Chapter 23 The first law of thermodynamics

23-1 Heat: energy in transit(传输 )

1. “HeatHeat is energy that flows between a system and its environment because of a temperature difference between them” We choose our sign convention so that Q is positive in the case that the internal energy of the system tends to be increased.

Like other forms of energy, heat can be expressed in the SI unit of Joules (J).

2. Misconceptions(误解 ) about heatNeither heat nor work is an intrinsic property of a system.

We cannot say that a system “contains” a certain amount of heat or work. They are not state functions.

Both heat and work associated with a “thermodynamic process”

3. The understanding of heat in historySee动画库 \力学夹 \5-03 作功与传热

23-2 The transfer of heat

1.Thermal conduction

Consider a thin slab of a homogeneous material of thickness and area A (Fig 23-2). One face is held at T and the other at a somewhat higher constant temperature .

Q

A

T

x

x

TT

TT

Fig 23-2

Experiment shows that is (23-1)H is the “rate of heat transfer” k is the “thermal conductivity” of the

materialA is the area of the slab is the thickness of the slab

)(t

QH

Δx

ΔTkAH

xConsidering the direction of H and infinitesimal thicknessof the slab, we have:

dx

dTkAH (23-4)

Sample Problem 23-2

A thin, cylindrical metal pipe is carrying steam at a temperature of Ts=1000c. The pipe has a diameter of 5.4 cm and is wrapped with a thickness of 5.2 cm of fiberglass(玻璃丝 ) insulation. A length D=6.2 m of the pipe passes through a room in which the temperature is TR=110c. At what rate does heat energy pass through the insulation?

2. Convection(对流 )

3. Radiation

Problem

Two identical rectangular rods of metal are welded end to end as shown in Fig. a), and 10 J of heat flows through the rods in 2.0 min. How long would it take for 30 J to flow throught the rods if they are welded as shown in Fig. b?

a)

b)

00C 1000C

1000C00C

23-3 The first law of thermodynamics

1.For a thermodynamic system, internal energy is the only type of energy the system may have. The law of conservation of energy of the system can be expressed as (First law of thermodynamics) (23-6)

(i) Q is the energy transferred (as heat) between the system and its environment because of a temperature different.

intEWQ

(ii) W is the work done on ( or by) the system by

forces that act through the system boundary.(a) (b) (c)

initial state process

final state

Fig 23-9

Environment

Q

W

Eint,i Eint,f

WQEEE if int,int,int

Boundary

(iii) In any thermodynamic process between equilibrium state i and f, the quantity Q+W has the same value. This quantity is equal to the change in the internal energy .

(iv) The first law of the thermodynamics is a general result that is thought to apply to every process in nature that proceeds between equilibrium states.

intE

23-4 Heat capacity (热容 )and specific heat（比热）

1. Heat capacity C： (23-7)2. Specific heat：The heat capacity per unit mass of a body

(23-8)

The heat capacity (C) is characteristic of a particular object,

but the specific heat (c) characterizes a kind of substance.

T

QC

Tm

Q

m

Cc

Usually both C and c depend on the temperature and condition under which the heat Q is added to the material.

3. Molar heat capacityIf we multiply the specific heat by the molar mass M, we obtain the “molar heat capacity”.

dTTcmdTTcmQ )()( (23-10)

Tn

Q

Tm

QMcmol

n--- the molar number

*4. Heats of Transformation (Latent heat(潜热 ))When heat enters a sample, the sample may change from one phase or state to another. In this case, the temperature of the sample does not rise. Vice versa.

Larger

3R

*Measured at room temperature and atmospheric pressure.

A cube of copper of mass mc=75 g is placed in an oven at a temperature of T0=3120C until it comes to thermal equilibrium. The cube is then dropped quickly into an insulated beaker(烧杯 ) containing a quantity of water of mass mw=220 g. The heat capacity of the beaker alone is Cb=190 J/K. Initially the water and the beaker are at a temperature of Ti=12.00c. What is the final equilibrium temperature Tf of the system consisting of the copper +water+beaker?

Sample Problem 23-3

23-5 Work done on or by an ideal gas

Fig 23-13

p

v

(b)

W

1v 2v

1. Work done on an ideal gas

VPW

(a)

x

ideal gasxF

dVVPdW )(

f

i

v

vdVVPW )(

Work done on gas in a more general form: (23-15) PdVW

(a) If PV relationship is known, the work done on the gas is equal to the area under the curve representing the process.

