Instructor Solution Manual - Home | Santa Rosa Junior Collegesrjcstaff.santarosa.edu/~lwillia2/41/41ch19hwkey.pdf∆th W S Q A + 0 + B − + 0 C 0 + + D − − − 19.10. Model: Process

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19-1

Conceptual Questions

19.1. sW W= − . (a) s0, 0W W< > . Work is done by the system; the area under the curve is positive. (b) s0, 0W W> < . Work is done on the system to compress it to a smaller volume. (c) s0, 0W W> < . More work is done on the system than by the system.

19.2. 3 1 2 4W W W W> = > . The amount of work done by the gas is the area inside the closed cycle loop (traversed in a clockwise direction).

19.3. No. outH

what you getwhat you had to pay

WQ

h = = . You cannot get out more than you put in.

19.4. (a) 1 (b) 3 (c) 3 (d) 2 In stage 1 the volume is fixed (so no work is done) but the temperature increases, so heat was added. In stage 2 on the isotherm work is done by the gas. In stage 3 work is done on the gas and heat is removed.

19.5. 3 1 2 4h h h h> = > . outH

WQ

h = 14

10h = 2

40100

h = 36

10h = 4

6100

h =

19.6. The thermal efficiency is larger for engine 1; the same amount of heat is added per cycle in both engines, but

the cycle for engine 1 has a larger outW due to the larger enclosed area: outH

WQ

h = .

19.7. It is an isothermal process with equal amounts of heat added to the system and work done by the system.

19.8. (a) No; cannot have out in15 J 10 J .>

(b) Yes, this is a heat engine with outH

4 0 4,10

WQ

h = = = . which is less than Carnot 0 5.h = .

(c) No, it isn’t possible to have Carnot:h h> outH

6 0 6,10

WQ

h = = = . but Carnot 0 5.h = .

HEAT ENGINES AND REFRIGERATORS

19

19-2 Chapter 19


19.9. (a) No, the purpose of a refrigerator is to remove heat from the cold reservoir, and this diagram dumps heat into the cold reservoir.

(b) Yes, Cin

10 0 5,20

QKW

= = = . which is less than Carnot 1.K =

(c) No, Cin

20 2,10

QKW

= = = which is less than Carnot 1.K =

19.10. No. The first law of thermodynamics (energy conservation) for a refrigerator or air conditioner requires H C .Q Q W= + There are no perfect refrigerators (second law), so 0>W (work must be done by the compressor) and

thus H C.Q Q> Because the “air conditioner” exhausts more heat into the room than it extracts from the room, the net effect is to increase the room temperature, not decrease it.

19.11. Yes, the first law says that energy is conserved, so we will never get more work out of the heat engine than heat energy is transferred to the system. In fact, the second law (informal statement #4) says that there are no perfect heat engines with 1,h = so there is always some waste heat exhausted to the cold reservoir.

Exercises and Problems

Section 19.1 Turning Heat into Work

Section 19.2 Heat Engines and Refrigerators

19.1. Solve: (a) The engine has a thermal efficiency of 40% 0 40h = = . and a work output of 100 J per cycle. The heat input is calculated as follows:

outH

H H

100 J0 40 250 JW QQ Q

h = ⇒ . = ⇒ =

(b) Because out H C,W Q Q= − the heat exhausted is

C H out 250 J 100 J 150 JQ Q W= − = − =

19.2. Solve: During each cycle, the work done by the engine is out 200 JW = and the engine exhausts C 400 JQ = of heat energy. By conservation of energy,

H out C 200 J 400 J 600 JQ W Q= + = + = Thus, the efficiency of the engine is

out

H

400 J 0 33600 J

WQ

h = = = .

19.3. Solve: (a) During each cycle, the heat transferred into the engine is H 55 kJ,Q = and the heat exhausted is

C 40 kJ.Q = The thermal efficiency of the heat engine is

C

H

40 kJ1 1 0 27 27%55 kJ

QQ

h = − = − = . =

(b) The work done by the engine per cycle is

out H C 55 kJ 40 kJ 15 kJW Q Q= − = − =

19.4. Solve: The coefficient of performance of the refrigerator is C H in

in in

600 J 200 J 2 0200 J

Q Q WKW W

− −= = = = .

Heat Engines and Refrigerators 19-3


19.5. Solve: (a) The heat extracted from the cold reservoir is calculated as follows:

C CC

in4 0 200 J

50 JQ QK QW

= ⇒ . = ⇒ =

(b) The heat exhausted to the hot reservoir is

H C in 200 J 50 J 250 JQ Q W= + = + =

19.6. Model: Assume that the car engine follows a closed cycle. Solve: (a) Since 2400 rpm is 40 cycles per second, the work output of the car engine per cycle is

outkJ 1 s kJ kJ500 12 5 13s 40 cycles cycle cycle

W = = . ≈

(b) The heat input per cycle is calculated as follows:

outH

H

12 5 kJ 62 5 kJ0 20

W QQ

h .= ⇒ = = ..

The heat exhausted per cycle is

C H in 62 5 kJ 12 5 kJ 50 kJQ Q W= − = . − . =

19.7. Solve: The amount of heat discharged per second is calculated as follows:

9out outC out

H C out

1 11 (900 MW) 1 1 913 10 W0 32

W W Q WQ Q W

hh

= = ⇒ = − = − = . × + .

That is, each second the electric power plant discharges 91 913 10 J. × of energy into the ocean. Since a typical Amer-ican house needs 42 0 10 J. × of energy per second for heating, the number of houses that could be heated with the waste heat is 9 4(1 913 10 J)/(2 0 10 J) 96,000.. × . × =

19.8. Solve: The amount of heat removed from the water in cooling it down in 1 hour is C water waterQ m c T= ∆ . The mass of the water is

3 3 3 3water water water (1000 kg/m )(1 L) (100 kg/m )(10 m ) 1 0 kgm Vr

−= = = = .

4C (1 0 kg)(4190 J/kg K)(20 C 5 C) 6 285 10 JQ = . ° − ° = . ×

The rate of heat removal from the refrigerator is 4

C6 285 10 J 17 46 J/s

3600 sQ . ×= = .

The refrigerator does work 8.0W = W 8.0 J/s= to remove this heat. Thus the performance coefficient of the refrig-erator is

17 46 J/s 2 28 0 J/s

K .= = ..

Section 19.3 Ideal-Gas Heat Engines

Section 19.4 Ideal-Gas Refrigerators

19.9. Model: Process A is isochoric, process B is isothermal, process C is adiabatic, and process D is isobaric. Solve: Process A is isochoric, so the increase in pressure increases the temperature and hence the thermal energy. Because th s s and 0 J, E Q W W Q∆ = − = increases for process A. Process B is adiabatic, so s0 J. Q W= is positive because of the increase in volume. Since s th th0 J , Q W E E= = + ∆ ∆ is negative for process B. Process C is isothermal,

19-4 Chapter 19


so T is constant and hence th 0 J.E∆ = The work done sW is positive because the gas expands. Because

s th , Q W E Q= + ∆ is positive for process B. Process D is isobaric, so the decrease in volume leads to a decrease in temperature and hence a decrease in the thermal energy. Due to the decrease in volume, sW is negative. Because

s th , Q W E Q= + ∆ also decreases for process D.

thE∆ SW Q A + 0 + B − + 0 C 0 + + D − − −

19.10. Model: Process A is adiabatic, process B is isochoric, and process C is isothermal. Solve: Process A is adiabatic, so 0 J.Q = Work sW is positive as the gas expands. Since s th th0 J, Q W E E= + ∆ = ∆ must be negative. The temperature falls during an adiabatic expansion. Process B is isochoric. No work is done

s( 0 J),W = and Q is positive as heat energy is added to raise the temperature th( positive).E∆ Process C is isother-mal so 0T∆ = and th 0 J.E∆ = The gas is compressed, so sW is negative. sQ W= for an isothermal process, so Q is negative. Heat energy is withdrawn during the compression to keep the temperature constant.

thE∆ sW Q A − + 0 B + 0 + C 0 − −

19.11. Solve: The work done by the gas per cycle is the area inside the closed p-versus-V curve. The area inside the triangle is

6 3 6 31out 2

56 31

2

(3 atm 1 atm)(600 10 m 200 10 m )

1 013 10 Pa2 atm (400 10 m ) 40 J1 atm

W − −

−

= − × − ×

. ×= × × =

19.12. Solve: The work done by the gas per cycle is the area enclosed within the pV curve. We have 3 3 51

max max2 6 32(60 J)60 J ( 100 kPa)(800 cm 200 cm ) 1 0 10 Pa

600 10 mp p−= − − ⇒ = − . ××

5max 3 0 10 Pa 300 kPap = . × =

19.13. Model: The heat engine follows a closed cycle, which consists of four individual processes. Solve: (a) The work done by the heat engine per cycle is the area enclosed by the p-versus-V graph. We get

6 3out (400 kPa 100 kPa)(100 10 m ) 30 JW

−= − × =

The heat energy leaving the engine is C 90 J 25 J 115 J.Q = + = The heat input is calculated as follows:

out H C H C out 115 J 30 J 145 J 0.15 kJW Q Q Q Q W= − ⇒ = + = + = ≈

(b) The thermal efficiency of the engine is

out

H

30 J 0 21145 J

WQ

h = = = .

Assess: Practical engines have thermal efficiencies in the range 0 1 0 4.h ≈ . − .



19.14. Model: The heat engine follows a closed cycle, starting and ending in the original state. The cycle consists of three individual processes. Solve: (a) The work done by the heat engine per cycle is the area enclosed by the p-versus-V graph. We get

6 31out 2 (200 kPa)(100 10 m ) 10 JW

−= × =

The heat energy transferred into the engine is H 30 J 84 J 114 J.Q = + = Because out H C,W Q Q= − the heat energy exhausted is

C H out 114 J 10 J 104 J 0 10 kJQ Q W= − = − = ≈ .


out

H

10 J 0 088114 J

h = = = .WQ

Assess: Practical engines have thermal efficiencies in the range 0 1 0 4.h ≈ . − .

