19.1. Model: The heat engine follows a closed cycle, starting and ending in the original state. The cycleconsists of three individual processes.Visualize: Please refer to Figure Ex19.1.Solve: (a) The work done by the heat engine per cycle is the area enclosed by the p-versus-V graph. We get

Wout3 kPa m 10 J= ( ) ×( ) =−1

26200 100 10

The heat energy transferred into the engine is QH = 120 J. Because W Q Qout H C= − , the heat energy exhausted is

Q Q WC H out 120 J 10 J 110 J= − = − =

(b) The thermal efficiency of the engine is

η = = =W

Qout

H

10 J

120 J0 0833.

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

19.2. Model: The heat engine follows a closed cycle, which consists of four individual processes.Visualize: Please refer to Figure Ex19.2.Solve: (a) The work done by the heat engine per cycle is the area enclosed by the p-versus-V graph. We get

Wout3 kPa 100 kPa m 30 J= −( ) ×( ) =−400 100 10 6

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

W Q Q Q Q Wout H C H C out 115 J 30 J 145 J= − ⇒ = + = + =

(b) The thermal efficiency of the engine is

η = = =W

Qout

H

30 J

145 J0 207.

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

19.3. Solve: (a) During each cycle, the heat transferred into the engine is QH 55 kJ= , and the heat exhausted isQC 40 kJ= . The thermal efficiency of the heat engine is

η = − = − =1 1 0 273Q

QC

H

40 kJ

55 kJ.

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

W Q Qout H C 55 kJ 40 kJ 15 kJ= − = − =

19.4. Solve: During each cycle, the work done by the engine is Wout 20 J= and the engine exhausts QC 30 J=of heat energy. Because W Q Qout H C= − ,

Q W QH out C 20 J 30 J 50 J= + = + =

Thus, the efficiency of the engine is

η = − = − =1 1 0 40Q

QC

H

30 J

50 J.

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

η = ⇒ = ⇒ =W

Q QQout

H HH

100 J250 J0 40.

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

Q Q WC H out 250 J 100 J 150 J= − = − =

19.6. Solve: The coefficient of performance of the refrigerator is

KQ

W

Q W

W= = − = − =C

in

H in

in

50 J 20 J

20 J1 50.

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

KQ

W

QQ= ⇒ = ⇒ =C

in

CC50 J

200 J4 0.

(b) The heat exhausted to the hot reservoir isQ Q WH C in J J 250 J= + = + =200 50

19.8. 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

Wout 500 kJ

s

1 s

40 cycles

kJ

cycle= × = 12 5.

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

η = ⇒ = =W

QQout

HH

kJ62.5 kJ

12 5

0 20

.

.

The heat exhausted per cycle isQ Q WC H in kJ kJ kJ= − = − =62 5 12 5 50. .

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

ηη

= =+

⇒ = −

= ( ) −

= ×W

Q

W

Q WQ Wout

H

out

C outC out MW W

11 900

1

0 321 1 913 109

..

That is, each second the electric power plant discharges 1 913 109. × J of energy into the ocean. Since a typicalAmerican house needs 2 0 104. × J of energy per second for heating, the number of houses that could be heatedwith the waste heat is 1 913 10 2 0 10 96 0009 4. . ,×( ) ×( ) = J J .

19.10. Solution: The amount of heat removed from the water in cooling it down in 1 hour is Q m c TC water water= ∆ .The mass of the water is

m Vwater water water3 3 31000 kg / m 1 L 100 kg / m m 1.0 kg= = ( )( ) = ( )( ) =−ρ 10 3

⇒ = ( )( ) ° − °( ) = ×QC 1.0 kg 4190 J / kg K 20 C 5 C J6 285 104.

The rate of heat removal from the refrigerator is

QC

J

3600 s17.46 J / s= × =6 285 104.

The refrigerator does work W W= =8 0 8 0. . J / s to remove this heat. Thus the performance coefficient of therefrigerator is

K = =17.46 J / s

8.0 J / s2 18.

19.11. Model: Process A is isochoric, process B is isothermal, process C is adiabatic, and process D is isobaric.Visualize: Please refer to Figure Ex19.11.Solve: Process A is isochoric, so the increase in pressure increases the temperature and hence the thermal energy.Because ∆E Q W Wth s s and J= − = 0 , Q increases for process A. Process B is isothermal, so T is constant andhence ∆Eth = 0 J. The work done Ws is positive because the gas expands. Because Q W E= +s th∆ , Q is positive forprocess B. Process C is adiabatic, so Q = 0 J. W s is positive because of the increase in volume. SinceQ W E= = +0 J s th∆ , ∆Eth is negative for process C. Process D is isobaric, so the decrease in volume leads to adecrease in temperature and hence a decrease in the thermal energy. Due to the decrease in volume, Ws is negative.Because Q W E= +s th∆ , Q also decreases for process D.

