Heat Engines, Entropy, and the Second Law of Thermodynamicspds7.egloos.com/pds/200805/26/22/SM_chapter18.pdf · the Second Law of Thermodynamics CHAPTER OUTLINE ... In even the most

491

Heat Engines, Entropy, andthe Second Law of

ThermodynamicsCHAPTER OUTLINE 18.1 Heat Engines and the

Second Law of Thermodynamics

18.2 Reversible and Irreversible Processes

18.3 The Carnot Engine 18.4 Heat Pumps and

Refrigerators 18.5 An Alternative Statement

of the Second Law 18.6 Entropy 18.7 Entropy and the Second

Law of Thermodynamics 18.8 Entropy Changes in

Irreversible Processes 18.9 Context ConnectionThe

Atmosphere as a Heat Engine

ANSWERS TO QUESTIONS Q18.1 First, the efficiency of the automobile engine cannot exceed the

Carnot efficiency: it is limited by the temperature of burning fuel and the temperature of the environment into which the exhaust is dumped. Second, the engine block cannot be allowed to go over a certain temperature. Third, any practical engine has friction, incomplete burning of fuel, and limits set by timing and energy transfer by heat.

Q18.2 No. Any heat engine takes in energy by heat and must also put out

energy by heat. The energy that is dumped as exhaust into the low- temperature sink will always be thermal pollution in the outside environment. So-called “steady growth” in human energy use cannot continue.

Q18.3 A higher steam temperature means that more energy can be extracted from the steam. For a

constant temperature heat sink at Tc , and steam at Th , the efficiency of the power plant is

determined by T T

TTT

h c

h

c

h

−= −1 and is maximized for a high Th .

Q18.4 No. The first law of thermodynamics is a statement about energy conservation, while the second is a

statement about stable thermal equilibrium. They are by no means mutually exclusive. For the particular case of a cycling heat engine, the first law implies Q W Qh eng c= + , and the second law

implies Qc > 0. Q18.5 Take an automobile as an example. According to the first law or the idea of energy conservation, it

must take in all the energy it puts out. Its energy source is chemical energy in gasoline. During the combustion process, some of that energy goes into moving the pistons and eventually into the mechanical motion of the car. Clearly much of the energy goes into heat, which, through the cooling system, is transferred into the atmosphere. Moreover, there are numerous places where friction, both mechanical and fluid, turns mechanical energy into heat. In even the most efficient internal combustion engine cars, less than 30% of the energy from the fuel actually goes into moving the car. The rest ends up as useless heat put into the atmosphere.

492 Heat Engines, Entropy, and the Second Law of Thermodynamics

Q18.6 Suppose the ambient temperature is 20°C. A gas can be heated to the temperature of the bottom of the pond, and allowed to cool as it blows through a turbine. The Carnot efficiency of such an engine

is about eT

Tch

= = =∆ 80373

22%.

Q18.7 No, because the work done to run the heat pump represents energy transferred into the house by

heat. Q18.8 A slice of hot pizza cools off. Road friction brings a skidding car to a stop. A cup falls to the floor and

shatters. Your cat dies. Any process is irreversible if it looks funny or frightening when shown in a videotape running backwards. The free flight of a projectile is nearly reversible.

Q18.9 (a) When the two sides of the semiconductor are at different temperatures, an electric potential

(voltage) is generated across the material, which can drive electric current through an external circuit. The two cups at 50°C contain the same amount of internal energy as the pair of hot and cold cups. But no energy flows by heat through the converter bridging between them and no voltage is generated across the semiconductors.

(b) A heat engine must put out exhaust energy by heat. The cold cup provides a sink to absorb

output or wasted energy by heat, which has nowhere to go between two cups of equally warm water.

Q18.10 The rest of the Universe must have an entropy change of +8.0 J/K, or more.

Q18.11 (a) For an expanding ideal gas at constant temperature, ∆ ∆S

QT

nRVV

= =FHGIKJln 2

1.

(b) For a reversible adiabatic expansion ∆Q = 0 , and ∆S = 0 . An ideal gas undergoing an

irreversible adiabatic expansion can have any positive value for ∆S up to the value given in part (a).

Q18.12 To increase its entropy, raise its temperature. To decrease its entropy, lower its temperature.

“Remove energy from it by heat” is not such a good answer, for if you hammer on it or rub it with a blunt file and at the same time remove energy from it by heat into a constant temperature bath, its entropy can stay constant.

Q18.13 An analogy used by Carnot is instructive: A waterfall continuously converts mechanical energy into

internal energy. It continuously creates entropy as the organized motion of the falling water turns into disorganized molecular motion. We humans put turbines into the waterfall, diverting some of the energy stream to our use. Water flows spontaneously from high to low elevation and energy spontaneously flows by heat from high to low temperature. Into the great flow of solar radiation from Sun to Earth, living things put themselves. They live on energy flow, more than just on energy. A basking snake diverts energy from a high-temperature source (the Sun) through itself temporarily, before the energy inevitably is radiated from the body of the snake to a low-temperature sink (outer space). A tree builds organized cellulose molecules and we build libraries and babies who look like their grandmothers, all out of a thin diverted stream in the universal flow of energy crashing down to disorder. We do not violate the second law, for we build local reductions in the entropy of one thing within the inexorable increase in the total entropy of the Universe. Your roommate’s exercise puts energy into the room by heat.

Chapter 18 493

Q18.14 Either statement can be considered an instructive analogy. We choose to take the first view. All processes require energy, either as energy content or as energy input. The kinetic energy that it possessed at its formation continues to make the Earth go around. Energy released by nuclear reactions in the core of the Sun drives weather on the Earth and essentially all processes in the biosphere. The energy intensity of sunlight controls how lush a forest or jungle can be and how warm a planet is. Continuous energy input is not required for the motion of the planet. Continuous energy input is required for life because energy tends to be continuously degraded, as heat flows into lower-temperature sinks. The continuously increasing entropy of the Universe is the index to energy-transfers completed.

Q18.15 Shaking opens up spaces between jellybeans. The smaller ones more often can fall down into spaces

below them. The accumulation of larger candies on top and smaller ones on the bottom implies a small increase in order, a small decrease in one contribution to the total entropy, but the second law is not violated. The total entropy increases as the system warms up, its increase in internal energy coming from the work put into shaking the box and also from a bit of gravitational energy loss as the beans settle compactly together.

SOLUTIONS TO PROBLEMS

Section 18.1 Heat Engines and the Second Law of Thermodynamics

P18.1 (a) eW

Qh= = =eng J

360 J25 0

0 069 4.

. or 6 94%.

(b) Q Q Wc h= − = − =eng J J J360 25 0 335.