19.15. Model: The heat engine follows a closed cycle. Solve: The work done by the gas per cycle is the area inside the closed p-versus-V curve. We get

3 3 3 6 31 1out 2 2(300 kPa 100 kPa)(600 cm 300 cm ) (200 10 Pa)(300 10 m ) 30 JW

−= − − = × × =

Because out H C,W Q Q= − the heat exhausted is

C H out (225 J 90 J) 30 J 315 J 30 J 285 JQ Q W= − = + − = − =

19.16. Model: The heat engine follows a closed cycle. Solve: (a) The work done by the gas per cycle is the area inside the closed p-versus-V curve. We get

3 3 3 6 31 1out 2 2(300 kPa 100 kPa)(600 cm 200 cm ) (200 10 Pa)(400 10 m ) 40 JW

−= − − = × × =

The heat exhausted is C 180 J 100 J 280 J.Q = + = The thermal efficiency of the engine is

out out

H C out

40 J 0 13280 J 40 J

W WQ Q W

h = = = = .+ +

(b) The heat extracted from the hot reservoir is H C out 320 J.Q Q W= + =

19.17. Model: The Brayton cycle involves two adiabatic processes and two isobaric processes. The adiabatic pro-cesses involve compression and expansion through the turbine. Solve: The thermal efficiency for the Brayton cycle is (1 )/B p1 ,r

γ γh −= − where P V/C Cγ = and pr is the pressure ratio. For a diatomic gas 1.4.γ = For an adiabatic process,

1 2 2 1 1 21 2 / ( / )p V p V p p V Vγ γ γ= ⇒ =

Because the volume is halved, 12 12V V= so

1 4p 2 1/ (2) 2 2 639r p p

γ .= = = = .

The efficiency is 0.4/1.4

B 1 (2.639) 0.24h−= − =

19.18. Model: The refrigerator follows a closed cycle. Solve: (a) The net work done on the refrigerator in one cycle is

in s 78 J 119 J 41 JW W= − = − + =

19-6 Chapter 19


This is the work needed to push the heat from the cold reservoir up to the hot reservoir. The heat exhausted to the hot reservoir is H 105 J.Q = From the first law of thermodynamics, the heat extracted from the cold reservoir is

in C H C H in 105 J 41 J 64 JW Q Q Q Q W+ = ⇒ = − = − =

(b) The coefficient of performance is

C

in

64 J 1.641 J

QKW

= = =

Assess: This is a reasonable value for the coefficient of performance for a refrigerator.

Section 19.5 The Limits of Efficiency

Section 19.6 The Carnot Cycle

19.19. Model: The efficiency of a Carnot engine Carnot( )h depends only on the temperatures of the hot and cold reservoirs. On the other hand, the thermal efficiency ( )h of a heat engine depends on the heats HQ and C.Q Solve: (a) According to the first law of thermodynamics, H out C.Q W Q= + For engine (a), H 500 J,Q = C 200 JQ = and out 300 J,W = so the first law of thermodynamics is obeyed. For engine (b), H C500 J, 200 JQ Q= = and

out 200 J,W = so the first law is violated. For engine (c) H C300 J, 200 JQ Q= = and out 100 J,W = so the first law of thermodynamics is obeyed. (b) For the three heat engines, the maximum or Carnot efficiency is

CCarnot

H

300 K1 1 0 50600 K

TT

h = − = − = .

Engine (a) has

C out

H H

300 J1 0 60500 J

Q WQ Q

h = − = = = .

This is larger than Carnot ,h thus violating the second law of thermodynamics. For engine (b),

outCarnot

H

200 J 0 40500 J

WQ

h h= = = . <

so the second law is obeyed. Engine (c) has a thermal efficiency of

Carnot100 J 0 33300 J

h h= = . <

so the second law of thermodynamics is obeyed.

19.20. Model: For a refrigerator H C in ,Q Q W= + and the coefficient of performance and the Carnot coefficient of performance are

C CCarnot

in H C,Q TK K

W T T= =

−

Solve: (a) For refrigerator (a) H C in (60 J 40 J 20 J),Q Q W= + = + so the first law of thermodynamics is obeyed. For refrigerator (b) 50 J 40 J 10 J,= + so the first law of thermodynamics is obeyed. For the refrigerator (c) 40 J 30 J 20 J,≠ + so the first law of thermodynamics is violated. (b) For the three refrigerators, the maximum coefficient of performance is

CCarnot

H C

300 K 3400 K 300 K

TKT T

= = =− −



For refrigerator (a), C

Carnotin

40 J 220 J

QK KW

= = = <

so the second law of thermodynamics is obeyed. For refrigerator (b),

CCarnot

in

40 J 410 J

QK KW

= = = >

so the second law of thermodynamics is violated. For refrigerator (c),

Carnot30 J 1 520 J

K K= = . <

so the second law is obeyed.

19.21. Model: The efficiency of a Carnot engine depends only on the absolute temperatures of the hot and cold reservoirs. Solve: The efficiency of a Carnot engine is

C CCarnot C

H1 0 60 1 280 K 7 C

(427 273) KT T TT

h = − ⇒ . = − ⇒ = = °+

Assess: A “real” engine would need a lower temperature than 7 C° to provide 60% efficiency because no real engine can match the Carnot efficiency.

19.22. Model: The efficiency of an ideal engine (or Carnot engine) depends only on the temperatures of the hot and cold reservoirs. Solve: (a) The engine’s thermal efficiency is

out out

H C out

10 J 0 40 40%15 J 10 J

W WQ Q W

h = = = = . =+ +

(b) The efficiency of a Carnot engine is Carnot C H1 / .T Th = − The minimum temperature in the hot reservoir is found as follows:

HH

293 K0 40 1 488 K 215 CTT

. = − ⇒ = = °

This is the minimum possible temperature. In a real engine, the hot-reservoir temperature would be higher than 215°C because no real engine can match the Carnot efficiency.

19.23. Model: Assume that the heat engine follows a closed cycle. Solve: (a) The engine’s thermal efficiency is

out out

H C out

200 J 0 25 25%600 J 200 J

W WQ Q W

h = = = = . =+ +

(b) The thermal efficiency of a Carnot engine is Carnot C H1 / .T Th = − For this to be 25%,

CC0 25 1 504 8 K 232 C(400 273) K

T T. = − ⇒ = . = °+

19.24. Solve: (a) The efficiency of the Carnot engine is

CCarnot

H

300 K1 1 0 40 40%500 K

TT

h = − = − = . =

(b) An engine with power output of 1000 W does out 1000 JW = of work during each 1 s.t∆ = A Carnot engine has a heat input that is

outin

Carnot

1000 J 2500 J0 40

WQh

= = =.

19-8 Chapter 19


during each 1 s.t∆ = The rate of heat input is 2500 J/s 2500 W.= (c) out in out| |,W Q Q= − so the heat output during 1 st∆ = is out in out| | 1500 JQ Q W= − = . The rate of heat output is thus 1500 J/s 1500 W.=

19.25. Model: We will use Equation 19.27 for the efficiency of a Carnot engine.

CCarnot

H1 T

Th = −

We are given H 673 KT = and the original efficiency Carnot 0 40.h = . Solve: First solve for CT .

C H Carnot(1 ) (673 K)(1 0 40) 404 KT T h= − = − . =

Solve for CT′ again with Carnot 0 60h′ = . .

C H Carnot(1 ) (673 K)(1 0 60) 269 KT T h′ ′= − = − . =

The difference of these CT values is 135 K, so the temperature of the cold reservoir should be decreased by 135°C to raise the efficiency from 40% to 60%. Assess: We expected to have to lower CT by quite a bit to get the better efficiency.

19.26. Model: The maximum possible efficiency for a heat engine is provided by the Carnot engine. Solve: The maximum efficiency is

Cmax Carnot

H

(273 20) K1 1 0 6644(273 600) K

TT

h h += = − = − = .+

Because the heat engine is running at only 30% of the maximum efficiency, max(0 30) 0 1993.h h= . = . The amount of heat that must be extracted is

outH

1000 J 5 0 kJ0 1993

WQh

= = = ..

19.27. Model: We are given H 773 KT = and C 273 K,T = therefore (by Equation 19.27) the Carnot efficiency is 273 K

Carnot 773 K1 0 647n = − = . We are also given Carnot0 60( ).h h= .

Solve: Rearrange Equation 19.5: H out (1 ).Q W h= − outW is the same for both engines, so it cancels.

H out Carnot

H Carnot out Carnot Carnot

(1 ) 1 0 60( ) 1 0 60(0 647) 1 7( ) (1 ) 1 1 0 647

Q WQ W

h hh h− − . − . .

= = = = .− − − .

Assess: This engine requires 1.7 times as much heat energy during each cycle as a Carnot engine to do the same amount of work.

19.28. Model: The coefficient of performance of a Carnot refrigerator depends only on the temperatures of the cold and hot reservoirs. Solve: (a) The Carnot performance coefficient of a refrigerator is

CCarnot

H C

( 20 273) K 6 325 6 3(20 273) K ( 20 273) K

TKT T

− += = = . ≈ .

− + − − +

(b) The rate at which work is done on the refrigerator is found as follows:

C Cin

in

200 J/s 32 J/s 32 W6 325

Q QK WW K

= ⇒ = = = =.

(c) The heat exhausted to the hot side per second is

H C in 200 J/s 32 J/s 232 J/s 0 23 kWQ Q W= + = + = ≈ .



19.29. Model: The minimum possible value of CT occurs with a Carnot refrigerator. Solve: (a) For the refrigerator, the coefficient of performance is

CC in

in(5 0)(10 J) 50 JQK Q KW

W= ⇒ = = . =

The heat energy exhausted per cycle is

H C in 50 J 10 J 60 JQ Q W= + = + =

(b) If the hot-reservoir temperature is 27°C 300 K,= the lowest possible temperature of the cold reservoir can be obtained as follows:

C CCarnot C

H C C5 0 250 K 23 C

300 KT TK T

T T T= ⇒ . = ⇒ = = − °

− −

19.30. Model: Equation 19.27 gives CCarnotH

1 .TTh = − We are given Carnot 1 3/h = .

Solve:

C CCarnot H C

H H

1 2 313 3 2

T T T TT T

h = − = ⇒ = ⇒ =

Equation 19.28 gives the coefficient of performance for the Carnot refrigerator.

C CCarnot 3 3

H C C C2 2

1 21

T TKT T T T

= = = =− − −

Assess: This result is in the ballpark for coefficients of performance.

