.00atm1.00"6"

"" VPW

Chapter 19

1930

Substitute numerical values and evaluate Tc:

$ % K313kJ4.00

LatmJ101.325Latm4.002

1K393c "

++++

,

-

.

.

.

.

/

0+,-

./0

616

!"T

78 •• Show that the coefficient of performance of a Carnot engine run as a refrigerator is related to the efficiency of a Carnot engine operating between the same two temperatures by C C c hCOP# 1 " T T . Picture the Problem We can use the definitions of the COP and #C to show that their relationship is hCCC COP TT"1# . Using the definition of the COP, relate the heat removed from the cold reservoir to the work done each cycle:

WQcCOP "

Apply energy conservation to relate Qc, Qh, and W:

WQQ !" hc

Substitute for Qc to obtain: W

WQ !" hCOP

Divide the numerator and denominator by Qh and simplify to obtain:

h

hh

1COP

QW

QW

WWQ

!"

!"

Because h

c

hC 1

TT

QW

!""# :

C

h

c

C

h

c

C

Cc

111COP

#### T

TTT

"++,

-../

0!!

"!

"

and

h

ccC COP

TT

"1#

1 77 •• A freezer has a temperature Tc = –23ºC. The air in the kitchen has a temperature Th = 27ºC. The freezer is not perfectly insulated and some heat


1931

leaks through the walls of the freezer at a rate of 50 W. Find the power of the motor that is needed to maintain the temperature in the freezer.

Picture the Problem We can use the definition of the COP to express the work the motor must do to maintain the temperature of the freezer in terms of the rate at which heat flows into the freezer. Differentiation of this expression with respect to time will yield an expression for the power of the motor that is needed to maintain the temperature in the freezer. Using the definition of the COP, relate the heat that must be removed from the freezer to the work done by the motor:

WQcCOP " 4

COPcQW "

Differentiate this expression with respect to time to express the power of the motor:

COPc dtdQ

dtdWP ""

Express the maximum COP of the motor:

TT)

" cmaxCOP

Substitute for COPmax to obtain:

c

c

TT

dtdQP )

"

Substitute numerical values and evaluate P: $ % W10

K250K50W50 "++,

-../

0"P

78 •• In a heat engine, 2.00 mol of a diatomic gas are taken through the cycle ABCA as shown in Figure 19-20. (The PV diagram is not drawn to scale.) At A the pressure and temperature are 5.00 atm and 600 K. The volume at B is twice the volume at A. The segment BC is an adiabatic expansion and the segment CA is an isothermal compression. (a) What is the volume of the gas at A? (b) What are the volume and temperature of the gas at B? (c) What is the temperature of the gas at C? (d) What is the volume of the gas at C? (e) How much work is done by the gas in each of the three segments of the cycle? (f) How much heat is absorbed or released by the gas in each segment of this cycle? Picture the Problem We can use the ideal-gas law to find the unknown temperatures, pressures, and volumes at points A, B, and C and then find the work done by the gas and the efficiency of the cycle by using the expressions for the work done on or by the gas and the heat that enters the system for the constant- pressure, adiabatic, and isothermal processes of the cycle.

Chapter 19

1932

(a) Apply the ideal-gas law to find the volume of the gas at A:

$ % $ %

L19.7

L 69.19m 10

L 1 m 10969.1

atmkPa101.325atm5.00

K600Kmol

J8.314mol2.00

3332

A

AA

"

"11"

1

+,-

./0

6"

"

!!

PnRTV

(b) We’re given that AB 2VV " .

Hence:

$ % L39.4L 38.39L19.692B """V

Apply the ideal-gas law to this constant-pressure process to obtain:

$ %

K1200

2K600A

A

A

BAB

"

""VV

VVTT

(c) Because the process C*A is isothermal:

K600AC "" TT

(d) Apply an equation for an adiabatic process (3 = 1.4) to find the volume of the gas at C:

1CC

1BB

!! " 33 VTVT 41

1

C

BBC

!

++,

-../

0"

3

TTVV

Substitute numerical values and evaluate VC: $ %

L223

L 77.222K600K1200L39.38

11.41

C

"

"++,

-../

0"

!