*P18.2 (a) The input energy each hour is

7 89 10 2 50060

1 18 103 9. .× = × J revolution rev minmin

1 h J he jb g

implying fuel input 1 18 101

29 49. .××

FHG

IKJ = J h

L4.03 10 J

L h7e j

(b) Q W Qh c= +eng . For a continuous-transfer process we may divide by time to have

Qt

W

tQ

tW

tQ

tQ

t

h c

h c

∆ ∆ ∆

∆ ∆ ∆

= +

= = −

= × − ×FHG

IKJ = ×

= × FHG

IKJ =

eng

eng

eng

Useful power output

Jrevolution

Jrevolution 1 min

min60 s

W

W1 hp

746 W hp

7 89 10 4 58 10 2 500 rev 11 38 10

1 38 10 185

3 35

5

. ..

.P

(c) PP

engeng J s

rev 60 s rev

2 rad N m= ⇒ = =

× FHG

IKJ = ⋅τω τ

ω π1 38 102 500

1527

5.b g

(d) Q

tc

∆= × F

HGIKJ = ×4 58 10 2 500

1 91 103

5..

Jrevolution

rev60 s

W


P18.3 (a) We have eW

QQ Q

QQQh

h c

h

c

h= =

−= − =eng 1 0 250.

with Qc = 8 000 J, we have Qh = 10 7. kJ

(b) W Q Qh ceng J= − = 2 667

and from P =W

teng

∆, we have ∆t

W= = =eng J

5 000 J s s

P2 667

0 533. .

*P18.4 The engine’s output work we identify with the kinetic energy of the bullet:

W K mv

eW

Q

QW

eQ W Q

h

h

h c

eng

eng

eng

eng

kg 320 m s J

J J

= = = =

=

= = = ×

= +

12

12

0 002 4 123

1230 011

1 12 10

2 2

4

.

..

b g

The energy exhaust is

Q Q W

Q mc T

TQmc

c h= − = × − = ×

=

= =× °

= °

eng4

4

J J 1.10 10 J

1.10 10 J kg C1.80 kg 448 J

C

1 12 10 123

13 7

4.

.

∆

∆

Section 18.2 Reversible and Irreversible Processes Section 18.3 The Carnot Engine P18.5 Tc = 703 K Th = 2 143 K

(a) eT

Tch

= ∆ = =1 4402 143

67 2%.

(b) Qh = ×1 40 105. J , W Qheng = 0 420.

P = = × =W

teng J

s kW

∆5 88 10

158 8

4..

Chapter 18 495

P18.6 When e ec= , 1 − =TT

W

Qc

h h

eng and W

tQ

t

c

hh

TT

eng

∆

∆

= −1

(a) Qt

h

Wt

TT

c

h

=FH IK

−=

×

−

eng W s∆ ∆

1

1 50 10 3 600

1

5

293773

.e jb g

Qh = × =8 70 10 8708. J MJ

(b) Q QW

ttc h= −

FHGIKJ = × − × = × =eng J MJ

∆∆ 8 70 10 1 50 10 3 600 3 30 10 3308 5 8. . .e jb g

P18.7 Isothermal expansion at Th = 523 K

Isothermal compression at Tc = 323 K

Gas absorbs 1 200 J during expansion.

(a) Q QTTc h

c

h=FHGIKJ =

FHGIKJ =1 200 741 J

323523

J

(b) W Q Qh ceng J J= − = − =1 200 741 459b g

P18.8 We use eTTc

c

h= −1

as 0 300 1573

.K= −

Th

From which, Th = = °819 546 K C

P18.9 The Carnot summer efficiency is eTTc s

c

h, .= − = −

++

=1 1273 20273 350

0 530a fa f

K K

And in winter, ec w, .= − =1283623

0 546

Then the actual winter efficiency is 0 3200 5460 530

0 330...

.FHGIKJ = or 33 0%.


P18.10 (a) eTT

c

hmax . .= − = − = × =−1 1

278293

5 12 10 5 12%2

(b) P = = ×W

teng J s

∆75 0 106.

Therefore, Weng J s s h J h= × = ×75 0 10 3 600 2 70 106 11. .e jb g

From eW

Qh= eng we find Q

W

eh = =×

×= × =−

eng J h J h TJ h

2 70 105 12 10

5 27 10 5 2711

212.

.. .

(c) As fossil-fuel prices rise, this way to use solar energy will become a good buy.

*P18.11 (a) eW W

Qe Q e Q

Qh

h h

h=

+=

+eng1 eng2

1

1 1 2 2

1

Now Q Q Q W Q e Qh c h h h2 1 1 1 1 1 1= = − = −eng .

So ee Q e Q e Q

Qe e e eh h h

h=

+ −= + −1 1 2 1 1 1

11 2 1 2

b g.

(b) e e e e eTT

TT

TT

TT

TT

TT

TT

TT

TT

TT

i

h

c

i

i

h

c

i

i

h

c

i

i

h

c

i

c

h

c

h

= + − = − + − − −FHGIKJ −FHGIKJ = − − − + + − = −1 2 1 2 1 1 1 1 2 1 1

The combination of reversible engines is itself a reversible engine so it has the Carnot efficiency.

(c) With W Weng2 eng1= , eW W

Q

W

Qe

h h=

+= =eng1 eng2 eng1

1 11

22

1 2 1

0 12

2

12

− = −FHGIKJ

− = −

= +

= +

TT

TT

TT

TT

T T T

T T T

c

h

i

h

c

h

i

h

i h c

i h cb g

(d) e e

TT

TT

T T T

T T T

i

h

c

i

i c h

i h c

1 2

2

1 2

1 1= = − = −

=

= b g

Chapter 18 497

P18.12 (a) First, consider the adiabatic process D A→ :

P V P VD D A Aγ γ= so P P

VVD A

A

D=FHGIKJ = F

HGIKJ =

γ

1 400 7125 3

kPa10.0 L15.0 L

kPa .

Also nRT

VV

nRTV

VD

DD

A

AA

FHGIKJ =FHGIKJ

γ γ

or T TVVD A

A

D=FHGIKJ = F

HGIKJ =

−γ 1 2 3

720 549 K10.015.0

K .

Now, consider the isothermal process C D→ : T TC D= = 549 K .

P PVV

PVV

VV

P V

V VC DD

CA

A

D

D

C

A A

C D

=FHGIKJ =FHGIKJ

LNMM

OQPPFHGIKJ = −

γ γ

γ 1

PC = =1 400 10 0

24 0 15 0445

5 3

2 3

kPa L

L L kPa

.

. .

a fa f

Next, consider the adiabatic process B C→ : P V P VB B C Cγ γ= .

But, PP V

V VCA A

C D

= −

γ

γ 1 from above. Also considering the isothermal process, P PVVB A

A

B=FHGIKJ .

Hence, PVV

VP V

V VVA

A

BB

A A

C DC

FHGIKJ =FHG

IKJ−

γγ

γγ

1 which reduces to VV V

VBA C

D= = =

10 015 0

16 0.

..

L 24.0 L L

La f

.

Finally, P PVVB A

A

B=FHGIKJ =

FHG

IKJ =1 400 875 kPa

10.0 L16.0 L

kPa .

State P(kPa) V(L) T(K) A

B C D

1 400 875 445 712

10.0 16.0 24.0 15.0

720 720 549 549

(b) For the isothermal process A B→ : ∆ ∆E nC TVint = = 0

so Q W nRTVV

B

A= − =

FHGIKJ = ⋅ F

HGIKJ = +ln . ln

.