19.31. Solve: The work done by the engine is equal to the change in the gravitational potential energy. Thus, 2

out grav (2000 kg)(9 8 m/s )(30 m) 588,000 JW U mgh= ∆ = = . =

The efficiency of this engine is

CCarnot

H

(273 20) K0 40 0 40 1 0 40 1 0 3484(273 2000) K

TT

h h +

= . = . − = . − = . +

The amount of heat energy transferred is calculated as follows:

6out outH

H

588,000 J 1 7 10 J0 3484

W WQQ

hh

= ⇒ = = = . ×.

19.32. Solve: The mass of the water is 3 3

33

10 m 1000 kg(100 10 L) 0 100 kg1 L m

−− × = .

The heat energy is removed from the water in three steps: (1) cooling from +15°C to 0°C, (2) freezing at 0°C, and (3) cooling from 0°C to −15°C. The three heat energies are

15

2 f

3

(0 100 kg)(4186 J/kg K)(15 K) 6279 J

(0 100 kg)(3 33 10 J/kg) 33,300 J(0 100 kg)(2090 J/kg K)(15 K) 3135 J

Q mc T

Q mLQ mc T

= ∆ = . =

= = . . × == ∆ = . =

C 1 2 3 42,714 JQ Q Q Q= + + =

Using the performance coefficient,

Cin

in in

42,714 J 42,714 J4 0 10,679 J4 0

QK WW W

= ⇒ . = ⇒ = =.

19-10 Chapter 19


The heat exhausted into the room is thus 4

H C in 42,714 J 10,679 J 5 3 10 JQ Q W= + = + = . ×

19.33. Solve: An adiabatic process has 0Q = and thus, from the first law, s th.W E= −∆ For any ideal-gas process,

th V ,E nC T∆ = ∆ so s V .W nC T= − ∆ We can use the ideal-gas law to find

f i f f i i( ) ( ) ( )pV pV pV pV p V p VT TnR nR nR nR

∆ − −= ⇒ ∆ = = =

Consequently, the work is

f f i i Vs V V f f i i( )

p V p V CW nC T nC p V p VnR R− = − ∆ = − = − −

Because P V ,C C R= + we can use the specific heat ratio γ to find

P V V V

V V V

/ 1 1/ 1

C C R C R CC C C R R

γγ

+ += = = ⇒ =

−

With this, the work done in an adiabatic process is

V f f i is f f i i f f i i

1( ) ( )1 1

C p V p VW p V p V p V p VR γ γ

−= − − = − − =

− −

19.34. Model: We are given H 323 KT = and C 253 K.T = See Figure 19.11. Solve: Every second, the refrigerator must draw enough heat from the cold reservoir to compensate for the heat lost through the stainless-steel panel. Therefore, the heat transferred from the cold reservoir to the system is

C(0.40 m)(0.40 m)(14 W/m K) [25 C ( 20 C)](1.0 s) 10,080 J

0.010 mAQ k T tL

= ∆ ∆ = ° − − ° =

Using the coefficient of performance for a Carnot refrigerator, we can find the energy required to operate for one second:

C H CCarnot in

H C in C

( ) (10,080 J)(70 K) 2.8 kJ253 K

C CT Q Q T TK WT T W T

−= = ⇒ = = =

−

The power required is therefore in = (2.8 kJ)(1.0 s) 2.8 kW.P W t∆ = = Assess: This is much more power than is required for the Brayton-cycle refrigerator of Example 19, which shows why refrigerators are insulated with more than simple steel doors.

19.35. Solve: For any heat engine, C H1 / .Q Qh = − For a Carnot heat engine, Carnot C H1 / .T Th = − Thus a property of the Carnot cycle is that C H C H/ / .Q Q T T= Consequently, the coefficient of performance of a Carnot refrigerator is

C C C H C H CCarnot

in H C C H C H H C

/ /1 / 1 /

Q Q Q Q T T TKW Q Q Q Q T T T T

= = = = =− − − −

19.36. Model: We are given H 298 KT = and C 273 K.T = See Figure 19.11. Solve: 5 6C f (10 kg)(3 33 10 J/kg) 3 33 10 J.Q mL= = . × = . ×

(a) For a Carnot cycle CCarnotH

1 TT

h = − but that must also equal CH

1 ,QQh = − so

C C

H H.Q T

Q T=

6 6HH C

C

298 K(3 33 10 J) 3 6 10 J273 K

TQ QT

= = . × = . ×

(b) 6 6 6 5

in H C 3 63 10 J 3 33 10 J 0 30 10 J 3 0 10 JW Q Q= − = . × − . × = . × = . × Assess: This is a reasonable amount of work to freeze 10 kg of water.



19.37. Model: We will use the Carnot engine to find the maximum possible efficiency of a floating power plant. Solve: The efficiency of a Carnot engine is

Cmax Carnot

H

(273 5) K1 1 0 0825 8 3%(273 30) K

TT

h h += = − = − = . ≈ .+

19.38. Model: The ideal gas in the Carnot engine follows a closed cycle in four steps. During the isothermal expan-sion at temperature H,T heat HQ is transferred from the hot reservoir into the gas. During the isothermal compression at C,T heat CQ is removed from the gas. No heat is transferred during the remaining two adiabatic steps. Solve: The thermal efficiency of the Carnot engine is

C out outCarnot out

H H

323 K1 1 436 J573 K 1000 J

T W W WT Q

h = − = ⇒ − = ⇒ =

Using H C out ,Q Q W= + we obtain

isothermal C H out 1000 J 436 J 0 56 kJQ Q Q W= = − = − ≈ .

19.39. Solve: Substituting into the formula for the efficiency of a Carnot engine,

C CCarnot C

H C1 0 25 1 240 K 33 C

80 KT T TT T

h = − ⇒ . = − ⇒ = = − °+

The hot-reservoir temperature is H C 80 K 320 K 47 C.T T= + = = °

19.40. Solve: From the thermal efficiency of the Carnot engine, we can find the work done each cycle:

C outCarnot out

H H

273 K1 1 (25 J) 10 J455 K

T W WT Q

h = − = ⇒ = − =

The work required to lift a 10 kg mass 10 m is 2(10 kg)(9.8 m/s )(10 m) = 980 J.W Fd= = At 10 J/cycle, the Carnot engine will have to cycle 98 times to do this work.

19.41. Model: Assume the soda is essentially made of water. We are given H 328 K,T = C 253 K,T = and

H 250 J.Q = See Figure 19.11. Solve: The total amount of heat to transfer from the soda is

C4 3 3

Total( ) (5.00 10 m )(1000 kg/m )(4190 J/kg K)(20 K) 41,900 JQ Mc T V c Tr−= ∆ = ∆ = × =

For a Carnot cycle CCarnotH

1 TT

h = − but also CH

1 ,QQ

h = − so C CH H

.Q TQ T

= Therefore, the heat extracted from the soda

each cycle is

CC H

H

253 K(250 J) 192.8 J328 K

TQ QT

= = =

To remove 41,900 J will therefore take

C Total

C

( ) 41,900 J 218 cycles192.8 J

QQ

= =

19.42. Solve: (a) 1Q is given as 1000 J. Using the energy transfer equation for the heat engine,

H C out 1 2 out 2 1 outQ Q W Q Q W Q Q W= + ⇒ = + ⇒ = −

The thermal efficiency of a Carnot engine is

C out

H 1

300 K1 1 0 50600 K

T WT Q

h = − = − = . =

2 1 1 1(1 ) (1000 J)(1 0 50) 500 JQ Q Q Qh h= − = − = − . =

19-12 Chapter 19


To determine 3Q and 4,Q we turn our attention to the Carnot refrigerator, which is driven by the output of the heat engine with in out .W W= The coefficient of performance is

C C 4 4

H C in out 1

4 1

400 K 4 0500 K 400 K

(4 0)(0 50)(1000 J) 2000 J

T Q Q QKT T W W Q

Q K Qh

h

= = = . = = =− −

= = . . =

Using now the energy transfer equation in 4 3,W Q Q+ = we have

3 out 4 1 4 (0 50)(1000 J) 2000 J 2500 JQ W Q Q Qh= + = + = . + =

(b) From part (a) 3 2500 JQ = and 1 1000 J,Q = so 3 1.Q Q> (c) Although 1 1000 JQ = and 3 2500 J,Q = the two devices together do not violate the second law of thermodynam-ics. This is because the hot and cold reservoirs are different for the heat engine and the refrigerator.

19.43. Solve: The work done by the Carnot engine powers the refrigerator, so out Carnot Eng in Refrigerator( ) ( ) .W W= We are given that H 350 KT = and C = 250 KT for both the Carnot engine and the refrigerator and H Carnot Eng( )Q = 10.0 J for the Carnot engine. The work done by the Carnot engine is

out Carnot EngCout Carnot Eng

H H Carnot Eng

( ) 250 K1 ( ) 1 (10.0 J) 2.857 J( ) 350 KWT W

T Qh = − = ⇒ = − =

The heat extracted from the cold reservoir by the refrigerator may be found from it coefficient of performance:

C CC

in Refrigerator out Carnot Eng(2.00)(2.857 J) 5.713 J

( ) ( )Q QK Q

W W= = ⇒ = =

The heat exhausted by the refrigerator to the hot reservoir may be found from the first law of thermodynamics:

in Refrigerator C H H( ) 2.857 J 5.713 J 8.57 JW Q Q Q+ = ⇒ = + =

Assess: The work done on the refrigerator is less than the heat exhausted to the hot reservoir, as expected.

19.44. Model: A heat pump is a refrigerator that is cooling the already cold outdoors and warming the indoors with its exhaust heat. Solve: (a) The coefficient of performance for this heat pump is C in5 0 / ,K Q W= . = where CQ is the amount of heat removed from the cold reservoir. HQ is the amount of heat exhausted into the hot reservoir. H C in ,Q Q W= + where

inW is the amount of work done on the heat pump. We have

C in H in in in5 0 5 0 6 0Q W Q W W W= . ⇒ = . + = .

If the heat pump is to deliver 15 kJ of heat per second to the house, then

H in in15 kJ15 kJ 6 0 2 5 kJ

6 0Q W W= = . ⇒ = = .

.