V

(e) The work done by the gas during the constant-pressure process AB is given by:

$ % $ %AA

AAAABAAB 2VP

VVPVVPW"

!"!"

Substitute numerical values and evaluate WAB:

$ %$ %

kJ9.98J 109754.9Latm

J101.325Latm98.45

L19.69atm5.00

3

AB

"1"

616"

"W


1933

Apply the first law of thermodynamics to express the work done on the gas during the adiabatic expansion BC: BC2

5

BCVBC int,

BC int,BC in,BCint,BC 0

TnR

TncEEQEW

)!"

)!")"

!)"!)"

Substitute numerical values and evaluate WBC:

$ % $ %

kJ24.9

J 10494.2K1200K600Kmol

J8.314mol2.00 425

BC

"

1"!+,-

./0

6!"W

The work done by the gas during the isothermal compression CA is:

$ % $ %

kJ2.24kJ20.24

L222.77L19.69lnK600

KmolJ8.314mol2.00ln

C

ACCA

!"!"

++,

-../

0+,-

./0

6"++

,

-../

0"

VVnRTW

(f) The heat absorbed during the constant-pressure expansion AB is:

$ % $ %

kJ9.34kJ92.34

K600K1200Kmol

J8.314mol2.00"" 27

BA27

BAPAB

""

!+,-

./0

6""" !! TnRTncQ

The heat absorbed during the adiabatic expansion BC is:

0BC "Q

Use the first law of thermodynamics to find the heat absorbed during the isothermal compression CA:

kJ2.24CACA int,CACA

!"

")&" WEWQ

because 0CA int, ")E for an isothermal

process. 79 •• [SSM] In a heat engine, 2.00 mol of a diatomic gas are carried through the cycle ABCDA shown in Figure 19-21. (The PV diagram is not drawn to scale.) The segment AB represents an isothermal expansion, the segment BC an adiabatic expansion. The pressure and temperature at A are 5.00 atm and 600 K. The volume at B is twice the volume at A. The pressure at D is 1.00 atm. (a) What is the pressure at B? (b) What is the temperature at C? (c) Find the total work done by the gas in one cycle.

Chapter 19

1934

Picture the Problem We can use the ideal-gas law to find the unknown temperatures, pressures, and volumes at points B, C, and D. We can then find the work done by the gas and the efficiency of the cycle by using the expressions for the work done on or by the gas and the heat that enters the system for the various thermodynamic processes of the cycle. (a) Apply the ideal-gas law for a fixed amount of gas to the isothermal process AB to find the pressure at B:

$ %

kPa253

kPa253.3atm1

kPa101.325atm2.50

2atm00.5

A

A

B

AAB

"

"1"

""V

VVVPP

(b) Apply the ideal-gas law for a fixed amount of gas to the adiabatic process BC to express the temperature at C:

BB

CCBC VP

VPTT " (1)

Use the ideal-gas law to find the volume of the gas at B:

$ % $ %

L39.39kPa253.3

K600Kmol

J8.314mol2.00

B

BB

"

+,-

./0

6"

"P

nRTV

Use the equation of state for an adiabatic process and 3 = 1.4 to find the volume occupied by the gas at C:

$ %

L75.78atm1.00atm2.50L39.39

1.411

C

BBC

"

++,

-../

0"++

,

-../

0"

3

PPVV

Substitute numerical values in equation (1) and evaluate TC:

$ % $ %$ %$ %$ %

K462

L39.39atm2.50L75.78atm1.00K600C

"

"T

(c) The work done by the gas in one cycle is given by:

DACDBCAB WWWWW &&&"


1935

The work done during the isothermal expansion AB is:

$ % $ % kJ6.915V

2VlnK600Kmol

J8.314mol2.00lnA

A

A

BAAB "++

,

-../

0+,-

./0

6"++

,

-../

0"

VVnRTW

The work done during the adiabatic expansion BC is:

$ % $ %

kJ5.737

K006K624Kmol

J8.314mol2.00"" 25

BC25

BCVBC

"

!+,-

./0

6!"!"!" TnRTCW

The work done during the isobaric compression CD is:

$ % $ %$ %

kJ5.680Latm

J101.325Latm56.09L75.78L19.7atm1.00CDCCD

!"6

16!"!"!" VVPW

Express and evaluate the work done during the constant-volume process DA:

0DA "W

Substitute numerical values and evaluate W: kJ97.6kJ972.6

0kJ5.680kJ5.737kJ915.6

""

&!&"W

80 •• In a heat engine, 2.00 mol of a monatomic gas are taken through the cycle ABCA as shown in Figure 19-20. (The PV diagram is not drawn to scale.) At A the pressure and temperature are 5.00 atm and 600 K. The volume at B is twice the volume at A. The segment BC is an adiabatic expansion and the segment CA is an isothermal compression. (a) What is the volume of the gas at A? (b) What are the volume and temperature of the gas at B? (c) What is the temperature of the gas at C? (d) What is the volume of the gas at C? (e) How much work is done by the gas in each of the three segments of the cycle? (f) How much heat is absorbed by the gas in each segment of the cycle? Picture the Problem We can use the ideal-gas law to find the unknown temperatures, pressures, and volumes at points A, B, and C and then find the work done by the gas and the efficiency of the cycle by using the expressions for the work done on or by the gas and the heat that enters the system for the isobaric, adiabatic, and isothermal processes of the cycle.

Chapter 19

1936

(a) Apply the ideal-gas law to find the volume of the gas at A:

$ % $ %

L19.7L19.69atm

kPa101.325atm5.00

K600Kmol

J8.314mol2.00

A

AA

""

1

+,-

./0

6"

"P

nRTV

(b) We’re given that:

$ %L39.4

L 39.39L19.6922 AB

"

""" VV

Apply the ideal-gas law to this isobaric process to find the temperature at B:

$ %

K1200

2K600A

A

A

BAB

"

""VV

VVTT

(c) Because the process CA is isothermal:

K600AC "" TT

(d) Apply an equation for an adiabatic process (3 = 5/3) to express the volume of the gas at C:

1CC

1BB

!! " 33 VTVT 41

1

C

BBC

!

++,

-../

0"

3

TTVV

Substitute numerical values and evaluate VC: $ %

L111L 4.111

K600K1200L39.39

23

C

""

++,

-../

0"V

(e) The work done by the gas during the isobaric process AB is given by:

$ % $ %AA

AAAABAAB 2VP

VVPVVPW"

!"!"

Substitute numerical values and evaluate WAB:

$ %$ %

kJ9.98kJ 975.9Latm

J101.325Latm98.45

L19.69atm5.00AB

""

616"

"W


1937

Apply the first law of thermodynamics to express the work done by the gas during the adiabatic expansion BC:

$ %BC2

3

BCVBC int,

BC int,

BC in,BC int,BC

0

TnRTncE

EQEW

)!"

)!")"

!)"

!)"

Substitute numerical values and evaluate WBC:

$ % $ %

kJ0.51

kJ 97.14K1200K600Kmol

J8.314mol2.0023

BC

"

"!+,-

./0

6!"W

The work done by the gas during the isothermal compression CA is:

$ % $ %

kJ3.17kJ29.17

L4.111L19.69lnK600

KmolJ8.314mol2.00ln

C

ACCA

!!"

++,

-../

0+,-

./0

6"++

,

-../

0"

VVnRTW

(f) The heat absorbed during the isobaric expansion AB is:

$ % $ %

kJ9.24

K600K1200Kmol

J8.314mol2.00"" 25

AB25

ABPAB in,

"

!+,-

./0

6""" TnRTncQ

Express and evaluate the heat absorbed during the adiabatic expansion BC:

0BC "Q

Use the first law of thermodynamics to express and evaluate the heat absorbed during the isothermal compression CA:

kJ3.17

" CACA int,CACA

!"

"&" WEWQ

because )Eint = 0 for an isothermal process.

81 •• In a heat engine, 2.00 mol of a monatomic gas are carried through the cycle ABCDA shown in Figure 19-21. (The PV diagram is not drawn to scale.) The segment AB represents an isothermal expansion, the segment BC an adiabatic expansion. The pressure and temperature at A are 5.00 atm and 600 K. The volume at B is twice the volume at A. The pressure at D is 1.00 atm.