..2 34 720

16 010 0

6 58 mol 8.314 J mol K K kJb ga f .

For the adiabatic process B C→ : Q = 0

∆E nC T TV C Bint mol32

J mol K K kJ= − = ⋅LNM

OQP − = −b g b g a f2 34 8 314 549 720 4 98. . .

and W Q E= − + = + − = −∆ int kJ kJ0 4 98 4 98. .a f .

continued on next page


For the isothermal process C D→ : ∆ ∆E nC TVint = = 0

and Q W nRTVV

D

C= − =

FHGIKJ = ⋅ F

HGIKJ = −ln . ln

.

..2 34 549

15 024 0

5 02 mol 8.314 J mol K K kJb ga f .

Finally, for the adiabatic process D A→ : Q = 0

∆E nC T TV A Dint mol32

J mol K K kJ= − = ⋅LNM

OQP − = +b g b g a f2 34 8 314 720 549 4 98. . .

and W Q E= − + = + = +∆ int kJ kJ0 4 98 4 98. . .

Process Q(kJ) W(kJ) ∆Eint (kJ) A B→ +6.58 –6.58 0 B C→ 0 –4.98 –4.98 C D→ –5.02 +5.02 0 D A→ 0 +4.98 +4.98 ABCDA +1.56 –1.56 0

The work done by the engine is the negative of the work input. The output work Weng is

given by the work column in the table with all signs reversed.

(c) eW

QWQh

ABCD

A B= =

−= =

→

eng kJ6.58 kJ1 56

0 237.

. or 23 7%.

eTTc

c

h= − = − =1 1

549720

0 237. or 23 7%.

Section 18.4 Heat Pumps and Refrigerators

*P18.13 COP refrigeratorb g = QW

c

(a) If Qc = 120 J and COP .= 5 00 , then W = 24 0. J

(b) Heat expelled = Heat removed + Work done.

Q Q Wh c= + = + =120 24 144 J J J

Chapter 18 499

*P18.14 COP .= =3 00QW

c . Therefore, WQc=3 00.

.

The heat removed each minute is

QtC = ° ° + ×

+ ° ° = ×

0 030 0 4 186 22 0 0 030 0 3 33 105. kg J kg C . C . kg . J kg

0.030 0 kg 2 090 J kg C 20.0 C 1.40 10 J min4

b gb ga f b ge jb gb ga f

or, Qt

c = 233 J s.

Thus, the work done per sec = = =P233

3 0077 8

J s W

.. .

P18.15 (a) 10 01055 1

3 60011

2 93. . Btu

h W J

1 Btu h

s W J s⋅

FHG

IKJFHG

IKJFHG

IKJFHGIKJ =

(b) Coefficient of performance for a refrigerator: COP refrigeratora f

(c) With EER 5, 510 000

Btu

h WBtu h

⋅=

P: P = = =

⋅

10 0002 000 2 005

Btu h W kW Btu

h W

.

Energy purchased is P ∆t = = ×2 00 1 500 3 00 103. kW h . kWha fb g

Cost kWh kWh= × =3 00 10 0 1003. . $ $300e jb g

With EER 10, 10 10 000Btu

h WBtu h

⋅=

P: P = = =

⋅

10 0001 000 1 0010 Btu

Btu h W kW

h W

.

Energy purchased is P ∆t = = ×1 00 1 50 103. . kW 1 500 h kWha fb g

Cost . kWh kWh= × =1 50 10 0 1003e jb g. $ $150

Thus, the cost for air conditioning is half as much with EER 10

P18.16 COP refriga f = = =TTc

∆27030 0

9 00.

.

P18.17 (a) For a complete cycle, ∆Eint = 0 and W Q Q QQ

Qh c ch

c= − = −

LNMM

OQPP

b g1 .

We have already shown that for a Carnot cycle (and only for a Carnot cycle) QQ

TT

h

c

h

c= .

Therefore, W QT T

Tch c

c=

−LNM

OQP

.



(b) We have the definition of the coefficient of performance for a refrigerator, COP =QW

c .

Using the result from part (a), this becomes COP =−

TT T

c

h c.

P18.18 COP heat pumpa f =+

= = =Q W

WT

Tc h

∆29525

11 8.

P18.19 COP COPCarnot cycle= 0 100.

or QW

QW

h h=FHGIKJ =

FHG

IKJ0 100 0 100

1. .

Carnot cycleCarnot efficiency

QW

TT T

h h

h c=

−FHG

IKJ = −

FHG

IKJ =0 100 0 100

293268

1 17. . . K

293 K K

FIG. P18.19

Thus, 1 17. joules of energy enter the room by heat for each joule of work done.

*P18.20 eWQh

= = 0 350. W Qh= 0 350.

Q W Qh c= + Q Qc h= 0 650.

COP refrigeratorb g = = =QW

QQ

c h

h

0 6500 350

1 86..

.

P18.21 COP Carnot refriga f = = = =TT

QW

c c

∆4 00289

0 013 8.

.

∴ =W 72 2. J per 1 J energy removed by heat.

*P18.22 (a) The TV set uses energy at the rate 40015

3 6 1060

100 106

3 J skWh

1 kWh. J

s

$0.min

$1. minFHGIKJ ×FHG

IKJFHG

IKJ = × − .

(b) An air conditioner is a refrigerator:

COP

QW

Wt COP

cQ

tW

t

Qt

c

c

= =

= = =

∆

∆

∆

∆400

4 5088 9

J s J s

..


Chapter 18 501

This is the rate at which extra electric energy is required by the air conditioner to keep the room cool, with cost

88 915

1 0001

6022 10 4.

$0.min

$2. minWkwh

k

h

FHGIKJFHGIKJFHG

IKJ = × − .

(c) Your father blames his extra output of 170 120 50W W W− = on the TV set. It must be

pumped out of the room by the air conditioner, with power expenditure

W

tQ tCOP

c

∆∆

= = FHGIKJFHGIKJFHG

IKJ = × −50

4 5015

1 0001

6078 10 5 W

kWhk

h

.$0.

min$2. min .

Section 18.5 An Alternative Statement of the Second Law No problems in this section

Section 18.6 Entropy Section 18.7 Entropy and the Second Law of Thermodynamics

P18.23 ∆SdQT

mcdTT

mcT

Ti

f

T

Tf

ii

f

= = =FHGIKJz z ln

∆S = ⋅° FHGIKJ = =250

353293

46 6 195 g 1.00 cal g C cal K J Kb g ln .

P18.24 For a freezing process,

∆ ∆S

QT

= =− ×

= −0 500 3 33 10

273610

5. . kg J kg

K J K

b ge j.