(b) The pressure force is not a conservative force.

A

B

C

DP

VFig 23-14iv fv

Two paths:A B DA C D

(a). Work done at constant volume (V is const.)

(23-16)0W

2. Several typical thermal processes

PdVW

(b). Work done at constant pressure (P is const.) (23-17) dVPdVVPW )(

constantPV

(c). Work done at constant temperature (T is const.)

namely “isothermal (等温 ) process”,

p

v

W=0

iv

fv W 0 p

vfi vv

ΔVP)vP(v if viv fv

p

P(V) curve is hyperbolic(双曲线 ).

The curve of PV=const. is called an “isotherm (等温线 )”.

P

V

W

Fig 23-15

iv fv

i

fv

v v

vnRT

v

dvnRTdv

v

nRTPdVW

f

i

ln (23-18)

v

nRTP

(d). Work done in thermal isolation

Thermal isolation is also called “adiabatic” process. Q=0. T can be changed.

(23-19)const.PV

const.pv

const.pv

iP

fP

iv fv

P

V

Fig 23-16

If we know , and the initialii V,P

constantVPPV i i

, we have

）8.1~1.1(

We can now find the adiabatic work:

V

VPP ii

]1)[()(γ

1γγ1γ1γ

γγ

f

iiifi

ii

γii

V

V γii

V

V

1γ

VPVV

1

VP

V

dVVPdV

V

VP

PdV-W

f

i

(23-20)

(23-21)

By further using , Eq(23-21) becomes

ffi VPVP i

)(1

1W iiff VPVP

(23-22)

Sample problem 23-4

A sample of gas consisting of 0.11 mol is compressed from a volume of to while its pressure increase from 10 to 40 Pa. Compare the work done along the three different paths shown in Fig 23-17.

Fig 23-171 2 3 4

1

2

3

10

20

30

40

P (Pa)( , )

( , )

iP

fP

iv

fv30.1 m30.4 m

)( 3mV

Solution:Path 1

Path 2 represents an isothermal process

JmmPaPi 30)0.40.1)(10()VV(W 33if1

JmPa

v

vVP

v

vnRTW

i

fii

i

f

550.4

0.1ln)0.4)(10(

lnln

3

2

Path 3 JPf 120)VV(0W if3

23-6 The internal energy of an ideal gas

1. Internal energy of ideal monatomic gas Translational kinetic energy is (23-23)

No potential energy. No rotational kinetic energy. So is the entire store of internal energy.

The total internal energy of n moles of an ideal monatomic gas is

(23-24)

KTK trans 2

3

KT2

3

nRTKTnNKnNE AtransA 2

3

2

3)(int

2. Internal energy of molecule consisting of two particles such as

X

y

z

Fig 23-18

2,2 NO CO,

The total kinetic energy of a diatomic molecule is (23-26)

22222

2

1

2

1

2

1

2

1

2

1yyxxzyx IImvmvmvK

d

U=0 Etot=K

3. Degrees of freedom The five terms in Eq23-26 represent

independent ways in which a molecule can absorb energy

and are called ‘degree of freedom”.

4. Equipartition of energy theorem (能量均分定理 )

Maxwell derived a theorem: “When the number of molecule is large, the average energy per molecule is for each independent degree of freedom”

KT2

1

See动画库 \力学夹 \4-13能量均分定理

5. The internal energy for different kinds of ideal gases:

(a) Monatomic ideal gas (23-27)(b) Diatomic gas (23-28)(c) A polyatomic gas generally has six degrees of freedom, (23-29)

nRTKTNE2

3)

2

3(int

nRTKTNE2

5)

2

5(int

nRTKTNE 3)2

6(int

(d) If the molecule is not a rigid one, there are also oscillating degrees of freedom.

Notes: 1* here dependents only on T and degree of freedom.intE

2* Equipartition of energy theorem only apply in classical physics.

23-7 Heat capacities of an ideal gas

1. Molar heat capacity at constant volumeConstant volume (23-30)If represents the molar heat capacity at constant volume, (23-31)

Using Eqs(23-27),(23-28),(23-29), we have

intEQ

vC

Tn

E

Tn

QCv

int

0W

2. Molar heat capacity at constant pressureFig 23-19 shows two isothermsdiffering in temperature by .

KMolJRCv 8.20

2

5

KMolJRCv 9.243 ( polyatomic gas)

( diatomic gas )

Fig 23-19

T

( monatomic gas)KMol

JRCv 5.122

3

Rigid model

A

B

C

V

P

n is fixed in the process.D

Along AC: (23-35) (23-36) (23-27)

Path AB is the constant-volume process; Path AC is a constant pressure process.