Chapter 19

1938

(a) What is the pressure at B? (b) What is the temperature at C? (c) Find the total work done by the gas in one cycle. Picture the Problem We can use the ideal-gas law to find the unknown temperatures, pressures, and volumes at points B, C, and D and then find the work done by the gas and the efficiency of the cycle by using the expressions for the work done on or by the gas and the heat that enters the system for the various thermodynamic processes of the cycle. (a) Apply the ideal-gas law for a fixed amount of gas to the isothermal process AB:

$ %

kPa253kPa 3.253

atm1kPa101.325atm2.50

2atm00.5

A

A

B

AAB

""

1"

""V

VVVPP

(b) Apply the ideal-gas law for a fixed amount of gas to the adiabatic process BC:

BB

CCBC VP

VPTT " (1)

Use the ideal-gas law to find the volume at B:

$ % $ %

L39.39kPa253.3

K600Kmol

J8.314mol2.00

B

BB

"

+,-

./0

6"

"P

nRTV

Use the equation of state for an adiabatic process and 3 = 5/3 to find the volume occupied by the gas at C:

$ %

L26.86atm1.00atm2.50L39.39

531

C

BBC

"

++,

-../

0"++

,

-../

0"

3

PPVV

Substitute numerical values in equation (1) and evaluate TC:

$ % $ %$ %$ %$ %

K416K 9.415

L39.39atm2.50L26.86atm1.00K600C

""

"T

(c) The work done by the gas in one cycle is given by:

DACDBCAB WWWWW &&&" (2)


1939

The work done during the isothermal expansion AB is:

$ % $ % kJ6.9152lnK600Kmol

J8.314mol2.00lnA

A

A

BAAB "++

,

-../

0+,-

./0

6"++

,

-../

0"

VV

VVnRTW

The work done during the adiabatic expansion BC is:

$ % $ %

kJ592.4

K006K9.154Kmol

J8.314mol2.00

""

23

BC23

BCVBC

"

!+,-

./0

6!"

!"!" TnRTCW

The work done during the isobaric compression CD is:

$ % $ %$ %

kJ920.4Latm

J101.325Latm56.84L26.86L19.7atm1.00CDCCD

!"6

16!"!"!" VVPW

The work done during the constant-volume process DA is:

0DA "W

Substitute numerical values in equation (2) to obtain: kJ59.6

0kJ920.4kJ592.4kJ915.6

"

&!&"W

82 •• Compare the efficiency of the Otto cycle to the efficiency of the Carnot cycle operating between the same maximum and minimum temperatures. (The Otto cycle is discussed in Section 19-1.) Picture the Problem We can express the efficiency of the Otto cycle using the result from Example 19-2. We can apply the relation constant1 "!3TV to the adiabatic processes of the Otto cycle to relate the end-point temperatures to the volumes occupied by the gas at these points and eliminate the temperatures at c and d. We can use the ideal-gas law to find the highest temperature of the gas during its cycle and use this temperature to express the efficiency of a Carnot engine. Finally, we can compare the efficiencies by examining their ratio.

Chapter 19

1940

The efficiency of the Otto engine is given in Example 19-2:

b

ad

TTTT

!!

!"c

Otto 1# (1)

where the subscripts refer to the various points of the cycle as shown in Figure 19-3.

Apply the relation constant1 "!3TV to the adiabatic process a*b to obtain:

1!

++,

-../

0"

3

b

aab V

VTT

Apply the relation constant1 "!3TV to the adiabatic process c*d to obtain:

1!

++,

-../

0"

3

c

ddc V

VTT

Subtract the first of these equations from the second to obtain:

11 !!

++,

-../

0!++

,

-../

0"!

33

b

aa

c

ddbc V

VTVVTTT

In the Otto cycle, Va = Vd and Vc = Vb. Substitute to obtain:

$ %1

11

!

!!

++,

-../

0!"

++,

-../

0!++

,

-../

0"!

3

33

b

aad

b

aa

b

adbc

VVTT

VVT

VVTTT

Substitute in equation (1) and simplify to obtain:

$ %

b

a

a

b

b

aad

ad

TT

VV

VVTT

TT

!"++,

-../

0!"

++,

-../

0!

!!"

!

!

11

1

1

1Otto

3

3#

Note that, while Ta is the lowest temperature of the cycle, Tb is not the highest temperature.