In other words, 2.5 kW of electric power is used by the heat pump to deliver 15 kJ/s of heat energy to the house. (b) The monthly heating cost in the house using an electric heater is

15 kJ 3600 s 1$(200 h) $270s 1 h 40 MJ

=

The monthly heating cost in the house using a heat pump is

2 5 kJ 3600 s 1$(200 h) $45s 1 h 40 MJ

. =



19.45. Visualize: We are given C 275 K,T = H 295 KT = . We are also given that in one second in 100 JW = and

C (1 s)(100 kJ/min)(1 min/60 s) 1667 JQ = = . Solve: The coefficient of performance of a refrigerator is given in Equation 19.8.

C

in

1667 J 16 67100 J

QKW

= = = .

However the coefficient of performance of a Carnot refrigerator is given in Equation 19.28.

CCarnot

H C

275 K 13 7520 K

TKT T

= = = .−

However, informal statement #8 of the second law says that the coefficient of performance cannot exceed the Carnot coefficient of performance, so the salesman is making false claims. You should not buy the DreamFridge. Assess: The second law imposes real-world restrictions.

19.46. Solve: The maximum possible efficiency of the heat engine is

Cmax

H

300 K1 1 0 40500 K

TT

h = − = − = .

The efficiency of the engine designed by the first student is

out1

H

110 J 0 44250 J

WQ

h = = = .

Because max ,h h1 > the first student has proposed an engine that would violate the second law of thermodynamics. His or her design will not work. The efficiency of the engine designed by the second student is 2 max90 J 250 J 0 36/h h= = . < in agreement with the second law of thermodynamics. Applying the first law of thermodynamics,

H C out 250 J 170 J 90 JQ Q W= + ⇒ = +

we see the first law is violated. This design will not work as claimed. The design by the third student satisfies the first law of thermodynamics because H C out 250 J.Q Q W= + = The thermal efficiency of this engine is 3 max0 36 ,h h= . < which satisfies the second law of thermodynamics. The data presented by students 1 and 2 are faulty. Only student 3 has an acceptable design.

19.47. Model: The power plant is to be treated as a heat engine. Solve: (a) Every hour 300 metric tons or 53 10 kg× of coal is burnt. The volume of coal is

5 333 10 kg m (24 h) 4800 m

1 h 1500 kg ×

=

The height of the room will be 48 m. (b) The thermal efficiency of the power plant is

8 8out

95 6H

7 50 10 J/s 7 50 10 J 0 32 32%2 333 10 J3 10 kg 28 10 J 1 h

1 h kg 3600 s

WQ

h . × . ×= = = = . = . ×× ×

Assess: An efficiency of 32% is typical of power plants.

19.48. Model: The power plant is treated as a heat engine. Solve: We are given that H 300°C 573 KT = = and C 25°C 298 K.T = = (a) The maximum possible thermal efficiency of the power plant is

Cmax

H

298 K1 1 0 48 48%573 K

TT

h = − = − = . =

19-14 Chapter 19


(b) In one second, the plant generates 6out 1000 10 JW = × of work and 6

H 3000 10 JQ = × of heat energy to replace the energy taken from the hot reservoir to heat the water. The plant’s actual efficiency is

6out

6H

1000 10 J 0 33 33%3000 10 J

WQ

h ×= = . =×

(c) Because H C out ,Q Q W= +

9 9 9C H out C 3 0 10 J/s 1 0 10 J/s 2 0 10 J/sQ Q W Q= − ⇒ = . × − . × = . ×

The mass of water that flows per second through the condenser is 3 3

8 43

L 1 hr 10 m 1000 kg1 2 10 (1s) 3 333 10 kgh 3600 s 1 L m

m− = . × = . ×

The change in the temperature as 9C 1 0 10 JQ = . × of heat is transferred to 43 333 10 kgm = . × of water is

9 4C 2 0 10 J (3 333 10 kg)(4186 J/kg K) 14 CQ mc T T T= ∆ ⇒ . × = . × ∆ ⇒ ∆ = °

The exit temperature is 18°C 14°C 32°C.+ =

19.49. Model: The power plant is treated as a heat engine. Solve: The mass of water per second that flows through the plant every second is

3 38 4

3L 1 hr 10 m 1000 kg1 0 10 2 778 10 kg/sh 3600 s 1 L m

m− = . × = . ×

The amount of heat transferred per second to the cooling water is thus 4 9

C (2 778 10 kg/s)(4186 J/kg K)(27 C 16 C) 1 279 10 J/sQ mc T= ∆ = . × ° − ° = . ×

The amount of heat per second input into the power plant is 9 9 9

H out C 0 750 10 J/s 1 279 10 J/s 2 029 10 J/sQ W Q= + = . × + . × = . ×

Finally, the power plant’s thermal efficiency is 9

out9

H

0 750 10 J/s 0 37 37%2 029 10 J/s

WQ

h . ×= = = . =. ×

19.50. Solve: (a) The energy supplied in one day is

9 13out

J 3600 s 24 h1 0 10 8 6 10 Js 1 h 1 d

W = . × = . ×

(b) The volume of water is 3 9 31 km 10 mV = = . The amount of energy is

9 3 15H 3

1000 kg(10 m ) (4190 J/kg K)(1 K) 4 10 Jm

Q mc T = ∆ = = ×

(c) While it’s true that the ocean contains vast amounts of thermal energy, that energy can be extracted to do useful work only if there is a cold reservoir at a lower temperature. That is, the ocean has to be the hot reservoir of a heat engine. But there’s no readily available cold reservoir, so the ocean’s energy cannot readily be tapped. There have been proposals for using the colder water near the bottom of the ocean as a cold reservoir, pumping it up to the sur-face where the heat engine is. Although possible, the very small temperature difference between the surface and the ocean depths implies that the maximum possible efficiency (the Carnot efficiency) is only a few percent, and the efficiency of any real ocean-driven heat engine would likely be less than 1%—perhaps much less. Thus the second law of thermodynamics prevents us from using the thermal energy of the ocean. Save your money. Don’t invest.



19.51. Visualize: If we do this problem on a “per-second” basis then in one second 5C (1 s)(5 0 10 J/min)Q = . × 3(1 min/60 s) 8 33 10 J= . × . 5 3H (1 s)(8 0 10 J/min)(1 min 60 s) 13 33 10 JQ /= . × = . × .

Solve: (a) Again, in one second 3 3 3

in H C 13 33 10 J 8 33 10 J 5 0 10 JW Q Q= − = . × − . × = . ×

Since this is per second, the power required by the compressor is 5.0 kW.P = (b) The coefficient of performance is

3C

3in

8 33 10 J 1 75 0 10 J

QKW

. ×= = = .

. ×

Assess: The result is typical for air conditioners.

19.52. Model: The heat engine follows a closed cycle with process 1 2→ and process 3 4→ being isochoric and process 2 3→ and process 4 1→ being isobaric. For a monatomic gas, 3V 2C R= and

5P 2C R= .

Visualize: Please refer to Figure P19.52. Solve: (a) The first law of thermodynamics is th S.Q E W= ∆ + For the isochoric process 1 2,→ S 1 2 0 J.W → = Thus,

( ) ( )

1 2 th V

3 3V 2 2

2 1 2

3750 J3750 J 3750 J 3750 J 301 K

(1 0 mol) (1 0 mol) (8 31 J/mol K)

300 8 K 300 8 K 300 K 601 K

Q E nC T

TnC R

T T T

→ = = ∆ = ∆

∆ = = = =. . .

− = . ⇒ = . + =

To find volume 2,V

3 312 1 5

1

(1 0 mol)(8 31 J/mol K)(300 K) 8 31 10 m3 0 10 Pa

nRTV Vp

−. .= = = = . ×. ×

The pressure 2p can be obtained from the isochoric condition as follows:

5 52 1 22 1

2 1 1

601 K (3 00 10 Pa) 6 01 10 Pa300 K

p p Tp pT T T

= ⇒ = = . × = . ×

With the above values of 2 2, ,p V and 2,T we can now obtain 3,p 3,V and 3.T We have 2 3

3 25

3 2

3 2 33 2 2

3 2 2

2 1 662 10 m

6 01 10 Pa

2 1202 K

V V

p pT T VT T TV V V

−= = . ×

= = . ×

= ⇒ = = =

For the isobaric process 2 3,→

( ) ( )5 52 3 P 3 22 25 3 3

S 2 3 3 3 2

th 2 3 S 2 3

(1 0 mol) ( ) (1 0 mol) (8 31 J/mol K)(601 K) 12,480 J

( ) (6 01 10 Pa)(8 31 10 m ) 4990 J12,480 J 4990 J 7490 J

Q nC T R T T

W p V VE Q W

→

−→

→ →

= ∆ = . − = . . =

= − = . × . × =

∆ = − = − =

We are now able to obtain 4,p 4,V and 4.T We have

2 34 3

54 1

54 3 4

4 3 54 3 3

1 662 10 m

3 00 10 Pa

3 00 10 Pa (1202 K) 600 K6 01 10 Pa

V V

p p

T T pT Tp p p

−= = . ×

= = . ×

. ×= ⇒ = = = . ×

19-16 Chapter 19


For isochoric process 3 4,→

( ) ( )3 33 4 V 4 32 2S 3 4 th 3 4 S 3 4

(1 0 mol) ( ) (1 0 mol) (8 31 J/mol K)( 602) 7500 J

0 J 7500 J

Q nC T R T T

W E Q W→

→ → →

= ∆ = . − = . . − = −

= ⇒ ∆ = − = −

For isobaric process 4 1,→ 5

4 1 P 25 3 3 2 3

S 4 1 4 1 4

th 4 1 S 4 1

(1 0 mol) (8 31 J/mol K)(300 K 600 K) 6230 J

( ) (3 00 10 Pa) (8 31 10 m 1 662 10 m ) 2490 J6230 J ( 2490 J) 3740 J

Q nC T

W p V VE Q W

→

− −→

→ →

= ∆ = . . − = −

= − = . × × . × − . × = −

∆ = − = − − − = −

S (J)W Q (kJ) th (kJ)E∆ 1 2→ 0 3750 3750 2 3→ 4990 12,480 7490 3 4→ 0 –7500 –7500 4 1→ –2490 –6230 –3740 Net 2500 2500 0

(b) The thermal efficiency of this heat engine is

out out

H 1 2 2 3

2500 J 0 15 15%3750 J 12,480 J

W WQ Q Q

h→ →

= = = = . =+ +

Assess: Note that more than two significant figures are retained in part (a) because the results are intermediate. For a closed cycle, as expected, s net net( )W Q= and th net( ) 0 JE∆ =

19.53. Model: The heat engine follows a closed cycle. For a diatomic gas, 5V 2C R= and 7

P 2C R= .