*P18.25 (a) The process is isobaric because it takes place under constant atmospheric pressure. As

described by Newton’s third law, the stewing syrup must exert the same force on the air as the air exerts on it. The heating process is not adiabatic (energy goes in by heat), isothermal (T goes up), isovolumetic (it likely expands a bit), cyclic (it is different at the end), or isentropic (entropy increases). It could be made as nearly reversible as you wish, by not using a kitchen stove but a heater kept always just incrementally higher in temperature than the syrup. The process would then also be eternal, and impractical for food production.

(b) The final temperature is

220 212 8 100 8032

104° = ° + ° = ° + ° − °− °

FHG

IKJ = °F F F C F

100 C212 F

C.



For the mixture,

Q m c T m c T= + = ⋅° + ⋅° ° − °

= × = ×

1 1 2 2

4 5

900 930 104 4 23

9 59 10 4 02 10

∆ ∆ g 1 cal g C g 0.299 cal g C C C

cal J

b ga f.

. .

(c) Consider the reversible heating process described in part (a):

∆SdQT

m c m c dTT

m c m cT

Ti

f

i

ff

i= =

+= +

= + ° FHGIKJ

°FHGIKJ

++

FHG

IKJ

= = ×

z z 1 1 2 21 1 2 2

3

900 1 930 0 2994 186 1 273 104

273 23

4 930 1 20 10

b g b g

a f a f b gb g

ln

..

ln

.

cal C J

1 calC

1 K

J K 0.243 J K

P18.26 We take data from Tables 17.1 and 17.2, and we assume a constant specific heat for each phase. As

the ice is warmed from –12°C to 0°C, its entropy increases by

∆

∆

∆

SdQT

mc dTT

mc T dT mc T

S

S

i

f

= = = =

= ⋅° − = ⋅° FHGIKJ

FHG

IKJ

=

z z z −ice

K

273 K

ice K

273 K

ice K273 K

kg 2 090 J kg C K K kg 2 090 J kg C

J K

261

1

261261

0 027 9 273 261 0 027 9273261

2 62

ln

. ln ln . ln

.

b ga f b g

As the ice melts its entropy change is

∆SQT

mL

Tf= = =

×=

0 027 9

27334 0

..

kg 3.33 10 J kg

K J K

5e j

As liquid water warms from 273 K to 373 K,

∆Smc dT

Tmc

T

Ti

ff

i= =

FHGIKJ = ⋅° F

HGIKJ =z liquid

liquid kg 4 186 J kg C J Kln . ln .0 027 9373273

36 5b g

As the water boils and the steam warms,

∆

∆

SmL

Tmc

T

T

S

v f

i= +

FHGIKJ

=×

+ ⋅° FHGIKJ = +

steam

6 kg 2.26 10 J kg

K kg 2 010 J kg C J K J K

ln

.. ln .

0 027 9

3730 027 9

388373

169 2 21e j b g

The total entropy change is

2 62 34 0 36 5 169 2 21 244. . . .+ + + + =a f J K J K .

For steam at constant pressure, the molar specific heat in Table 17.3 implies a specific heat of

35 41

1 970. J mol K mol

0.018 kg J kg K⋅

FHG

IKJ = ⋅b g , nearly agreeing with 2 010 J kg K⋅ .

Chapter 18 503

P18.27 (a) A 12 can only be obtained one way: 6 6+

(b) A 7 can be obtained six ways: 6 1+ , 5 2+ , 4 3+ , 3 4+ , 2 5+ , 1 6+

P18.28 (a) The table is shown below. On the basis of the table, the most probable result of a toss is

2 heads and 2 tails .

(b) The most ordered state is the least likely state. Thus, on the basis of the table this is

either all heads or all tails .

(c) The most disordered is the most likely state. Thus, this is 2 heads and 2 tails .

Result Possible Combinations Total All heads HHHH 1 3H, 1T THHH, HTHH, HHTH, HHHT 4 2H, 2T TTHH, THTH, THHT, HTTH, HTHT, HHTT 6 1H, 3T HTTT, THTT, TTHT, TTTH 4 All tails TTTT 1 P18.29 (a) Result Possible Combinations Total All red RRR 1 2R, 1G RRG, RGR, GRR 3 1R, 2G RGG, GRG, GGR 3 All green GGG 1 (b) Result Possible Combinations Total All red RRRRR 1 4R, 1G RRRRG, RRRGR, RRGRR, RGRRR, GRRRR 5 3R, 2G RRRGG, RRGRG, RGRRG, GRRRG, RRGGR, RGRGR, GRRGR, RGGRR, GRGRR, GGRRR 10 2R, 3G GGGRR, GGRGR, GRGGR, RGGGR, GGRRG, GRGRG, RGGRG, GRRGG, RGRGG, RRGGG 10 1R, 4G RGGGG, GRGGG, GGRGG, GGGRG, GGGGR 5 All green GGGGG 1

Section 18.8 Entropy Changes in Irreversible Processes

P18.30 ∆SQT

QT

= − = −FHG

IKJ =2

2

1

1

1 000290

1 0005 700

3 27 J K J K.

P18.31 The car ends up in the same thermodynamic state as it started, so it undergoes zero changes in

entropy. The original kinetic energy of the car is transferred by heat to the surrounding air, adding to the internal energy of the air. Its change in entropy is

∆SmvT

= = =12

2 2750 20 0293

1 02.

.a f

J K kJ K .


P18.32 c iron J kg C= ⋅°448 ; cwater J kg C= ⋅°4 186

Q Qcold hot= − : 4 00 10 0 1 00 448 900. . . kg 4 186 J kg C C kg J kg C C⋅° − ° = − ⋅° − °b gd i b gb gd iT Tf f

which yields Tf = ° =33 2 306 2. .C K

∆

∆

∆

∆

Sc m dT

Tc m dT

T

S c m c m

S

S

= +

= FHGIKJ +

FHGIKJ

= ⋅ + ⋅ −

=

z zwater water

K

306.2 Kiron iron

K

306.2 K

water water iron iron

J kg K kg J kg K kg

J K

283 1 173

306 2283

306 21 173

4 186 4 00 0 078 8 448 1 00 1 34

718

ln.

ln.

. . . .b gb gb g b gb ga f

P18.33 Sitting here writing, I convert chemical energy, in ordered molecules in food, into internal energy

that leaves my body by heat into the room-temperature surroundings. My rate of energy output is equal to my metabolic rate,

2 5002 500 10 4 186

1203

kcal d cal

86 400 s J

1 cal W=

× FHG

IKJ =

..

My body is in steady state, changing little in entropy, as the environment increases in entropy at the

rate

∆∆ ∆

∆St

Q Tt

Q tT

= = = =1200 4 1

W293 K

W K W K. ~ .

When using powerful appliances or an automobile, my personal contribution to entropy production

is much greater than the above estimate, based only on metabolism.

P18.34 ∆S nRV

VRf

i=FHGIKJ = =ln ln .2 5 76 J K

There is no change in temperature .

