Chapter 19: The Second Law of Thermodynamics

Directions of thermodynamic processes

Irreversible and reversible processes

• Thermodynamic processes that occur in nature are all irreversible processes which proceed spontaneously in one direction but not the other.• In a reversible process the system must be capable of being returned to its original state with no other change in the surroundings.• A reversible process proceeds slowly through equilibrium states.

Heat engines

Heat engine

• Any device that transforms heat partly into work or mechanical energy is called a heat engine.• A quantity of matter inside the engine undergoes inflow and outflow of heat, expansion and compression, and sometimes change of phase.• This matter inside the engine is called the working substance of the engine.• The simplest kind of engine to analyze is one in which the working substance undergoes a cyclic process.• All heat engines absorb heat from a source at a relatively high temperature, perform some mechanical work, and discard or reject some heat at a lower temperature.

WQWQUUU 012

For a cyclic process

Heat engines

Energy flow diagram for a heat engine

HT

0HQ

CT

0CQ

Temperature of hot reservoir

Temperature of cold reservoir

Heat flow from hot reservoirper cycle

Heat flow from cold reservoirper cycle

Net heat Q absorbed per cycle:

CHCH QQQQQ Net work done W per cycle:

CHCH QQQQQW

waste!

Heat engines (cont’d)

Energy flow diagram for a heat engine (cont’d)

Net heat Q absorbed per cycle:

CHCH QQQQQ Net work done W per cycle:

CHCH QQQQQW

Thermal efficiency

H

C

H

C

H Q

Q

Q

Q

Q

W 11

Heat engines (cont’d)

Steam engines (external combustion)

Internal-combustion engines

Gasoline engine (a heat engine)

• First a mixture of air and gasoline vapor flows into a cylinder through an open intake valve while the piston descends, increasing the volume of the cylinder from a minimum of V to a maximum of rV (r: compression ratio 8-10).• At the end of intake stroke, the intake valve closes and the mixture is compressed, approximately adiabatically, back to volume V during the compression stroke.• Then the mixture is ignited by the spark plug, and the heated gas expands, approximately adiabatically, back to volume rV, pushing on the piston-power stroke.

Internal-combustion engines (cont’d)

The Otto cycle

:ba

:cb

:dc

:ad

compression stroke(adiabatic compression)

ignite fuel (heating at constant volume)power stroke (adiabatic expansion)

Reject heat to environment(cooling at constant volume)

0)(: bcVH TTnCQcb

0)(: daVC TTnCQad

bc

dabc

H

CH

TT

TTTT

Q

QQ

11)(: VTrVTba ba11)(: VTrVTdc cd

1

1

11

11

)(

)1)((

rTT

rTT

rTrT

TTrTrT

ad

ad

ad

daad

11

1 11

1

rr

rthermal efficiencyin Otto cycle

r=8,=1.4->=56%

Internal-combustion engines (cont’d)

The Otto cycle

Internal-combustion engines (cont’d)

The Diesel cycle

:ba

:cb

:dc

:ad

compression stroke(adiabatic compression)

ignite fuel (heating at constant pressure)power stroke (adiabatic expansion)

Reject heat to environment(cooling at constant volume)

)(

),(,0:

abVabab

abVabab

TTnCUW

TTnCUQba

))((

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bcVpbcbcbc

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)(

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1

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a

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V

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Internal-combustion engines (cont’d)

The Diesel cycle

Refrigerators Refrigerator

A refrigerator works as a reversed heat engine.

Refrigerators Energy flow of a refrigerator

HH QQWW ,

A refrigerator takes heat from lowtemperature reservoir and gives itoff to high temperature reservoir.

From the first law of thermodynamics

WQQWQQ CHCH 0

WQQ CH Coefficient of performance

CH

C

CH

CC

TT

T

QQ

Q

W

QK

( Carnot refrigerator)

Refrigerators Air conditioner

A air conditioner works on exactlythe same principle as a refrigerator.