Apply the ideal-gas law to c and b to obtain an expression for the cycle’s highest temperature Tc:

b

b

c

c

TP

TP" 4 b

b

cbc T

PPTT C"


1941

The efficiency of a Carnot engine operating between the maximum and minimum temperatures of the Otto cycle is given by:

c

a

TT

!"1Carnot#

Express the ratio of the efficiency of a Carnot engine to the efficiency of an Otto engine operating between the same temperatures:

11

1

Otto

Carnot C!

!"

b

a

c

a

TTTT

## because Tc > Tb.

Hence, OttoCarnot ## C

83 ••• [SSM] A common practical cycle, often used in refrigeration, is the Brayton cycle, which involves (1) an adiabatic compression, (2) an isobaric (constant pressure) expansion,(3) an adiabatic expansion, and (4) an isobaric compression back to the original state. Assume the system begins the adiabatic compression at temperature T1, and transitions to temperatures T2, T3 and T4 after each leg of the cycle. (a) Sketch this cycle on a PV diagram. (b) Show that the

efficiency of the overall cycle is given by # " 1!T4 ! T1$ %T3 ! T2$ % . (c) Show that this

efficiency, can be written as # " 1! r 1!3$ % 3 , where r is the pressure ratio Phigh/Plow of the maximum and minimum pressures in the cycle. Picture the Problem The efficiency of the cycle is the ratio of the work done to the heat that flows into the engine. Because the adiabatic transitions in the cycle do not have heat flow associated with them, all we must do is consider the heat flow in and out of the engine during the isobaric transitions. (a) The Brayton heat engine cycle is shown to the right. The paths 1*2 and 3*4 are adiabatic. Heat Qh enters the gas during the isobaric transition from state 2 to state 3 and heat Qc leaves the gas during the isobaric transition from state 4 to state 1. 1

2 3

4

P

V

;

;

hQ

cQ

(b) The efficiency of a heat engine is given by: in

ch

in QQQ

QW !

""# (1)

Chapter 19

1942

During the constant-pressure expansion from state 1 to state 2 heat enters the system:

$ %23PPh23 " TTnCTnCQQ !"""

During the constant-pressure compression from state 3 to state 4 heat enters the system:

$ %41PPc41 " TTnCTnCQQ !!"!"!"

Substituting in equation (1) and simplifying yields:

$ % $ %$ %$ %

$ % $ %$ %$ %$ %23

14

23

4123

23P

41P23P

1TTTT

TTTTTT

TTnCTTnCTTnC

!!

!"

!!&!

"

!!!!!

"#

(c) Given that, for an adiabatic transition, constant 1 "!3TV , use the ideal-gas law to eliminate V and obtain:

constant 1 "!3

3

PT

Let the pressure for the transition from state 1 to state 2 be Plow and the pressure for the transition from state 3 to state 4 be Phigh. Then for the adiabatic transition from state 1 to state 2:

1high

21

low

1!! " 3

3

3

3

PT

PT

4 2

1

high

low1 T

PPT

33 !

++,

-../

0"

Similarly, for the adiabatic transition from state 3 to state 4:

3

1

high

low4 T

PPT

33 !

++,

-../

0"

Subtract T1 from T4 and simplify to obtain:

$ %23

1

high

low

2

1

high

low3

1

high

low14

TTPP

TPPT

PPTT

!++,

-../

0"

++,

-../

0!+

+,

-../

0"!

!

!!

33

33

33


1943

Dividing both sides of the equation by T3 ! T2 yields:

33 1

high

low

23

14

!

++,

-../

0"

!!

PP

TTTT

Substitute in the result of Part (b) and simplify to obtain:

$ % 33

33

33

#

!

!!

!"

++,

-../

0!"+

+,

-../

0!"

1

1

low

high

1

high

low

1

11

r

PP

PP

where low

high

PP

r "

84 ••• Suppose the Brayton cycle engine (see Problem 83) is run in reverse as a refrigerator in your kitchen. In this case, the cycle begins at temperature T1 and expands at constant pressure until its temperature T4. Then the gas is adiabatically compressed, until its temperature is T3. And then it is compressed at constant pressure until its temperature T2. Finally, it adiabatically expands until it returns to its initial state at temperature T1. (a) Sketch this cycle on a PV diagram. (b)

Show that the coefficient of performance isCOPB=T4 ! T1$ %

T3 ! T2 ! T4 & T1$ % .