The change in internal energy is the same for path AB and AC.

ACAB EE int,int,

WQE AC int,

VPW TnCQ P

Along AB: TnCE VAB int,

A

B

C

V

P

Using the ideal gas law (23-38)

or (23-39)

( monatomic gas) (23-40)

( diatomic gas ) (23-41)

( polyatomic gas) (23-42)

TnRVPW

TnCTnRTnCE VPAC int,

RCC VP

KMolJRCP 8.20

2

5

KMolJRCP 1.29

2

7

KMolJRCP 3.334

Sample problem 23-7

An ideal gas with 0.11 mole begins at the initial point with volume and pressure . Let us compress the gas adiabatically until volume is . Find the change in internal energy of the gas, assuming it to be helium ( a monatomic gas with ).

30.4 mVi PaPi 10

30.1 mV f

66.1

Solution:Using Eq(23-19)

The initial and final temperature are

The change in internal energy is

Pam

mPaVP i

f 100)0.1(

)0.4)(10(

V

P66.13

66.13

f

i

KnR

VT i

i 44Pi K

nR

VT f

f 109Pf

JTnRE 892

3int

23-8 Applications of the first law of thermodynamics

Let’s derive: .constPV (23-19)

0Q

1. Adiabatic process ( )0Q

WE int (23-47)

PdVdWdE int

(for ideal gases)

can always be expressed as:

)0(,int WdTnCdE v

dTnCPdV v (23-48)

nRTPV

)()( nRTdPVd nRdTVdPPdV (23-49)

intdE

Take the ratio between Eqs(23-49) and (23-48)

vv C

R

dTnC

nRdT

PdV

VdPPdV

v

P

v

V

C

C

C

RC

PdV

VdPV

dV

P

dP

f

i

f

i

V

V

P

P V

dV

P

dP i

f

i

f

V

V

P

Plnln

fVPVP fii (23-51)

v

P

C

C

)ln(lni

f

i

f

V

V

P

P

Since i and f are arbitrary points, then we obtain (23-52)We can rewrite these results in terms of temperature, using the ideal gas equation of state (PV=nRT) (23-53) or (23-54)

2. Isothermal process or (23-55)

.constPV

.1 consTV

1)(

f

iif V

VTT

0int E0WQ

3. Constant-volume processes

(23-56)In this case all the heat that enters the gas (Q>0) is stored as internal energy.

4. Cyclical processes

In a cyclical process, the system

restores to its initial state (A-B-C-A).

0WQ

QE int0W

0int EFig 23-21

1

2

3

P

Vor (23-57)

A

B CW>0

5. Free expansionThe gas in Fig(23-22) is initially in one side of the container, and when the stopcock is opened, the gas expands into the previously evacuated half.

In this process, no work is done. The container is insulated, so the process is adiabatic.

gas vacu.

insulatingStopcock (隔离阀 )

Fig 23-22

Hence (23-58)The initial state ( all gas on one side ) is an equilibrium state, as is the final state.

But the intermediate process during intial and final state is nonequilibrium. T and P do not have unique value, and we can not plot this process on a PV diagram.

0W 0Q

0int E )0( T

Summary of typical thermal processes:

Sample problem 23-9

The cycle shown in Fig 23-21 consists of three

processes, starting at point A: A->B: constant-volume; B->C: constant-pressure; C->A: isothermal compression. n=0.75mol diatomic gas with , , . Find Q,W and .

PaPA3102.3

PaPB3102.1

321.0 mVA

intE V

PA

B C

1

2

3

Fig 23-21

AP

CB PP

cVBA VV

Solution:

at point C:

process 1 (A->B)

CAA

A TKnR

VPT 108

356.0 mP

nRTV

C

CC

JTTnCQ ABV 1060)(1 01 W

JE 10601int,

KnR

VPT BB

B 48

process 2 (B->C)

process 3 C->A

JTTnCQ BCP 1480)(2

JVVPW BC 420)(2

JWQE 1060222int,

JV

VnRTW

C

AC 660ln3

03int, E

JWEQ 66033int,3

For the cycle, we have:

JQQQQ 240321

JWWWW 240321

0int WQE