For air conditioners, the quantities ofgreatest practical importance are therate of heat removal (the heat currentH from the region being cooled) andthe power input P=W/t.

P

H

Pt

Ht

QQ

Q

W

QK

CH

CC

H in Btu/h, P in watts

H/P (energy efficiency rating EER)in (Btu/h)/W. Note 1 W=3.413 Btu/h.EER~7-10.

The second law of thermodynamics The second law of thermodynamics

It is impossible for any system to undergo a process in which it absorbsheat from a reservoir at a single temperature and converts the heatcompletely into mechanical work, with the system ending in the samestate in which it began (Kelvin-Planck statement)

• The kinetic and potential (due to interactions between molecules) energies associated with random motion constitute the internal energy.• When a body sliding on a surface comes to rest as a result of friction, the organized motion of the body is converted to random motion of molecules in the body and in the surface.• Since we can’t control the motions of individual molecules, we can’t convert this random motion COMPLTELY back to organized motion.

It is impossible for any process to have as its sole result the transfer ofheat from a cooler to a hotter body (Clausius statement).

The second law of thermodynamics (cont’d)

Proof : Kelvin-Planck statement = Clausius statement

(a)A workless refrigerator violates 2nd law (K-P statement). If it existed, it could be used to make a 100% efficient engine, which violate K-P statement.

(b) An impossible 100% efficient engine violates C statement. If it existed, it could be used to make a workless refrigerator, which violates C state-

ment.

K-P statement C-statement

impossible impossible

The second law of thermodynamics (cont’d) The second law of thermodynamics

• The conversion of work to heat, as in friction or viscous fluid flow, and heat flow from hot to cold across a finite temperature gradient, are irreversible processes.• Two equivalent statements of the 2nd law state that these processes can be only partially reversed• Gases, for example, always seep spontaneously through an opening from a region of high pressure to a region of low pressure.• The 2nd law is an expression of the inherent one-way aspect of this and many other irreversible processes.

• In any process in which the temperature of the working substance of the engine is intermediate between TH and TC , there must be NO heat transfer between the engine and either reservoir because such heat transfer could not be reversible.• Any process in which the temperature of the working substance changes must be adiabatic.• As heat flow through a finite temperature drop is an irreversible, during heat transfer, there must be NO finite temperature difference.

• To have the maximum efficiency of a heat engine, we must avoid all irreversible processes. Answer: Carnot cycle

The Carnot cycle

The Carnot cycle

reversible expansion

reversible compression

adiabatic expansionadiabatic compression

The Carnot cycle

The Carnot cycle : ideal gas approximation 0,0: WQQba H

0,0: WQQdc C

)(a

bHabH V

VnnRTWQ

)(0)()( dcd

cC

c

dCcdC VV

V

VnnRT

V

VnnRTWQ

dcabdcabaCaHcCbH

H

C

ab

dc

H

C

H

C

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T

T

VVn

VVn

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T

Q

Q

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)/(

)/()(

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H

C

H

C

T

T

Q

Q

H

CH

H

C

T

TT

T

T 1 thermal efficiency

of a Carnot engine

isothermalexpansion

isothermalexpansion

Adiabatic processes:

The Carnot cycle (cont’d)

The Carnot refrigerator

A reversed Carnot engine = a Carnot refrigerator

H

C

H

C

HC

HC

CH

CC

T

T

Q

Q

QQ

QQ

QQ

Q

W

QK

,

/1

/

CH

C

TT

TK

coefficient of performance of a Carnot refrigerator

When the temperature difference is small, K is much larger thanunity, in which case a lot of heat can be pumped from the lowerto the higher temperature with only a little expenditure of work.

The Carnot cycle (cont’d)

The Carnot cycle and the second law

impossibleMore efficientthan a Carnotengine

A Carnot refrigerator

Since each step in the Carnotcycle is reversible, the entirecycle may be reversed. If yourun the entire backward, theengine becomes a refrigerator.