Visualize: Please refer to Figure P19.53. Solve: (a) Since 1 293 K,T = the number of moles of the gas is

5 6 341 1

1

(0 5 1 013 10 Pa)(10 10 m ) 2 08 10 mol(8 31 J/mol K)(293 K)

p VnRT

−−. × . × ×= = = . ×

.

At point 2, 2 1 2 14 and 3 .V V p p= = The temperature is calculated as follows:

1 1 2 2 2 22 1

1 2 1 1(3)(4)(293 K) 3516 Kp V p V p VT T

T T p V= ⇒ = = =

At point 3, 3 2 1 3 14 and .V V V p p= = = The temperature is calculated as before:

3 33 1

1 1(1)(4)(293 K) 1172 Kp VT T

p V= = =

For process 1 2,→ the work done is the area under the p-versus-V curve. That is, 3 3 3 31

s 25

6 3

(0 5 atm)(40 cm 10 cm ) (1 5 atm 0 5 atm)(40 cm 10 cm )

1 013 10 Pa(30 10 m )(1 atm) 3 04 J1 atm

W

−

= . − + . − . −

. ×= × = .

The change in the thermal energy is 4 5

th V 2(2 08 10 mol) (8 31 J/mol K)(3516 K 293 K) 13 93 JE nC T−∆ = ∆ = . × . − = .

The heat is s th 16 97 J.Q W E= + ∆ = . For process 2 3,→ the work done is s 0 JW = and

( )5th V 3 224 5

2

( )

(2 08 10 mol) (8 31 J/mol K)(1172 K 3516 K) 10 13 J

Q E nC T n R T T−

= ∆ = ∆ = −

= . × . − = − .



For process 3 1,→ 3 3 5 6 3

s4 5

th V 2

(0 5 atm)(10 cm 40 cm ) (0 5 1 013 10 Pa)( 30 10 m ) 1 52 J

(2 08 10 mol) (8 31 J/mol K)(293 K 1172 K) 3 80 J

W

E nC T

−

−

= . − = . × . × − × = − .

∆ = ∆ = . × . − = − .

The heat is th s 5 32 J.Q E W= ∆ + = − .

s (J)W Q (J) thE∆ 1 2→ 3.04 16.97 13.93 2 3→ 0 −10.13 −10.13 3 1→ −1.52 −5.32 −3.80 Net 1.52 1.52 0

(b) The efficiency of the engine is net

H

1 52 J 0 090 9 0%16 97 J

WQ

h .= = = . = ..

(c) The power output of the engine is

netrevolutions 1 min 500500 (1 52 J/s) 13 Wmin 60 s revolution 60

W = . =

Assess: Note that more than two significant figures are retained in part (a) because the results are intermediate. For a closed cycle, as expected, s net net( )W Q= and th net( ) 0 J.E∆ =

19.54. Model: For the closed cycle, process 1 2→ is isothermal, process 2 3→ is isobaric, and process 3 1→ is isochoric. Visualize: Please refer to Figure P19.54. Solve: (a) We first need to find the conditions at points 1, 2, and 3. We can then use that information to find SW and Q for each of the three processes that make up this cycle. Using the ideal-gas equation the number of moles of the gas is

5 6 31 1

1

(1 013 10 Pa)(600 10 m ) 0 0244 mol(8 31 J/mol K)(300 K)

p VnRT

−. × ×= = = .

.

We are given that 1.25,γ = which means this is not a monatomic or a diatomic gas. The specific heats are

V P V4 51RC R C C R R

γ= = = + =

−

At point 2, process 1 2→ is isothermal, so we can find the pressure 2p as follows:

4 351

1 1 2 2 2 1 1 14 32

6 00 10 m 3 3 atm 3 039 10 Pa2 00 10 m

Vp V p V p p p pV

−

−. ×

= ⇒ = = = = = . ×. ×

At point 3, process 2 3→ is isobaric, so we can find the temperature 3T as follows:

4 32 3 3

3 2 2 24 32 3 2

6 00 10 m 3 900 K2 00 10 m

V V VT T T TT T V

−

−. ×

= ⇒ = = = =. ×

Point P (Pa) 3 (m )V T (K)

1 51.0 atm 1.013 10= × 46.00 10−× 300 2 53.0 atm 3.039 10= × 42.00 10−× 300 3 53.0 atm 3.039 10= × 46.00 10−× 900

19-18 Chapter 19


Process 1 2→ is isothermal:

S 12 1 1 2 1( ) ln( / ) 66 8 JW p V V V= = − . 12 S 12( ) 66 8 JQ W= = − .

Process 2 3→ is isobaric:

S 23 2 2 3 2( ) ( ) 121 6 JW p V p V V= ∆ = − = . 23 P P 3 2( ) 608 3 JQ nC T nC T T= ∆ = − = .

Process 3 1→ is isochoric: S 31( ) 0 JW = 31 V V 1 3( ) 486 7 JQ nC T nC T T= ∆ = − = − .

We find that

S cycle( ) 66 8 J 121 6 J 0 J 54 8 JW = − . + . + = . cycle 66 8 J 608 3 J 486 7 J 54 8 JQ = − . + . − . = .

These are equal, as they should be. Knowing that the work done is out S cycle( ) 54.8 J/cycle,W W= = an engine operat-ing at 20 cycles/s has a power output of

out54 8 J 20 cycle J1096 1096 W 1 10 kWcycle s s

P . = = = ≈ .

(b) Only 23Q is positive, so in 23 608 J.Q Q= = Thus, the thermal efficiency is

out

in

54 8 J 0 0901 9 01%608 3 J

WQ

h .= = = . = ..

19.55. Model: For the closed cycle of the heat engine, process 1 2→ is isobaric, process 2 3→ is isochoric, and process 3 1→ is adiabatic. 3V 2C R= and

5P 2C R= for a monatomic gas, so 5/3.γ =

Visualize: Please refer to Figure P19.55. Solve: (a) We can use the adiabat 3 1→ to calculate 1p as follows:

5/333

1 3 1 31 3 31

600 cm(100 kPa) 1981 kPa100 cm

Vp V p V p pV

γγ γ = ⇒ = = =

1T can be determined by taking the ratio of the ideal-gas equation applied to points 1 and 2. This gives

1 1 1

2 2 23

11 2 3

2

100 cm(600 K) 100 K600 cm

p V Tp V T

VT TV

=

= = =

where we have used the fact that 1 2.p p= Applying the same strategy at point 3 gives

2 2 2

3 3 3

33 2

2

100 kPa(600 K) 30 3 K1981 kPa

p V Tp V T

pT Tp

=

= = = .

where we have used the fact that 2 3.V V= Before we calculate the work and heat exchanged for each cycle, we need to know the number of moles. This may be calculated by applying the ideal gas law at any point on the cycle:

6 31 1

1

(1981 kPa)(100 10 m ) 0.238 mol(8.31 J/mol K)(3600 K)

p VnRT

−×= = =

Now we can calculate s ,W Q, and thE∆ for the three processes involved in the cycle. For process 1 2,→

( )3th,1 2 V 2 1 2 12( ) ( ) 1 486 kJE nC T T n R T T→∆ = − = − = . ( )51 2 P 2 1 2 12( ) ( ) 2 476 kJQ nC T T n R T T→ = − = − = .



The work done s 1 2W → is the area under the p-versus-V graph. We have

6 3 6 3s 1 2 (1981 kPa)(600 10 m 100 10 m ) 0 991 kJW

− −→ = × − × = .

For process 2 3,→ s 2 3 0 JW → = and

( )3th, 2 3 2 3 V 3 2 3 22( ) ( ) 1 693 kJE Q nC T T n R T T→ →∆ = = − = − = − . For process 3 1,→ 3 1 0 JQ → = and

( )3th, 3 1 V 1 3 1 32( ) ( ) 0 2072 kJE nC T T n R T T→∆ = − = − = . Because thE W Q∆ = + and s ,W W= − s 3 1 th,3 1 0 2072 kJW E→ →= −∆ = − . for process 3 1.→

s (kJ)W Q (kJ) th (kJ)E∆ 1 2→ 0.991 2.476 1.486 2 3→ 0 −1.693 −1.693 3 1→ −0.207 0 0.207 Net 0.783 0.783 0


out

H

783 J 0 32 32%2476 J

WQ

h = = = . =

Assess: Note that more than two significant figures are retained in part (a) because the results are intermediate. As expected for a closed cycle, s net net( )W Q= and th net( ) 0 J.E∆ =

19.56. Model: For the closed cycle of the heat engine, process 1 2→ is isochoric, process 2 3→ is adiabatic, and process 3 1→ is isothermal. For a diatomic gas 5V 2C R= and

75 .γ =

Solve: (a) From the graph 32 1000 cm .V = The pressure 2p lies on the adiabat from 2 3.→ We can find the pressure as follows:

7/535 53

2 3 2 32 3 32

4000 cm(1 00 10 Pa) 6 964 10 Pa 696 kPa1000 cm

Vp V p V p pV

γγ γ = ⇒ = = . × = . × ≈

The temperature 2T can be obtained from the ideal-gas equation relating points 1 and 2:

51 1 2 2 2 2

2 1 51 2 1 1

6 964 10 Pa(300 K) (1) 522 3 K 522 K4 00 10 Pa

p V p V p VT TT T p V

. ×= ⇒ = = = . ≈ . ×

(b) The number of moles of the gas is 5 3 3

1 1

1

(4 00 10 Pa)(1 00 10 m ) 0 1604 mol(8 31 J/mol K)(300 K)

p VnRT

−. × . ×= = = .

.

For isochoric process s1 2, 0 JW→ = and

( )5th V 2 741 1 JQ E nC T n R T= ∆ = ∆ = ∆ = . For adiabatic process 2 3, 0 JQ→ = and

( )5th V 3 22 ( ) 741 1 JE nC T n R T T∆ = ∆ = − = − . Using the first law of thermodynamics, th s ,E W Q∆ = + which means s th 741.1 J.W E= −∆ = + sW can also be deter-mined from

( )( )

433 3 2 2 3 2

s 25

J/K (300 K 522 3 K)( ) 741 1 J1 1

p V p V nR T TWγ γ

− .− −= = = = .