FIG. P18.34

P18.35 ∆S nRV

VRf

i=FHGIKJ =ln . ln0 044 0 2 2b ga f

∆S = =0 088 0 8 314 2 0 507. . ln .a f J K

FIG. P18.35

Chapter 18 505

Section 18.9 Context ConnectionThe Atmosphere as a Heat Engine

P18.36 The Earth presents a circular projected area of π 6 37 106 2. × me j perpendicular to the direction of

energy flow in sunlight. The power incident on the Earth is then

P = = ×LNM

OQP = ×IA 1 370 6 37 10 1 75 106 2 17 W m m W2e j e jπ . . .

The power absorbed in the atmosphere is 0 64 1 75 10 1 1 1017 17. . .× = × W We j . The mechanical wind

power is 0 008 1 1 10 9 1017 14. . × = × W We j .

P18.37 (a) The volume of the air is π r h2 and its mass is m V r h= =ρ ρπ 2 . Then its kinetic energy is

K mv r hv

K

K

= =

= FHG

IKJ

= ×

12

12

12

1 20 300 000 11 00060 0003 600

5 2 10

2 2 2

22

17

ρ π

π.

.

kg m m m m s

J

3e j a f a f

(b) We can suppose the light shines perpendicularly down onto the hurricane. Its intensity is

IA

EA t

tEAI

= = = = × ⋅ ⋅

×= ×P

∆∆:

..

5 2 10

3 10 101 8 10

17

5 2 3

3 J m s

m J s

2

π e j e j.

Additional Problems

P18.38 For the Carnot engine, eTTc

c

h= − = − =1 1

3000 600

K750 K

. .

Also, eW

Qch

= eng .

so QW

ehc

= = =eng J J

1500 600

250.

.

and Q Q Wc h= − = − =eng J J J250 150 100 .

FIG. P18.38

(a) QW

ehS

= = =eng J0.700

J150

214

Q Q Wc h= − = − =eng J J J214 150 64 3.

(b) Qh, .net J J J= − = −214 250 35 7

Qc , . .net J J J= − = −64 3 100 35 7

The net flow of energy by heat from the cold to the hot reservoir without work input, is impossible.

FIG. P18.38(b)



(c) For engine S: Q Q WW

eWc h

S= − = −eng

engeng .

so WQc

eS

eng J

J=−

=−

=1 10 7001100

1233

.

.

and Q Q Wh c= + = + =eng J J J233 100 333 .

(d) Qh, .net J J J= − =333 250 83 3

Wnet J J J= − =233 150 83 3.

Qc ,net = 0

The output of 83.3 J of energy from the heat engine by work in a cyclic process without any exhaust by heat is impossible.

FIG. P18.38(d)

(e) Both engines operate in cycles, so ∆ = ∆ =S SS Carnot 0 .

For the reservoirs, ∆ = −SQTh

h

h and ∆ = +S

QTc

c

c.

Thus, ∆ = ∆ + ∆ + ∆ + ∆ = + − + = −S S S S SS h ctotal CarnotJ

750 K K J K0 0

83 3 0300

0 111.

. .

A decrease in total entropy is impossible. P18.39 As in Section 6.6, let TET represent the energy transferred by electrical transmission. Then

(a) Pelectric =T

tET

∆ so if all the electric energy is converted into internal energy, the steady-state

condition of the house is described by T QET = .

Therefore, Pelectric W= =Qt∆

5 000

(b) For a heat pump, COPK

27 KCarnota f = = =T

Th

∆295

10 92.

Actual COP = = = =0 6 10 92 6 55. . .a f QW

Q tW t

h h ∆∆

Therefore, to bring 5 000 W of energy into the house only requires input power

Pheat pump COP W

6.56 W= = = =W

tQ th

∆∆ 5 000

763

Chapter 18 507

P18.40 The conversion of gravitational potential energy into kinetic energy as the water falls is reversible. But the subsequent conversion into internal energy is not. We imagine arriving at the same final state by adding energy by heat, in amount mgy, to the water from a stove at a temperature infinitesimally above 20.0°C. Then,

∆SdQT

QT

mgyT

= = = = = ×z 5 000 1 000 9 80 50 0

2938 36 106

m kg m m s m

K J K

3 3 2e je ja f. .. .

P18.41 Q mc T mL mc Tc = + + =∆ ∆

Q

Q

QW

TT T

WQ T T

T

c

c

cc

c

h c

c h c

c

= ⋅° ° + × + ⋅° °

= ×

= =−

=−

=× ° − − °

−=

0 500 10 0 500 0 500 20

2 08 10

2 08 10 20 0 20 0

273 20 032 9

5

5

. . .

.

. . .

..

kg 4 186 J kg C C kg 3.33 10 J kg kg 2 090 J kg C C

J

COP refrigerator

J C C

K kJ

5b ga f e j b ga f

b g

b g e j a fa f

*P18.42 (a) Let state i represent the gas before its compression and state f afterwards, VV

fi=

8. For a

diatomic ideal gas, C Rv = 52

, C Rp = 72

, and γ = =C

Cp

V1 40. . Next,

PV P V

P PVV

P P

PV nRT

P VPV

PV nRT nRT

i i f f

f ii

fi i

i i i

f fi i

i i i f

γ γ

γ

=

=FHGIKJ = =

=

= = = =

8 18 4

18 48

2 30 2 30

1.40 .

.. .

so T Tf i= 2 30.

∆ ∆E nC T n R T T nR T PVV f i i i iint . .

. . . .

= = − = =

= × × =−

52

52

1 3052

1 30

52

1 30 1 013 10 0 12 10 39 45 3 3

d i

e j N m m J2

Since the process is adiabatic, Q = 0 and ∆E Q Wint = + gives W = 39 4. J .

(b) The moment of inertia of the wheel is I MR= = = ⋅12

12

5 1 0 085 0 018 42 2. . . kg m kg m2a f . We

want the flywheel to do work 39.4 J, so the work on the flywheel should be –39.4 J:



K W K

I

i f

i

i

rot rot

2

J

J kg m

rad s

+ =

− =

=⋅

FHG

IKJ =

12

39 4 0

2 39 40 018 4

65 4

2

1 2

ω

ω

.

..

.a f

(c) Now we want W K i= 0 05. rot

39 4 0 0512

0 018 4

2 789

0 018 4293

2

1 2

. . .

.

J i= ⋅

=⋅

FHG

IKJ =

kg m

J

kg mrad s

2

2

ω

ω a f

P18.43 (a) For an isothermal process, Q nRTVV

=FHGIKJln 2

1

Therefore, Q nR Ti1 3 2= b g ln

and Q nR Ti312

= FHGIKJb g ln

For the constant volume processes, Q E nR T Ti i232

3= = −∆ int, 2 b g

and Q E nR T Ti i432

3= = −∆ int, 4 b g

The net energy by heat transferred is then

Q Q Q Q Q= + + +1 2 3 4

FIG. P18.43

or Q nRTi= 2 2ln .

(b) A positive value for heat represents energy transferred into the system.

Therefore, Q Q Q nRTh i= + = +1 4 3 1 2lna f

Since the change in temperature for the complete cycle is zero,

∆Eint = 0 and W Qeng =

Therefore, the efficiency is eW

QQQc

h h= = =

+=eng 2 2

3 1 20 273

lnln

.a f

P18.44 ∆Shot J

600 K=

−1 000

∆Scold J

350 K= +750

(a) ∆ ∆ ∆S S SU = + =hot cold J K0 476.