The Carnot cycle (cont’d)

The Carnot cycle and the second law (cont’d)

• The refrigerator does work W=-|W|, takes in heat QC from the cold reservoir, and expels hear |QH| to the hot reservoir.• The superefficient heat engine expels heat |QC|, but to do so, it takes in a greater amount of heat QH+Its work output is then W+.• The net effect of the two machines together is to take a quantity of heat and convert it completely into work, which violates the 2nd law.

• No engine can be more efficient than a Carnot engine.• All Carnot engines operating between the same two temperatures have the same efficiency, irrespective of the nature of the working substance.

Entropy

Entropy and disorder

Entropy provides a quantitative measure of disorder.

Define the infinitesimal entropy change dS during an infinitesimalreversible process at absolute temperature T as:

T

dQdS

If a total amount of heat Q is added during a reversible isothermalprocess at absolute temperature T, the total entropy change is givenby:

T

QSSS 12

• Higher temperature means greater randomness of motion• Adding heat Q causes a substantial fractional increase in molecular motion and randomness

Entropy (cont’d) Entropy and disorder (cont’d)

Generalization of definition of entropy change is to include ANYreversible process leading from one state to another, whetherit is isothermal or not: Let us represent the process as a series of infinitesimal reversible steps. During a step, an infinitesimal quantity of heat dQ is added to the system at absolute temperature T. Then the change of entropy for the entire process is:

2

1 T

dQS entropy change in a

reversible process

• The change in entropy does not depend on the path leading from the initial to the final state but is the same for all possible processes.• Since entropy is a function only of the current state of the system, we can compute entropy change in irreversible processes using the above formula. ( we need to invent a path connecting the given initial and final state that consist entirely of reversible, equilibrium processes and compute the total entropy change as for the actual path).

Entropy (cont’d)

Entropy in cyclic processes

For any reversible process(e.g. Carnot cycle):

0T

dQ

b

aII

a

bII

b

aI

a

bII

b

aI

T

dQ

T

dQ

T

dQ

T

dQ

T

dQ0

The entropy of a system ina given state is independentof the path taken to get there,and is thus a state variable.

Entropy (cont’d) Entropy in irreversible processes

Consider a combined system with a Carnot engine anda cyclic system.

heat reservoir TR

RdQ

Carnotengine revdW

TCyclicsystem

sysdW

dQ

com

bin

ed c

yclic

sys

tem

CsysrevR dUdWdWdQ )(

CdW

RRR

R dQT

dQT

T

dQ

T

dQ Carnot

cycle

CRC dUT

dQTdW

)(0, cyclicdUdUT

dQTW CCRC

Entropy (cont’d) Entropy in irreversible processes

Consider a combined system with a Carnot engine anda cyclic system.

heat reservoir TR

RdQ

Carnotengine revdW

TCyclicsystem

sysdW

dQ

com

bin

ed c

yclic

sys

tem

)(0, cyclicdUdUT

dQTW CCRC

T

dQTW RC

If WC > 0, a cyclic device exchangingheat with a single heat reservoir andproducing an equivalent amount of work.- Violation of K-P statement of 2nd law-

00T

dQWC

-the 2nd law-The equality is true for a reversible process.Clausius inequality

Entropy (cont’d) Entropy in irreversible processes (cont’d)

T

S

A

B

Processes: aAb and aBb reversibleProcess : aCb irreversible

C

• A reversible heat transfer causes changes in entropy of both the system and the reservoir (at least one needed except for adiabatic process)• In an irreversible heat transfer process, there is a finite source of energy instead of an energy reservoir, and the temperature difference during the heat transfer process is finite.

T

dQdS

T

dQSSS

subtractedT

dQ

T

dQ

T

dQ

T

dQ

T

dQ

T

dQ

T

dQ

bCax

baba

bBaaAbbCax

bBa

bCax

aAbaAbCax

)0(0

00

(> for irreversible and = for reversible process)

For an isolated systemdQ=0, so dS>=0.