− − −

19-20 Chapter 19


For isothermal process 3 1,→ th 0 JE∆ = and

1s 1

3ln 554 5 JVW nRT

V= = − .

Using the first law of thermodynamics, th s ,E W Q∆ = − + s 554 5 J.Q W= = − .

th (J)E∆ s (J)W Q (J) 1 2→ 741.1 0 741.1 2 3→ −741.1 741.1 0 3 1→ 0 −554.5 −554.5 Net 0 186.6 186.6

(c) The work per cycle is 187 J and the thermal efficiency is s

H

186 6 J 0 25 25%741 1 J

WQ

h .= = = . =.

19.57. Model: For this heat engine, process 1 2→ is adiabatic, process 2 3→ is isothermal, and process 3 1→ is isobaric. For a diatomic gas, 5V 2 ,=C R

7P 2 ,=C R and 1 40.γ = .

Solve: (a) From the graph, 1 100 kPa.=p Point 1 is connected by an adiabatic process to point 2, where 2 400 kPa=p

and 32 1000 cm .=V So 1 21 2p V p Vγ γ= and thus

1/ 1/1 403 3 32

1 21

400 kPa(1000 cm ) 2692 cm 2690 cm100 kPa

pV Vp

γ . = = = ≈

Point 1 is connected by an isochoric process to point 3, where 33 4000 cm=V and 3 400 K.=T Thus 3

11 2 3

2

2692 cm(400 K) 269 K4000 cm

= = =

VT TV

Altogether, 1 100 kPa,=p 3

1 2690 cm ,=V and 1 269 K= .T (b) For the adiabatic process 1 2,→ 0=Q and s th.W E= −∆ For any process, th V .∆ = ∆E nC T The number of moles can be found from point 2:

5 3(4 0 10 Pa)(0 0010 m ) 0 120 mol(8 31 J/mol K)(400 K)

pVnRT

. × .= = = .

.

Thus

( )5th s2(0 120 mol) (400 K 269 K) 327 J WE R∆ = . − = = − For the isothermal process 2 3,→ th 0∆ =E and s.=Q W The work done is

3s 2

2ln 553 JVW nRT Q

V

= = =

For the isobaric process 3 1,→ 5 3 3

s (1 0 10 Pa)(0 00269 m 0 00400 m ) 131 JW p V= ∆ = . × . − . = − The heat exhausted to the cold reservoir is

( )7P 2(0 120 mol) (269 K 400 K) 458 JQ nC T R= ∆ = . − = − Finally, from the first law, th s 327 J.∆ = − = −E Q W

thE∆ sW Q 1 → 2 327 −327 0 2 → 3 0 553 553 3 → 1 −327 −131 −458 Net 0 95 95



(c) The work per cycle is out 95 J.=W Heat is input only during process 2 3,→ so H 553 J=Q and the engine’s thermal efficiency is

out

H

95 J 0 17 17553 J

WQ

h = = = . = ,

Assess: As expected for a closed cycle, th net( ) 0∆ =E and s net net( ) .=W Q

19.58. Model: For the closed cycle of the refrigerator, process 1 → 2 is isochoric, process 2 → 3 is adiabatic, pro-cess 3 → 4 is isochoric, and process 4 → 1 is adiabatic. For a monatomic gas 3V 2C R= and

53 .γ =

Visualize: Please refer to the figure below.

Solve: (a) The number of moles of gas may be found by applying the ideal-gas equation to point 2. The result is 4 3

2 2

2

150 kPa (1.00 10 m ) 7.22 10 mol(8.31 J/mol K)(250 K)

p VnRT

−−3( ) ×= = = ×

The temperatures at points 3 and 4 may be found using Table 19.1: 2/31 3

23 2 3

3

100 cm(250 K) 461 K40 cm

VT TV

γ − = = =

2/31 31

4 1 34

100 cm(200 K) 368 K40 cm

VT TV

γ − = = =

Only the adiabatic segments do work, so the total work done by the system is

( )( )

3S S,2 3 S,4 1 v 2 3 4 1 3 2 1 42

3 32

( ) ( )

(7.22 10 mol) (8.31 J/molK)(461 K 250 K 200 K 369 K) 3.97 J

W W W nC T T n R T T T T→ → → →−

= + = − ∆ + ∆ = − − + −

= − × − + − = −

Thus, the work done on the system is in S 3.79 J.W W= − = − During the adiabatic segments, no heat is exchanged with the heat reservoirs, so heat is exchanged only during the isochoric segments. For a refrigerator, the heat exchanged with the cold reservoir is the heat that is put into the system (i.e., > 0), which occurs in segment 1 → 2. With the help of Table 19.1, this is

( ) ( )33 31 2 2 12 2( ) (7.22 10 mol) (8.31 J/molK)(250 K 200 K) 4.50 JC VQ nC T n R T T −→= ∆ = − = × − = Thus, the coefficient of performance is

C

in

4.50 J 1.193.71 J

QKW

= = =

(b) The power needed to run the refrigerator is 1

incycles( cycle ) 60 227 W

sP W − = =

19-22 Chapter 19


19.59. Model: Process 1 → 2 of the cycle is isochoric, process 2 → 3 is isothermal, and process 3 → 1 is isobaric. For a monatomic gas, 3V 2=C R and

5P 2 R= .C

Visualize: Please refer to Figure P19.59. Solve: (a) At point 1, the pressure 51 1 atm 1.013 10 Pap = = × and the volume

6 3 3 31 1000 10 m 1 10 m .

− −= × = ×V The number of moles is

0 120 g 0 03 mol4 g/mol

.= = .n

Using the ideal-gas law, 5 3 3

1 11

(1 013 10 Pa)(1 0 10 m ) 406 K 0 4 kK(0 030 mol)(8 31 J/mol K)

−. × . ×= = = ≈ .

. .p VTnR

At point 2, the pressure 52 5 atm 5 06 10 Pa= = . ×p and 3 3

2 1 10 m .−= ×V The temperature is

5 3 32 2

2(5 06 10 Pa)(1 0 10 m ) 2030 K 2 kK

(0 030 mol)(8 31 J/mol K)

−. × . ×= = = ≈

. .p VTnR

At point 3, the pressure is 53 1 atm 1 013 10 Pa= = . ×p and the temperature is 3 2 2030 K.= =T T The volume is

3 3 3 323 2

3

5 atm(1 10 m ) 5 10 m1 atm

− − = = × = ×

pV Vp

(b) For the isochoric process 1 → 2, W1→2 = 0 J and

( )31 2 V 2(0.030 mol) (2030 K 406 K) 607 JQ nC T R→ = ∆ = − = For the isothermal process 2 → 3, th 2 3 0 JE →∆ = and

3 33

2 3 2 3 2 3 32

5 0 10 mln (0 030 mol)(8 31 J/mol K)(2030 K)ln 815 J1 0 10 m

−

→ → −

. ×= = = . . = . ×

VQ W nRTV

For the isobaric process 3 → 1, 5 3 3 3 3

3 1 3 (1 013 10 Pa)(1 0 10 m 5 0 10 m ) 405 J− −

→ = ∆ = . × . × − . × = −W p V

( )53 1 P 2(0.030 mol) (8.31 J/mol K)(406 K 2030 K) 1012 JQ nC T→ = ∆ = − = − The total work done is net 1 2 2 3 3 1 410 J.→ → →= + + =W W W W The total heat input is H 1 2 2 3 1422 J.→ →= + =Q Q Q The thermal efficiency of the engine is

net

H

410 J 291422 J

h = = =WQ

,

(c) The maximum possible efficiency of a heat engine that operates between Tmax and Tmin is

minmax

max

406 K1 1 802030 K

h = − = − =TT

,

Assess: The actual efficiency of an engine is less than the maximum possible efficiency.

19.60. Model: The process 2 → 3 of the heat engine cycle is isochoric and the process 3 → 1 is isobaric. For a monatomic gas 3V 2C R= and

5P 2 .C R=

Solve: (a) The three temperatures are 5 3

1 11

(4.0 10 Pa)(0.025 m ) 601.7 K 0.60 kK(2.0 mol)(8.31 J/mol K)

p VTnR

×= = = ≈



5 32 2

2(6.0 10 Pa)(0.050 m ) 1805.1 K 1.8 kK(2.0 mol)(8.31 J/mol K)

p VTnR

×= = = ≈

5 33 3

3(4.0 10 Pa)(0.050 m ) 1203.4 K 1.2 kK(2.0 mol)(8.31 J/mol K)

p VTnR

×= = = ≈

(b) For process 1 → 2, the work done is the area under the p-versus-V graph. The work and the change in internal energy are

( )( )

5 5 3 3 5 3 31s 2

4

3th V 2 12

432

(6.0 10 Pa 4.0 10 Pa)(0.050 m 0.025 m ) (4.0 10 Pa)(0.050 m 0.025 m )

1.25 10 J

(2.0 mol) ( )

(2.0 mol) (8.31 J/mol K)(1805.1 K 601.7 K) 3.00 10 J

W

E nC T R T T

= × − × − + × −

= ×

∆ = ∆ = −

= − = ×

The heat input is 4s th 4.25 10 J.Q W E= + ∆ = × For isochoric process 2 → 3, s 0 JW = and

43th V 2(2.0 mol) (8.31 J/mol K)(1203.4 K 1805.1 K) 1.50 10 JQ E nC T= ∆ = ∆ = − = − ×

For isobaric process 3 → 1, the work done is the area under the p-versus-V curve. Hence, 5 3 3 4

s (4.0 10 Pa)(0.025 m 0.050 m ) 1.0 10 JW = × − = − ×

( ) 43 3th V 1 32 2( ) (2.0 mol) (8.31 J/mol K)(601.7 K 1203.4 K) 1.5 10 JE nC T n R T T∆ = ∆ = − = − = − × The heat input is 4S th 2.50 10 J.Q W E= + ∆ = − ×

th (J)E∆ S (J)W Q (J) 1 → 2 43.0 10× 41.25 10× 44.25 10× 2 → 3 4. 10−1 5× 0 41.50 10− × 3 → 1 4. 10−1 5× 4.0 10−1 × 42.50 10− × Net 0 32.5 10× 32.5 10×