(c) Suppose your 'Brayton cycle refrigerator' is run as follows. The cylinder containing the refrigerant (a monatomic gas) has an initial volume and pressure of 60 mL and 1.0 atm. After the expansion at constant pressure, the volume and temperature are 75 mL and –25°C. The pressure ratio r = Phigh/Plow for the cycle is 5.0. What is the coefficient of performance for your refrigerator? (d) To absorb heat from the food compartment at the rate of 120 W, what is the rate at which electrical energy must be supplied to the motor of this refrigerator? (e) Assuming the refrigerator motor is actually running for only 4.0 h each day, how much does it add to your monthly electric bill. Assume 15 cents per kWh of electric energy and thirty days in a month. Picture the Problem The efficiency of the Brayton refrigerator cycle is the ratio of the heat that enters the system to the work done to operate the refrigerator. Because the adiabatic transitions in the cycle do not have heat flow associated with them, all we must do is consider the heat flow in and out of the refrigerator during the isobaric transitions.

Chapter 19

1944

(a) The Brayton refrigerator cycle is shown to the right. The paths 1*2 and 3*4 are adiabatic. Heat Qc enters the gas during the constant-pressure transition from state 1 to state 4 and heat Qh leaves the gas during the constant-pressure transition from state 3 to state 2. 1

2 3

4

P

V

;

;

hQ

cQ

(b) The coefficient of performance of the Brayton cycle refrigerator is given by:

ch

hhBCOP

QQQ

WQ

!"" (1)

During the constant-pressure compression from state 3 to state 2 heat leaves the system:

$ %32PPh32 " TTnCTnCQQ !!"!"!"

During the constant-pressure expansion from state 1 to state 4 heat enters the system:

$ %14PPc14 " TTnCTnCQQ !"""

Substituting in equation (1) and simplifying yields:

$ %$ % $ %$ %

$ % $ %

1423

14

3214

14

32P14P

14PBCOP

TTTTTT

TTTTTT

TTnCTTnCTTnC

&!!!

"

!!!!!

"

!!!!!

"

(c) The COPB requires the temperatures corresponding to states 1, 2, 3, and 4. We’re given that the temperature in state 4 is:

K 248K 273C254 "&(!"T

For the constant-pressure transition from state 1 to state 4, the quotient T/V is constant:

4

4

1

1

VT

VT" 4 4

4

11 T

VVT ++,

-../

0"


$ % K 198K 248mL 75mL 60

1 "+,-

./0"T


1945

Given that, for an adiabatic transition, constant 1 "!3TV , use the ideal-gas law to eliminate V and obtain:

constant 1 "!3

3

PT

For the adiabatic transition from state 4 to state 3:

14

41

3

3!! " 3

3

3

3

PT

PT

4 4

1

4

33 T

PPT

33 !

++,

-../

0"


$ % $ % K 473K 2485 67.1167.1

3 ""!

T

Similarly, for the adiabatic transition from state 2 to state 1:

$ % $ %

K378

K 9815 67.1167.1

1

1

1

22

"

"++,

-../

0"

!!

TPPT

33

Substitute numerical values in the expression derived in Part (a) and evaluate COPB: 1.1

K 198K 248K 378K 473K 198K 248COPB

"&!!

!"

(d) From the definition of COPB: BCOP

cQW "

The rate at which energy must be supplied to this refrigerator is given by:

dtdQ

dtdW c

BCOP1

"

or, if the frequency of the AC power input is f,

B

c

COPfQ

dtdW

"

Express the heat Qc that is drawn from the cold reservoir:

$ %14PPc " TTnCTnCQ !""

Substituting for Qc yields: $ %B

14P

COPTTfnC

dtdW !

"

Use the ideal-gas law to express the number of moles of the gas:

4

44

RTVPn "

Chapter 19

1946

Because the gas is monatomic, RC 2

5P " . Substitute for n and CP to

obtain:

$ %

$ %$ % 4B

144425

B

144

4425

COP

COP

TTTVfP

TTRRT

VPf

dtdW

!"

!"