00 aAb bBaaAbBa T

dQ

T

dQ

T

dQ

Entropy (cont’d) Entropy and the second law

When all systems taking part in a process are included, the entropyeither remains constant or increases. ORNo process is possible in which the total entropy decreases, when allsystems taking part in the process are included.

Exercises

.1400500

350)2000( J

K

KJ

T

TQQ

H

CHC

.600)1400(2000 JJJQQW CH

Problem 1

Solution

A Carnot engine takes 2000 J of heat from a reservoir at 500 K, doessome work, and discards some heat to a reservoir at 350 K.How much work does it do, how much heat is discarded, and whatis the efficiency?

The heat discarded by the engine is:

From the first law, the work done by the engine is:

The thermal efficiency is:

30.0500

35011

K

K

T

T

H

C

ExercisesProblem 2

Solution

One kilogram of water at 0oC is heated to 100oC. Compute itschange in entropy.

In practice, the process described would be done irreversibly, perhapsby setting a pan of water on an electric range whose cooking surface ismaintained at 100oC. But the entropy change of the water depends onlyon the initial and final states of the system, and is the same whetherthe process is reversible or not. Hence we can imagine that the temp.of the water is increased reversibly in a series of infinitesimal steps,in each of which the temp is raised by an infinitesimal amount dT.

2

1

./1031.1 3

1

22

112

T

TKJ

T

Tnmc

T

dTmc

T

dQSSS

mcdTdQ

Example: Refrigerator

• A refrigerator pumps heat from the inside of the freezer (-5°C) to the room (25°C). What is the maximum coefficient of performance?

KCT

KCTo

H

oL

29827325

2682735

9.8268298

268

KK

K

TT

TK

CH

Cideal

• i.e. 8.9 Joules of heat would be pumped from the freezer for every Joule of work done by the compressor. (typical K = 3-5)

ExercisesProblem 3

Solution

What is the entropychange in a freeexpansion process, when the volume isdoubled.

The work done by n moles of ideal gas in an isothermal expansionfrom V1 to V2 is: )/( 12 VVnnRTW

.22

nnRTV

VnnRTWQ

Therefore the entropy change for n=1 is:

./76.5)2)](/(314.8)[1(2 KJnKmolJmolnnRT

QS

ExercisesProblem 4

Solution

A physics student immerses one end of a copper rod in boiling water atat 100oC and the other end in an ice-water mixture at 0oC. The sides of therod are insulated. After steady-state conditions have been achieved in therod, 0.160 kg of ice melts in a certain time interval. For this time interval find(a) the entropy change of the boiling water; (b) the entropy change of theice-water mixture; (c) the entropy change of the copper rod; (d) the totalentropy change of the entire system.

(a) S=Q/T=-mLf/T=-(0.160 kg)(334x103 J/kg)/(373.15 K)=-143 J/K.

(b) S=Q/T=mLf/T=(0.160 kg)(334x103 J/kg)/(273.15 K)= 196 J/K.(c) From the time equilibrium has been reached, there is no net heat exchange between the rod and its surroundings, so the entropy change of the copper rod is zero.(d) 196 J/K-143 J/K=53 J/K.

ExercisesProblem 5

Solution

An experimental power plant at the Natural Energy Lab generateselectricity from the temperature gradient of the ocean. The surface anddeep-water temperatures are 27oC and 6oC, respectively. (a) What isthe maximum theoretical efficiency of this power plant? (b) If the powerplant is to produce 210 kW of power, at what rate must heat be extractedfrom the warm water? At what rate must heat be absorbed by the coldwater? (c) The cold water that enters the plant leaves it at a temperatureof 10oC. What must be the flow rate of cold water through the system?

(a) %0.715.300

15.2791

K

K

(b) .8.2)210)(1/1(2100.3,0.3070.0/210/ MWkWkWMWMWkWPout

(c) ./106/106)4)](/(4190[

)/3600)(108.2(/ 556

hhkgKKkgJ

hsW

Tc

dtQd

dt

dm