(c) The thermal efficiency is 3

net4

H

2.5 10 J 5.94.25 10 J

WQ

h ×= = =×

,

19.61. Model: The closed cycle in this heat engine includes adiabatic process 1 → 2, isobaric process 2 → 3, and isochoric process 3 → 1. For a diatomic gas, 5 7 7V P2 2 5, , and 1.4.C R C R γ= = = =

Visualize: Please refer to Figure P19.61. Solve: (a) We can find the temperature 2T from the ideal-gas equation as follows:

5 3 32 2

2(4.0 10 Pa)(1.0 10 m ) 2407 K 2.4 kK(0.020 mol)(8.31 J/mol K)

p VTnR

−× ×= = = ≈

We can use the equation 2 12 1p V p Vγ γ= to find 1,V

1 1.41 53 3 3 32

1 2 51

4.0 10 Pa(1.0 10 m ) 2.692 10 m1.0 10 Pa

pV Vp

γ //− − ×= = × = × ×

The ideal-gas equation can now be used to find 1,T 5 3 3

1 11

(1.0 10 Pa)(2.692 10 m ) 1620 K 1.6 kK(0.020 mol)(8.31 J/mol K)

p VTnR

−× ×= = = ≈

19-24 Chapter 19


At point 3, 3 1V V= so we have 5 3 3

3 33

(4 10 Pa)(2.692 10 m ) 6479 K 6.5 kK(0.020 mol)(8.31 J/mol K)

p VTnR

−× ×= = = ≈

(b) For adiabatic process 1 → 2, Q = 0 J, th s ,E W∆ = − and

2 2 1 1 2 1s

( ) (0.020 mol)(8.31 J/mol K)(2407 K 1620 K) 327.0 J1 1 (1 1.4)


− − −= = = = −

− − −

For isobaric process 2 → 3, ( )7 7P 2 2( ) (0.020 mol) (8.31 J/mol K)(6479 K 2407 K) 2369 JQ nC T n R T= ∆ = ∆ = − =

( )5th V 2 1692 JE nC T n R T∆ = ∆ = ∆ = The work done is the area under the p-versus-V graph. Hence,

5 3 3 3 3s (4.0 10 Pa)(2.692 10 m 1.0 10 m ) 677 JW

− −= × × − × =

For isochoric process 3 → 1, s 0 JW = and

( )5th V 2(0.020 mol) (8.31 J/mol K)(1620 K 6479 K) 2019 JE Q nC T∆ = = ∆ = − = −

th (J)E∆ S (J)W Q (J) 1 → 2 327 −327 0 2 → 3 1692 677 2369 3 → 1 −2019 0 −2019 Net 0 350 350

(c) The engine’s thermal efficiency is net

H

350 J 0.15 152369 J

WQ

h = = = = ,

19.62. Model: For the closed cycle of the heat engine, process 1 → 2 is isothermal, process 2 → 3 is isobaric, and process 3 → 1 is isochoric. For a diatomic gas 5V 2C R= and

75 .γ =

Visualize: Please refer to the figure below.

Solve: (a) Begin by expressing the pressure, volume, and temperature in terms of the pressure, volume, and temperature at point 1. In the isothermal expansion 1 → 2, the volume is halved so the pressure must double (ideal gas equation). Therefore 2 12 .p p= Because 2 → 3 is isobaric, 3 2 12 .p p p= = We are given that 2 1/2V V= and that 3 1.V V= Finally, we know that 2 1T T= because they are on the same isotherm, and the ideal gas equation gives

3 3 1 13 1

2 2p V p VT TnR nR

= = =



The table below summarizes:

1p 1V 1T

2 12p p= 2 1/2V V= 2 1T T=

3 12p p= 3 1V V= 3 12T T=

With the help of Table 19.1, we can find expressions for the work and heat for each segment in the cycle. The results are given in the table below.

SW Q

1 → 2 1 2 1 1ln( / ) ln 2nRT V V nRT= − 1 ln 2nRT−

2 → 3 2 1 2 1 1 1 1 1( ) (2 )( /2)p V V p V p V nRT− = = = 7P 3 2 12( )nC T T nRT− =

3 → 1 0 5V 1 3 12( )nC T T nRT− = −

The heat transferred from the hot reservoir into the heat engine (> 0) is done in the isochoric segment 2 → 3. The total work done by the system is

out 1 1 1ln 2 (1 ln 2)W nRT nRT nRT= − + = − The thermal efficiency is therefore

out 17

H 12

(1 ln 2) 0.088 8.8W nRTQ nRT

h −= = = = ,

(b) The thermal efficiency of a Carnot engine operating between 1T and 3T is

C 1Carnot

31 1 0.5 50

H

T TT T

h = − = − = = ,

Assess: The efficiency is much less than the Carnot efficiency.

19.63. Model: The closed cycle of the heat engine involves the following four processes: isothermal expansion, isochoric cooling, isothermal compression, and isochoric heating. For a monatomic gas 3V 2 .C R=

Visualize:

Solve: Using the ideal-gas law, 51

1 3 31

(0.20 mol)(8.31 J/mol K)(600 K) 4.986 10 Pa2.0 10 m

nRTpV −

= = = ××

At point 2, because of the isothermal conditions, 2 1 600 KT T= = and

3 35 51

2 1 3 32

2.0 10 m(4.986 10 Pa) 2.493 10 Pa4.0 10 m

Vp pV

−

−

×= = × = × ×

19-26 Chapter 19


At point 3, because it is an isochoric process, 33 2 4000 cmV V= = and

5 533 2

2

300 K(2.493 10 Pa) 1.247 10 Pa600 K

Tp pT

= = × = ×

Likewise at point 4, 4 3 300 KT T= = and

3 35 53

4 3 3 34

4.0 10 m(1.247 10 Pa) 2.493 10 Pa2.0 10 m

Vp pV

−

−

×= = × = × ×

Let us now calculate net 1 2 2 3 3 4 4 1.W W W W W→ → → →= + + + For the isothermal processes,

21 2 1

1ln (0.20 mol)(8.31 J/mol K)(600 K)ln(2) 691.2 JVW nRT

V→= = =

( )4 13 4 3 23

ln (0.20 mol)(8.31 J/mol K)(300 K)ln 345.6 JVW nRTV→

= = = −

For the isochoric processes, 2 3 4 1 0 J.W W→ →= = Thus, the work done per cycle is net 345.6 J 350 J.W = ≈ Because

S th ,Q W E= + ∆

1 2 1 2 th 1 2( ) 691.2 J 0 J 691.2 JQ W E→ → →= + ∆ = + = For the first isochoric process,

( )32 3 V 3 2232

(0.20 mol) ( )

(0.20 mol) (8.31 J/mol K)(300 K 600 K) 747.9 K

Q nC T R T T→ = ∆ = −

= − = −

For the second isothermal process

3 4 3 4 th 3 4( ) 345.6 J 0 J 345.6 JQ W E→ → →= + ∆ = − + = −

For the second isochoric process,

( )( )

34 1 V 1 42

32

( )

(0.20 mol) (8.31 J/mol K)(600 K 300 K) 747.9 K

Q nC T n R T T→ = ∆ = −

= − =

Thus, H 1 2 4 1 1439.1 J.Q Q Q→ →= + = The thermal efficiency of the engine is

net

H

345.6 J 0.24 241439.1 J

WQ

h = = = = ,

19.64. Model: Processes 2 → 1 and 4 → 3 are isobaric. Processes 3 → 2 and 1 → 4 are isochoric. Visualize:

Solve: (a) Except in an adiabatic process, heat must be transferred into the gas to raise its temperature. Thus heat is transferred in during processes 4 → 3 and 3 → 2. This is the reverse of the heat engine in Example 19.2.



(b) Heat flows from hot to cold. Since heat energy is transferred into the gas during processes 4 → 3 and 3 → 2, which end with the gas at temperature 2700 K, the reservoir temperature must be T > 2700 K. This is the hot reservoir, so the heat transferred is H.Q Similarly, heat energy is transferred out of the gas during processes 2 → 1 and 1 → 4. This requires that the reservoir temperature be T < 300 K. This is the cold reservoir, and the energy trans-ferred during these two processes is C.Q (c) The heat energies were calculated in Example 19.2, but now they have the opposite signs.

5 5 5H 43 32

5 5 5C 21 14

7.09 10 J 15.19 10 J 22.28 10 J

21.27 10 J 5.06 10 J 26.33 10 J

Q Q Q

Q Q Q

= + = × + × = ×

= + = × + × = ×

(d) For a counterclockwise cycle in the pV diagram, the work is in .W Its value is the area inside the curve, which is inW = 3 5( ) 2 101,300 Pa (2 m ) 4.05 10 J.p V(∆ ) ∆ = ( × ) = × Note that in C ,W Q QΗ= − as expected from energy conservation.

(e) No. A refrigerator uses work input to transfer heat energy from the cold reservoir to the hot reservoir. This device uses work input to transfer heat energy from the hot reservoir to the cold reservoir.