Chapter 18 509

(b) eTTc = − =1 0 4171

2.

W e Qc heng J J= = =0 417 1 000 417. b g

(c) Wnet J J J= − =417 250 167

T SU1 350 167∆ = = K 0.476 J K Jb g

P18.45 eTT

W

Qcc

h h

Wt

Qth

= − = =1 engeng

∆

∆

: Q

tT

T Th

TT

h

h cc

h∆

=−

=−

P P

1e j

Q W Qh c= +eng : Q

tQ

t

W

tc h

∆ ∆ ∆= − eng

Q

tT

T TT

T Tc h

h c

c

h c∆=

−− =

−P

PP

Q mc Tc = ∆ : Q

tmt

c TT

T Tc c

h c∆∆∆

∆= FHGIKJ =

−P

∆∆ ∆mt

TT T c T

c

h c=

−P

b g

∆∆mt

=×

⋅° °= ×

1 00 10 300

200 6 005 97 10

94

.

..

W K

K 4 186 J kg C C kg s

e ja fb ga f

P18.46 eTT

W

Qcc

h h

Wt

Qth

= − = =1 engeng

∆

∆

Q

tT

T Th

TT

h

h cc

h∆

=−

=−

P P

1e j

Q

tQ

tT

T Tc h c

h c∆ ∆=FHGIKJ − =

−P

P

Q mc Tc = ∆ , where c is the specific heat of water.

Therefore, Q

tmt

c TT

T Tc c

h c∆∆∆

∆= FHGIKJ =

−P

and ∆∆ ∆mt

TT T c T

c

h c=

−P

b g .

We test for dimensional correctness by identifying the units of the right-hand side: W C

C J kg C C

J s kg

Jkg s

⋅°° ⋅° °

= =b gb g

, as on the left hand side. Think of yourself as a power-company

engineer arranging to have enough cooling water to carry off your thermal pollution. If the plant power P increases, the required flow rate increases in direct proportion. If environmental regulations require a smaller temperature change, then the required flow rate increases again, now in inverse proportion. If you can run the reactor core or firebox hotter, since Th is in the bottom of the fraction, the required flow rate decreases! If the turbines take in steam at higher temperature, they can be made more efficient to reduce waste heat output.


P18.47 We want to evaluate dPdV

for the function implied by PV nRT= = constant , and also for the different

function implied by PV γ = constant . We can use implicit differentiation:

From PV = constant PdVdV

VdPdV

+ = 0 dPdV

PV

FHGIKJ = −

isotherm

From PV γ = constant P V VdPdV

γ γ γ− + =1 0 dPdV

PV

FHGIKJ = −

adiabat

γ

Therefore, dPdV

dPdV

FHGIKJ = FHG

IKJadiabat isotherm

γ . The theorem is proved.

*P18.48 (a) 35 059

35 0 32 0 1 67 273 15 274 82. . . . . .° = − ° = + =F C K Ka f a f

98 659

98 6 32 0 37 0 273 15 310 15

453 6 1 00 453 6310 15274 82

54 86

453 6 1 00310 15 274 82

310 1551 67

54 86 51 67 3 19

274.82

310 15

. . . . . .

. . . ln..

.

. .. .

..

. . .

.

° = − ° = + =

= = ⋅ × = FHG

IKJ =

= − = −−

= −

= − =

z zF C K K

g cal g K cal K

cal K

cal K

ice water

bodybody

system

a f a f

b gb g

a fa f a f∆

∆

∆

SdQT

dTT

SQ

T

S

(b) 453 6 1 274 82 70 0 10 1 310 153. . . .a fa fb g e ja fb gT TF F− = × −

Thus,

70 0 0 453 6 10 70 0 310 15 0 453 6 274 82 103 3. . . . . .+ × = + ×b g a fa f b ga fTF

and TF = = ° = °309 92 36 77 98 19. . . K C F

∆

∆

′ = FHG

IKJ =

′ = − × FHG

IKJ = −

S

S

ice water

body

cal K

cal K

453 6309 92274 82

54 52

70 0 10310 15309 92

51 933

. ln..

.

. ln..

.e j

∆ ′ = − =Ssys cal K54 52 51 93 2 59. . . which is less than the estimate in part (a).

Chapter 18 511

P18.49 (a) For the isothermal process AB, the work on the gas is

W P VVV

W

W

AB A AB

A

AB

AB

= −FHGIKJ

= − × × FHGIKJ

= − ×

−

ln

. . ln..

.

5 1 013 10 10 0 1050 010 0

8 15 10

5 3

3

Pa m

J

3e je j

where we have used 1 00 1 013 105. .atm Pa= ×

and 1 00 1 00 10 3. .L m3= × −

FIG. P18.49

W P VBC B= − = − × − × = + ×−∆ 1 013 10 10 0 50 0 10 4 05 105 3 3. . . . Pa m J3e j a f

WCA = 0 and W W WAB BCeng J kJ= − − = × =4 11 10 4 113. .

(b) Since AB is an isothermal process, ∆E ABint, = 0

and Q WAB AB= − = ×8 15 103. J

For an ideal monatomic gas, CR

V = 32

and CR

P = 52

T TP VnR R RB AB B= = =

× ×= ×

−1 013 10 50 0 10 5 05 105 3 3. . .e je j

Also, TP VnR R RCC C= =

× ×= ×

−1 013 10 10 0 10 1 01 105 3 3. . .e je j

Q nC T RRCA V= = F

HGIKJ

× − ×FHG

IKJ =∆ 1 00

32

5 05 10 1 01 106 08

3 3

.. .

. kJ

so the total energy absorbed by heat is Q QAB CA+ = + =8 15 6 08 14 2. . . kJ kJ kJ .

(c) Q nC T nR T P VBC P B BC= = =∆ ∆ ∆52

52

a f

QBC = × − × = − × = −−52

1 013 10 10 0 50 0 10 1 01 10 10 15 3 4. . . . .e j a f J kJ

(d) eW

Q

W

Q Qh AB CA= =

+= ×

×=eng eng

4 J

1.42 10 J4 11 10

0 2893.

. or 28 9%.


P18.50 Like a refrigerator, an air conditioner has as its purpose the removal of energy by heat from the cold reservoir.

Its ideal COP is COPK

20 KCarnot =−

= =T

T Tc

h c

28014 0.

(a) Its actual COP is 0 400 14 0 5 60. . .a f = =−

=−

QQ Q

Q t

Q t Q tc

h c

c

h c

∆∆ ∆

5 60 5 60. .Q

tQ

tQ

th c c

∆ ∆ ∆− =

5 60 10 0 6 60. . . kWa f = Qtc

∆ and

Qtc

∆= 8 48. kW

(b) Q W Qh c= +eng : W

tQ

tQ

th ceng kW kW kW

∆ ∆ ∆= − = − =10 0 8 48 1 52. . .