Substitute numerical values and evaluate dW/dt:

$ %$ % $ %

$ %$ %kW 21.0

W207K 2481.11

K 198K 248L

m 10mL 75kPa 325.101s 6033

125

"

"!++

,

-../

01

"

!!

dtdW

(e) The monthly cost of operation is given by

monthper days ofnumber nconsumptiodaily ratenConsumptioPower Power ofPer Unit Cost CostMonthly

11"1"

Substitute numerical values and evaluate the monthly cost of operation:

4$d 03d

h 4.0 W207kWh

15.0$CostMonthly 2111"

85 ••• Using $ % $ %v 2 1 2 1ln ln) " !S C T T nR V V (Equation 19-16) for the entropy change of an ideal gas, show explicitly that the entropy change is zero for a quasi-static adiabatic expansion from state (V1, T1) to state (V2, T2). Picture the Problem We can use VP CCnR !" , VP CC"3 , and

1!3TV = a constant to show that the entropy change for a quasi-static adiabatic expansion that proceeds from state (V1,T1) to state (V2,T2) is zero. Express the entropy change for a general process that proceeds from state 1 to state 2:

++,

-../

0&++,

-../

0")

1

2

1

2V lnln

VVnR

TTCS

For an adiabatic process: 1

2

1

1

2

!

++,

-../

0"

3

VV

TT


1947

Substitute for 1

2

TT and simplify to obtain:

$ %$ %8 9V

1

2

2

1

2

1V

1

2

1

2

1

2

1V

1

2

1

2

1

2

1V

1lnln

ln1ln

ln

lnlnlnln

CnRVV

VV

VVC

nRVV

VVVVC

nRVV

VVnR

VVCS

!!++,

-../

0"

=====

>

?

@@@@@

A

B

!

++,

-../

0!

&++,

-../

0"

=====

>

?

@@@@@

A

B++,

-../

0

&++,

-../

0"++

,

-../

0&++

,

-../

0")

!

!

33

3

3

Use the relationship between CP and CV to obtain:

VP CCnR !"

Substituting for nR and 3 and simplifying yields:

0

1ln VV

pVP

1

2

"

=>

?@A

B++,

-../

0!!!++

,

-../

0") C

CC

CCVVS

86 ••• (a) Show that if the refrigerator statement of the second law of thermodynamics were not true, then the entropy of the universe could decrease. (b) Show that if the heat-engine statement of the second law were not true, then the entropy of the universe could decrease. (c) A third statement of the second law is that the entropy of the universe cannot decrease. Have you just proved that this statement is equivalent to the refrigerator and heat-engine statements? Picture the Problem (a) Suppose the refrigerator statement of the second law is violated in the sense that heat Qc is taken from the cold reservoir and an equal mount of heat is transferred to the hot reservoir and W = 0. The entropy change of the universe is then )Su = Qc/Th ! Qc/Tc. Because Th > Tc, Su < 0, i.e., the entropy of the universe would decrease. (b) In this case, is heat Qh is taken from the hot reservoir and no heat is rejected to the cold reservoir; that is, Qc = 0, then the entropy change of the universe is )Su = !Qh/Th + 0, which is negative. Again, the entropy of the universe would decrease.

Chapter 19

1948

(c) The heat-engine and refrigerator statements of the second law only state that some heat must be rejected to a cold reservoir and some work must be done to transfer heat from the cold to the hot reservoir, but these statements do not specify the minimum amount of heat rejected or work that must be done. The statement )Su D 0 is more restrictive. The heat-engine and refrigerator statements in conjunction with the Carnot efficiency are equivalent to )Su D 0. 87 ••• Suppose that two heat engines are connected in series, such that the heat released by the first engine is used as the heat absorbed by the second engine as shown in Figure 19-22. The efficiencies of the engines are #1 and #2, respectively. Show that the net efficiency of the combination is given by

2121net ##### !&" . Picture the Problem We can express the net efficiency of the two engines in terms of W1, W2, and Qh and then use #1 = W1/Qh and #2 = W2/Qm to eliminate W1, W2, Qh, and Qm. Express the net efficiency of the two heat engines connected in series:

h

21net Q

WW &"#

Express the efficiencies of engines 1 and 2: h

11 Q

W"# and

m

22 Q

W"#

Solve for W1 and W2 and substitute to obtain:

2h

m1

h

m2h1net #####

QQ

QQQ

&"&

"

Express the efficiency of engine 1 in terms of Qm and Qh: h

m1 1

QQ

!"# 4 1h

m 1 #!"QQ

Substitute for Qm/Qh and simplify to obtain:

$ % 2121211net 1 ######## !&"!&"

88 ••• Suppose that two heat engines are connected in series, such that the heat released by the first engine is used as the heat absorbed by the second engine, as shown in Figure 19-22. Suppose that each engine is an ideal reversible heat engine. Engine 1 operates between temperatures Th and Tm and Engine 2 operates between Tm and Tc, where Th > Tm > Tc. Show that the net efficiency of

the combination is given by cnet

h

1# " !TT

. (Note that this result means that two


1949

reversible heat engines operating !in series! are equivalent to one reversible heat engine operating between the hottest and coldest reservoirs.) Picture the Problem We can express the net efficiency of the two engines in terms of W1, W2, and Qh and then use #1 = W1/Qh and #2 = W2/Qm to eliminate W1, W2, Qh, and Qm. Finally, we can substitute the expressions for the efficiencies of the ideal reversible engines to obtain hcnet 1 TT!"# . Express the efficiencies of ideal reversible engines 1 and 2:

h

m1 1

TT

!"# (1)

and

m

c2 1

TT

!"# (2)

The net efficiency of the two engines connected in series is given by:

h

21net Q

WW &"# (3)

Express the efficiencies of engines 1 and 2: h

11 Q

W"# and

m

22 Q

W"#

Solve for W1 and W2 and substitute in equation (3) to obtain:

2h

m1

h

m2h1net #####

QQ

QQQ

&"&

"

Express the efficiency of engine 1 in terms of Qm and Qh: h

m1 1

QQ

!"# 4 1h

m 1 #!"QQ

Substitute for h

m

QQ to obtain:

$ % 211net 1 #### !&"

Substitute for #1 and #2 and simplify to obtain:

h

c

h

c

h

m

h

m

m

c

h

m

h

mnet

11

11

TT

TT

TT

TT

TT

TT

TT

!"!&!"

++,

-../

0!++

,

-../

0&!"#

89 ••• [SSM] The English mathematician and philosopher Bertrand Russell (1872-1970) once said that if a million monkeys were given a million typewriters and typed away at random for a million years, they would produce all of Shakespeare’s works. Let us limit ourselves to the following fragment of Shakespeare (Julius Caesar III:ii):

Chapter 19

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Friends, Romans, countrymen! Lend me your ears. I come to bury Caesar, not to praise him. The evil that men do lives on after them, The good is oft interred with the bones. So let it be with Caesar. The noble Brutus hath told you that Caesar was ambitious, And, if so, it were a grievous fault, And grievously hath Caesar answered it . . . Even with this small fragment, it will take a lot longer than a million years! By what factor (roughly speaking) was Russell in error? Make any reasonable assumptions you want. (You can even assume that the monkeys are immortal.) Picture the Problem There are 26 letters and four punctuation marks (space, comma, period, and exclamation point) used in the English language, disregarding capitalization, so we have a grand total of 30 characters to choose from. This fragment is 330 characters (including spaces) long; there are then 30330 different possible arrangements of the character set to form a fragment this long. We can use this number of possible arrangements to express the probability that one monkey will write out this passage and then an estimate of a monkey’s typing speed to approximate the time required for one million monkeys to type the passage from Shakespeare. Assuming the monkeys type at random, express the probability P that one monkey will write out this passage:

330301

"P

Use the approximation 5.110100030 "2 to obtain: $ %$ %

4954953305.1 10

101

101 !"""P

Assuming the monkeys can type at a rate of 1 character per second, it would take about 330 s to write a passage of length equal to the quotation from Shakespeare. Find the time T required for a million monkeys to type this particular passage by accident:

$ %$ %

$ %

y10

s103.16y1s1030.3

1010s330

484

7491

6

495

2

++,

-../

01

1"

"T

Express the ratio of T to Russell’s estimate:

4786

484

Russell

10y10y10""

TT

or Russell

47810 TT 2

Physics Chapt 19

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Transcript of Physics Chapt 19