19.65. Solve: (a) If you wish to build a Carnot engine that is 80% efficient and exhausts heat into a cold reservoir at 0°C, what temperature (in °C) must the hot reservoir be? (b)

3H

H H

(0 C 273) 2730.80 1 0.20 1.1 10 C( 273) 273

TT T° +

= − ⇒ = ⇒ = × °+ +

19.66. Solve: (a) A refrigerator with a coefficient of performance of 4.0 exhausts 100 J of heat in each cycle. What work is required each cycle and how much heat is removed each cycle from the cold reservoir? (b) We have C in C in4.0 / 4 .Q W Q W= ⇒ = This means

HH C in in in in in

100 J4 5 20 J5 5

QQ Q W W W W W= + = + = ⇒ = = =

Hence, C H in 100 J 20 J 80 J.Q Q W= − = − =

19.67. Solve: (a) A heat engine operates at 20% efficiency and produces 20 J of work in each cycle. What is the net heat extracted from the hot reservoir and the net heat exhausted in each cycle? (b) We have C H0.20 1 / .Q Q= − Using the first law of thermodynamics,

out H C C H20 J 20 JW Q Q Q Q= − = ⇒ = −

Substituting into the definition of efficiency,

HH

H H H

20 J 20 J 20 J 20 J0.20 1 1 1 100 J0.20

Q QQ Q Q−

= − = − + = ⇒ = =

The heat exhausted is C H 20 J 100 J 20 J 80 J.Q Q= − = − =

19.68. Solve: (a)

19-28 Chapter 19


In this heat engine, 400 kJ of work is done each cycle. What is the maximum pressure? (b)

5 3 5 5max max

1 ( 1.0 10 Pa)(2.0 m ) 4.0 10 J 5.0 10 Pa 500 kPa2

p p− × = × ⇒ = × =

19.69. Model: The heat engine follows a closed cycle, starting and ending in the original state. Visualize: The figure indicates the following seven steps. First, the pin is inserted when the heat engine has the ini-tial conditions. Second, heat is turned on and the pressure increases at constant volume from 1 to 3 atm. Third, the pin is removed. The flame continues to heat the gas and the volume increases at constant pressure from 350 cm to

3100 cm . Fourth, the pin is inserted and some of the weights are removed. Fifth, the container is placed on ice and the gas cools at constant volume to a pressure of 1 atm. Sixth, with the container still on ice, the pin is removed. The gas continues to cool at constant pressure to a volume of 350 cm . Seventh, with no ice or flame, the pin is inserted back in and the weights returned bringing the engine back to the initial conditions and ready to start over. Solve: (a)

(b) The work done per cycle is the area inside the curve: 6 3

out ( )( ) (2 101,300 Pa)(50 10 m ) 10 JW p V−= ∆ ∆ = × × =

(c) Heat energy is input during processes 1 → 2 and 2 → 3, so H 12 23.Q Q Q= + This is a diatomic gas, with 3

V 2C R= and 5

P 2 .C R= The number of moles of gas is

6 31 1

1

(101,300 Pa)(50 10 m ) 0.00208 mol(8.31 J/mol K)(293 K)

p VnRT

−×= = =

Process 1 → 2 is isochoric, so 2 2 1 1 1( / ) 3 879 K.T p p T T= = = Process 2 → 3 is isobaric, so

3 3 2 2 2( / ) 2 1758 K.T V V T T= = = Thus

5 512 V 2 12 2( ) (0.00208 mol)(8.31 J/mol K)(586 K) 25.32 JQ nC T nR T T= ∆ = − = =

Similarly, 7 7

23 P 3 22 2( ) (0.00208 mol)(8.31 J/mol K)(879 K) 53.18 JQ nC T nR T T= ∆ = − = =

Thus H 25.32 J 53.18 J 78.50 JQ = + = and the engine’s efficiency is

out

H

10.13 J 0.13 1378.50 J

WQ

h = = = = ,

19.70. Model: System 1 undergoes an isochoric process and system 2 undergoes an isobaric process. Solve: (a) Heat will flow from system 1 to system 2 because system 1 is hotter. Because there is no heat input from (or loss to) the outside world, we have 1 2 0 J.Q Q+ = Heat 1,Q which is negative, will change the temperature of system 1. Heat 2Q will both change the temperature of system 2 and do work by lifting the piston. But these conse-quences of heat flow don’t change the fact that 1 2 0 J.Q Q+ = System 1 undergoes constant volume cooling from



1i f600 K to .T T= System 2, whose pressure is controlled by the weight of the piston, undergoes constant pressure heating from 2i f300 K to .T T= Thus,

( ) ( )3 51 2 1 V f 1i 2 P f 2i 1 f 1i 2 f 2i2 20 J ( ) ( ) ( ) ( )Q Q n C T T n C T T n R T T n R T T+ = = − + − = − + − Solving this equation for fT gives

1 1i 2 2if

1 2

3 5 3(0.060 mol)(600 K) 5(0 030 mol)(300 K) 464 K3 5 3(0 060 mol) 5(0 030 mol)

n T n TTn n

+ + .= = =

+ . + .

(b) Knowing f ,T we can compute the heat transferred from system 1 to system 2:

( )52 2 P f 2i 2 f 2i2( ) ( ) 102 JQ n C T T n R T T= − = − = (c) The change of thermal energy in system 2 is

( )3 3th 2 V 2 f 2i 22 5( ) 61.2 JE n C T n R T T Q∆ = ∆ = − = = According to the first law of thermodynamics, 2 S th.Q W E= + ∆ Thus, the work done by system 2 is

S th 102.0 J 61.2 J 40.8 J.W Q E= − ∆ = − = The work is done to lift the weight of the cylinder and the air above it by a

height ∆y. The weight of the air is 2 2 2air (101.3 10 N/m ) (0.050 m) 795.6 N.w pA p rp p= = = × = Therefore,

ss cyl air 2

cyl air

40.8 J( ) 0.050 m( ) (2 0 kg)(9.8 m/s ) 795.6 N

WW w w y yw w

= + ∆ ⇒ ∆ = = =+ . +

(d) The fraction of heat converted to work is

s

2

40.8 J 0.40 40102.0 J

WQ

= = = ,

19.71. Model: Process 1 → 2 and process 3 → 4 are adiabatic, and process 2 → 3 and process 4 → 1 are isochoric. Visualize: Please refer to Figure CP19.71. Solve: (a) For adiabatic process 1 → 2, 12 0 JQ = and

2 2 1 1 2 112

( )1 1


− −= =

− −

For isochoric process 2 → 3, 23 23 V 3 20 J and ( ).W Q nC T T= = − For adiabatic process 3 → 4, 34 0 JQ = and

4 4 3 3 4 334

( )1 1


− −= =

− −

For isochoric process 4 → 1, 41 41 V 1 40 J and ( ).W Q nC T T= = − The work done per cycle is

2 1 4 3net 12 23 34 41 2 1 4 3

( ) ( )0 J 0 J ( )1 1 1

nR T T nR T T nRW W W W W T T T Tγ γ γ− −

= + + + = + + + = − + −− − −

(b) The thermal efficiency of the heat engine is

out C 41 V 4 1

H H 23 V 3 2

| | ( )1 1 1| | ( )

W Q Q nC T TQ Q Q nC T T

h −= = − = − = −−

The last step follows from the fact that 3 2 4 1 and .T T T T> > We will now simplify this expression further as follows: 1

1 1 1 1 111 2 2 1 11 2

2

VpV pVV nRTV nRTV nRT V T T T rV

γγ γ γ γ γ γ

−− − − − − = = ⇒ = ⇒ = =

Similarly, 13 4 .T T rγ −= The equation for thermal efficiency now becomes

4 11 1 1

4 1

11 1T TT r T r rγ γ γ

h − − −−

= − = −−

19-30 Chapter 19


(c)

19.72. Model: For the Diesel cycle, process 1 → 2 is an adiabatic compression, process 2 → 3 is an isobaric expan-sion, process 3 → 4 is an adiabatic expansion, and process 4 → 1 is isochoric. Visualize: Please refer to CP19.72. Solve: (a) It will be useful to do some calculations using the compression ratio, which is

3max 1

3min 2

1050 cm 2150 cm

V VrV V

= = = =

The number of moles of gas is 5 6 3

1 1

1

(1 013 10 Pa)(1050 10 m ) 0.0430 mol(8 31 J/mol K)[(25 273) K]

p VnRT

−. × ×= = =

. +

For an adiabatic process,

1.40 611 2 2 1 11 2

221 1atm 71.0 atm 7.19 10 PaVp V p V p p r p

V

γγ γ γ = ⇒ = = = × = = ×

11 1 1 0 411

1 2 2 1 11 22

21 298 K 1007 KVTV T V T T r TV

γγ γ γ

−− − − . = ⇒ = = = × =

Process 2 → 3 is an isobaric heating with Q = 1000 J. Constant pressure heating obeys

PP

QQ nC T TnC

= ∆ ⇒ ∆ =

The gas has a specific heat ratio 1.40 7/5,γ = = thus 5V 2C R= and 7

P 2 .C R= Knowing p,C we can calculate first

∆T = 800 K and then 3 2 1807 K.T T T= + ∆ = Finally, for an isobaric process we have

6 3 6 32 3 33 2

2 3 2

1807 K (50 10 m ) 89.7 10 m1007 K

V V TV VT T T

− −= ⇒ = = × = ×

Process 3 → 4 is an adiabatic expansion to 4 1.V V= Thus, 1 46 3

6 533 4 4 33 4 6 3

4

0 41 6 31 1 3

3 4 4 33 4 6 34

89.7 10 m (7.19 10 Pa) 2.30 10 Pa 2.27 atm1050 10 m

89.7 10 m (1807 K) 675 K1050 10 m

Vp V p V p pV

VT V T V T TV

γγ γ

γγ γ

.−

−

.− −− −

−

×= ⇒ = = × = × = ×

×= ⇒ = = = ×

Point P V T 1 51.00 atm 1.013 Pa= ×10 6 31050 10 m−× 298 K 25 C= °

2 671.0 atm 7.19 Pa= ×10 6 350.0 10 m−× 1007 K 734 C= °

3 671.0 atm 7.19 Pa= ×10 6 389.7 10 m−× 1807 K 1534 C= °

4 52.27 atm 2.30 Pa= ×10 6 31050 10 m−× 675 K 402 C= °



(b) For adiabatic process 1 → 2,

2 2 1 1s 12( ) 633 J1

p V p VWγ

−= = −

−

For isobaric process 2 → 3,

s 23 2 2 3 2( ) ( ) 285 JW p V p V V= ∆ = − =

For adiabatic process 3 → 4,

4 4 3 3s 34( ) 1009 J1

p V p VWγ

−= =

−

For isochoric process 4 → 1, s 41( ) 0 J.W = Thus,

s cycle out s 12 s 23 s 34 s 41( ) ( ) ( ) ( ) ( ) 661 JW W W W W W= = + + + =

(c) The efficiency is

out

H

661 J 66.1 661000 J

WQ

h = = = % ≈ %

(d) The power output of one cylinder is 661 J 2400 cycle 1 min J26,440 26.4 kWcycle min 60 sec s

= =

For an 8-cylinder engine the power will be 211 kW or 283 horsepower.

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Instructor Solution Manual - Home | Santa Rosa Junior Collegesrjcstaff.santarosa.edu/~lwillia2/41/41ch19hwkey.pdf∆th W S Q A + 0 + B − + 0 C 0 + + D − − − 19.10. Model: Process