(c) The air conditioner operates in a cycle, so the entropy of the working fluid does not change.

The hot reservoir increases in entropy by

QT

h

h=

×= ×

10 0 10 3 600

3001 20 10

35

..

J s s

K J K

e jb g

The cold room decreases in entropy by

∆SQT

c

c= − = −

×= − ×

8 48 10 3 600

2801 09 10

35

..

J s s

K J K

e jb g

The net entropy change is positive, as it must be:

+ × − × = ×1 20 10 1 09 10 1 09 105 5 4. . . J K J K J K

(d) The new ideal COP is COPK

25 KCarnot =−

= =T

T Tc

h c

28011 2.

We suppose the actual COP is 0 400 11 2 4 48. . .a f =

As a fraction of the original 5.60, this is 4 485 60

0 800..

.= , so the fractional change is to

drop by 20.0% .

Chapter 18 513

P18.51 At point A, PV nRTi i i= and n = 1 00. mol

At point B, 3PV nRTi i B= so T TB i= 3

At point C, 3 2P V nRTi i Cb gb g = and T TC i= 6

At point D, P V nRTi i D2b g = so T TD i= 2

The heat for each step in the cycle is found using CR

V = 32

and

CR

P = 52

:

Q nC T T nRT

Q nC T T nRT

Q nC T T nRT

Q nC T T nRT

AB V i i i

BC P i i i

CD V i i i

DA P i i i

= − =

= − =

= − = −

= − = −

3 3

6 3 7 50

2 6 6

2 2 50

b gb gb gb g

.

.

Q3

Q4

B C

DA

FIG. P18.51

(a) Therefore, Q Q Q Q nRTh AB BC ientering = = + = 10 5.

(b) Q Q Q Q nRTc CD DA ileaving = = + = 8 50.

(c) Actual efficiency, eQ Q

Qh c

h=

−= 0 190.

(d) Carnot efficiency, eTT

TTc

c

h

i

i= − = − =1 1

60 833.

P18.52 ∆SdQT

nC dTT

nC T dT nC T nC T T nCT

Ti

fP

i

f

Pi

f

P TT

P f i Pf

ii

f= = = = = − =FHGIKJz z z −1 ln ln ln lnd i

∆S nCPV

nRnRPV

nCPf

iP=

FHG

IKJ =ln ln3

P18.53 (a) The ideal gas at constant temperature keeps constant

internal energy. As it puts out energy by work in expanding it must take in an equal amount of energy by heat. Thus its entropy increases. Let Pi , Vi , Ti represent the state of the gas before the isothermal expansion. Let PC , VC , Ti represent the state after this process, so that PV P Vi i C C= . Let Pi , 3Vi , Tf represent

the state after the adiabatic compression.

Then P V P VC C i iγ γ= 3b g

A B

C

Pi

PC

Vi 3ViV

P

FIG. P18.53

Substituting PPVVCi i

C=



gives PV V P Vi i C i iγ γ γ− =1 3e j

Then V VC iγ γ γ− −=1 13 and

VV

C

i= −3 1γ γb g

The work output in the isothermal expansion is

W PdV nRT V dV nRTVV

nRT nRTi

C

ii

C

iC

ii i= = =

FHGIKJ = =

−FHGIKJz z − −1 13

13ln ln lnγ γ γ

γb ge j

This is also the input heat, so the entropy change is

∆SQT

nR= =−FHGIKJ

γγ 1

3ln

Since C C C RP V V= = +γ

we have γ − =1b gC RV , CR

V =−γ 1

and CR

P =−

γγ 1

Then the result is ∆S nCP= ln3

(b) The pair of processes considered here carry the gas from the initial state in Problem 18.52 to

the final state there. Entropy is a function of state. Entropy change does not depend on path. Therefore the entropy change in Problem 18.52 equals ∆ ∆S Sisothermal adiabatic+ in this problem. Since ∆Sadiabatic = 0, the answers to Problems 18.52 and 18.53 (a) must be the same.

*P18.54 Simply evaluate the maximum (Carnot) efficiency.

eT

TCh

= = =∆ 4 000 014 4

..

K277 K

The proposal does not merit serious consideration. P18.55 (a) Use the equation of state for an ideal gas

VnRT

P

V

V

A

C

=

=×

= ×

=×

= ×

−

−

1 00 8 314 600

25 0 1 013 101 97 10

1 00 8 314 400

1 013 1032 8 10

53

53

. .

. ..

. .

..

a fa fe ja fa f

m

m

3

3

Since AB is isothermal, P V P VA A B B=

FIG. P18.55

and since BC is adiabatic, P V P VB B C Cγ γ=


Chapter 18 515

Combining these expressions, VPP

VVB

C

A

C

A=FHGIKJ

LNMM

OQPP = FHG

IKJ

×

×

L

NMMM

O

QPPP

− −

−

γ γ1 1 3 1.40

3

1 0 400

1 0025 0

32 8 10

1 97 10

b g b ge j.

.

.

.

.

m

m

3

3

VB = × −11 9 10 3. m3

Similarly, VPP

VVD

A

C

A

C=FHGIKJ

LNMM

OQPP = FHG

IKJ

×

×

L

NMMM

O

QPPP

− −

−

γ γ1 1 3 1.40

3

1 0 400

25 01 00

1 97 10

32 8 10

b g b ge j.

.

.

.

.

m

m

3

3

or VD = × −5 44 10 3. m3

Since AB is isothermal, P V P VA A B B=

and P PVVB A

A

B=FHGIKJ =

××

FHG

IKJ =

−

−25 011 9 10

4 143..

. atm1.97 10 m

m atm

3 3

3

Also, CD is an isothermal and P PVVD C

C

D=FHGIKJ =

××

FHG

IKJ =

−

−1 005 44 10

6 033

3..

. atm32.8 10 m

m atm

3

3

Solving part (c) before part (b):

(c) For this Carnot cycle, eTTc

c

h= − = − =1 1

4000 333

K600 K

.

(b) Energy is added by heat to the gas during the process AB. For the isothermal process,

∆Eint = 0 .

and the first law gives Q W nRTVVAB AB h

B

A= − =

FHGIKJln

or Q Qh AB= = ⋅ FHGIKJ =1 00 600

11 91 97

8 97. ln.

.. mol 8.314 J mol K K kJb ga f

Then, from eW

Qh= eng

the net work done per cycle is W e Qc heng kJ kJ= = =0 333 8 97 2 99. . .a f .

P18.56 Q

WT

T Tc c

h c

Qt

Wt

c

= =−

=COP refrigeratorC b g ∆

∆

0 150 260

0 150 23 1

.

. .

W K40.0 K

W40.0 K260 K

mW

Wt

Wt

∆

∆

=

= = FHG

IKJ =P


P18.57 Define T1 5 00= = ° =Temp Cream . C 278 K . Define T2 60 0= = ° =Temp Coffee C 333 K. . The final

temperature of the mixture is: TT T

f =+

= ° =( . ) ( )

.20 0 200

22055 0 3281 2g g

gC K . The entropy change due

to this mixing is ∆Sc dT

Tc dT

TV

T

T VT

Tf f= +z z20 0 2001 2

. g gb g b g

∆

∆

ST

T

T

T

S

f f=FHGIKJ +

FHGIKJ =

FHGIKJ +

FHGIKJ

= +

84 0 840 84 0328278

840328333

1 18

1 2. ln ln . ln ln

.

J K J K K J J K

J K

b g b g b g b g

P18.58 No energy is transferred by heat during the adiabatic processes. Energy Qh is added by heat during

process BC and energy Qh is exhausted during process DA. We have Q nC T Th V C B= −b g and

Q nC T TC V D A= −b g . So eQ

QT TT T

C

h

D A

C B= − = −

−−

1 1 . For the adiabatic processes, P V P VC D2 1γ γ= and

P V P VB A2 1γ γ= .

From the gas law PnRT

VCC=

2 P

nRTVD

D=1

PnRT

VBB=

2 P

nRTVA

A=1

Then nRT V nRT VC D21

11γ γ− −= and T V T VB A2

11

1γ γ− −= . Thus, TT

VV

rC

D=FHGIKJ =

−−1

2

11

γγ and

TT

VV

rB

A=FHGIKJ =

−−1

2

11

γγ . Then e

T TT r T r r

rD A

D A

= −−−

= − = −− − −−1 1

111 1 1

1γ γ γ

γ .

ANSWERS TO EVEN PROBLEMS P18.2 (a) 29 4. L h; (b) 185 hp; (c) 527 N m⋅ ; (d) 1 91 105. × W P18.4 13 7. °C P18.6 (a) 870 MJ; (b) 330 MJ P18.8 546°C P18.10 (a) 5 12%. ; (b) 5 27. TJ h; (c) As conventional energy sources become

more expensive, or as their true costs are recognized, alternative sources become economically viable.

P18.12 (a) State P (kPa) V (L) T (K) A 1 400 10.0 720 B 875 16.0 720 C 445 24.0 549 D 712 15.0 549

(b) Process Q (kJ) W (kJ) ∆Eint (kJ) A→B 6.58 –6.58 0 B→C 0 –4.98 –4.98 C→D −5.02 5.02 0 D→A 0 4.98 4.98 ABCDA 1.56 –1.56 0 P18.14 77 8. W P18.16 9.00 P18.18 11.8 P18.20 1.86 P18.22 (a) $1. min00 10 3× − ; (b) $2. min22 10 4× − ;

(c) $2. min78 10 5× − P18.24 −610 J K

Chapter 18 517

P18.26 244 J K P18.28 (a) 2 heads and 2 tails ; (b) either all heads or all tails ; (c) 2 heads and 2 tails P18.30 3 27. J K P18.32 718 J K P18.34 5 76. J K , no change in temperature P18.36 9 1014× W P18.38 (a) 214 J, 64.3 J; (b) –35.7 J, –35.7 J. The net effect is the

transport of energy by heat from the cold to the hot reservoir without expenditure of external work.

(c) 333 J, 233 J; (d) 83.3 J, 83.3 J, 0. The net effect is

converting energy, taken in as heat, entirely into energy output by work in a cyclic process.

(e) −0 111. J K , the entropy of the Universe has decreased.

P18.40 8 36 106. × J K P18.42 (a) 39 4. J ; (b) 65 4. rad s ; (c) 293 rad s P18.44 (a) 0 476. J K ; (b) 417 J ; (c) W T SUnet J= =1 167∆

P18.46 mt

TT T c T

c

h c=

−P

b g ∆

P18.48 (a) 3 19. cal K; (b) 98 19. °F, 2 59. cal K P18.50 (a) 8 48. kW ; (b) 1 52. kW ; (c) 1 09 104. × J K ; (d) COP drops by 20.0% P18.52 nCP ln3 P18.54 No, the device cannot meet its

specifications, according to the second law of thermodynamics.

P18.56 23 1. mW P18.58 see the solution

CONTEXT 5 CONCLUSION SOLUTIONS TO PROBLEMS CC5.1 We take 214 K as the tropopause temperature and − °6 5. C km as the lapse rate. Then

214 6 5 13 84

298 K

K C km km Csurface

surface

− = − ° = − °

=

T

T

.b g

CC5.2 The lapse rate is

∆∆

=−

−

=−

= −− °

=∆∆

Ty

T T

y

yT T

Ty

tropopause surface

tropopause

tropopausetropopause surface K K

C km km

0

200 7328 8

60.


CC5.3 (a) The power balance equation for the planet as a whole is

P Pabsorbed emitted absorbing emitting

2 2 4

2

2 4

W m W m K

W m

W m K K

= =

= × ⋅

=× ⋅

=

−

−

:

. .

.

.

e I A e A T

R R T

T

vis s ir

E E

σ

π π

14

2 8 21

4

1 8

4

0 70 1 370 1 5 67 10 4

0 70 1370

4 5 67 10255

e j e jd ie j

e j

(b) The power balance equation for layer 1, required for the constancy of its temperature, is

e R T e R Tir E ir Eσ π σ π8 421

4 22

4d i d i= or 2 14

24T T= . Similarly, for layer 2, 2 2

41

43

4T T T= + . Again,

2 234

24

44 4

14 4T T T T T TN N S= + = +−,K . Thus, T T2

412= .

By substitution, 4 14

14

34T T T= + T T3

413=

6 214

14

44T T T= + T T T NTN4

41

414= =,K

and T N TS = + 1 1 41a f /

(c) From part (b), TS = + =2 1 255 3361 4a f a f/ K K .

(d) The troposphere and stratosphere are too thick to be accurately modeled as having uniform

temperatures. (e) Venus is at 1 08 1011. × m from the Sun, instead of 1 496 1011. × m like the Earth. The intensity

of solar radiation incident on Venus is then 1 3701 4961 08

2 63 102

3 W m W m2 2e j ..

.FHGIKJ = × . The

overall power balance equation for Venus is e I A e A Tvis s irabsorbing emitting= σ 14

1 0 77 2 63 10 1 5 67 10 4

0 23 2 63 10

4 5 67 10227

3 2 8 21

4

1

3

8

4

− × = × ⋅

=×

× ⋅=

−

−

. . .

. .

.

a fe jd i e jd ie j

e j

W m W m K

W m

W m K K

2 2 4

2

2 4

π πR R T

T

V V

(f) T N T NTTS

S= + =FHGIKJ − = FHG

IKJ − =1 1

732227

1 1071 41

1

4 4

a f / :

(g) The multi-layer model should be better for Venus than for Earth. There are many layers, so

the temperature of each can be reasonably uniform.

ANSWERS TO EVEN CONTEXT 5 CONCLUSION PROBLEMS CC5.2 60 km

Documents

Heat Engines, Entropy, and the Second Law of Thermodynamicspds7.egloos.com/pds/200805/26/22/SM_chapter18.pdf · the Second Law of Thermodynamics CHAPTER OUTLINE ... In even the most