∆Eth Ws QA + 0 +B 0 + +C − + 0D − − −

19.12. Model: Process A is adiabatic, process B is isochoric, process C is isothermal, and process D is isobaric.Visualize: Please refer to Figure Ex19.12.Solve: Process A is adiabatic, so Q = 0 J. Work Ws is positive as the gas expands. Since Q = Ws + ∆Eth = 0 J, ∆Eth

must be negative. The temperature falls during an adiabatic expansion. Process B is isochoric. No work is done(Ws = 0 J), and Q is negative as heat energy is removed to lower the temperature (∆Eth negative). Process C isisothermal, so ∆T = 0 and ∆Eth = 0 J. The gas is compressed, so Ws is negative. Q = Ws for an isothermal process, soQ is negative. Heat energy is withdrawn during the compression to keep the temperature constant. Finally, work Ws

is positive during the isobaric expansion of process D. Temperature increases, so ∆Eth is also positive. This makesQ = Ws + ∆Eth positive.

∆Eth Ws QA – + 0B – 0 –C 0 – –D + + +

19.13. Visualize: Please refer to Figure Ex.19.13.Solve: The work done by the gas per cycle is the area inside the closed p-versus-V curve. The area inside thetriangle is

Wout3 3

3

3 atm 1 atm m m

2 atm Pa

1 atm m J

= −( ) × − ×( )= × ×

×( ) =

− −

−

12

6 6

12

56

600 10 200 10

1 013 10400 10 40 5

..

19.14. Visualize: Please refer to Figure Ex19.14.Solve: The work done by the gas per cycle is the area enclosed within the pV curve. We have

60 J kPa 800 cm 200 cm12 max

3 3= −( ) −( )p 100 ⇒ ( )×

= − ×−

2

600 101 0 106

560 J

m Pa3 maxp .

⇒ = × =pmax Pa 300 kPa3 0 105.

19.15. Model: The heat engine follows a closed cycle.Visualize: Please refer to Figure Ex19.15.Solve: (a) The work done by the gas per cycle is the area inside the closed p-versus-V curve. We get

Wout3 3300 kPa 100 kPa 600 cm 300 cm= −( ) −( )1

2 = ×( ) ×( )−12

3 6200 10 300 10 Pa m3 = 30 J

The thermal efficiency is

η = =+

=W

Qout

H

30 J

225 J J900 0952.

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

Q Q WC H out 315 J 30 J 285 J= − = − =

19.16. Model: The heat engine follows a closed cycle.Visualize: Please refer to Figure Ex19.16.Solve: (a) The work done by the gas per cycle is the area inside the closed p-versus-V curve. We get

Wout3 3300 kPa 100 kPa 600 cm 200 cm= −( ) −( )1

2 = ×( ) ×( )−12

3 6200 10 400 10 Pa m3 = 40 J

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

η = =+

=+

=W

Q

W

Q Wout

H

out

C out

40 J

280 J J400 125.

(b) The heat extracted from the hot reservoir is QH = 320 J.

19.17. Model: The Brayton cycle involves two adiabatic processes and two isobaric processes.

Solve: From Equation 19.21, the efficiency of a Brayton cycle is η γ γB p= − −( )1 1r , where rp is the pressure ratio

pmax/pmin. The specific heat ratio for a diatomic gas is

γ = = =C

CP

V

72

52

R

R1 4.

Solving the above equation for rp,

1 1−( ) = −( )η γ γB rp

/ ⇒ = −( ) = −( ) =−−

rp 1 1 0 60 24 711 4

0 4ηγγ

B . ..

.

19.18. Model: The Brayton cycle involves two adiabatic processes and two isobaric processes. The adiabaticprocesses involve compression and expansion through the turbine.

Solve: The thermal efficiency for the Brayton cycle is η γ γB = − −( )1 1rp

/ , where γ = C CP V/ and rp is the pressure

ratio. For a diatomic gas γ = 1.4. For an adiabatic process,

p V p V p p V V1 1 2 2 2 1 1 2γ γ γ= ⇒ = ( )/ /

Because the volume is halved, V V212 1= and hence

r p pp = = ( ) = =2 11 42 2 2 639/ ..γ

The efficiency is

ηB = − ( ) =−

1 2 639 0 2420 4

1 4. ..

.

19.19. Model: The efficiency of a Carnot engine (ηCarnot) depends only on the temperatures of the hot and coldreservoirs. On the other hand, the thermal efficiency (η) of a heat engine depends on the heats QH and QC.Visualize: Please refer to Figure Ex19.19.Solve: (a) According to the first law of thermodynamics, Q W QH out C= + . For engine (a), QH = 50 J, QC = 20 Jand Wout = 30 J, so the first law of thermodynamics is obeyed. For engine (b), QH = 10 J, QC = 7 J and Wout = 4 J, sothe first law is violated. For engine (c) the first law of thermodynamics is obeyed.(b) For the three heat engines, the maximum or Carnot efficiency is

ηCarnotC

H

300 K

600 K= − = − =1 1 0 50

T

T.

Engine (a) has

η = − = = =1 0 60Q

Q

W

QC

H

out

H

30 J

50 J.

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

η η= = = <W

Qout

HCarnot

4 J

10 J0 40.

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

η η= = <10 J

30 J Carnot0 333.

so the second law of thermodynamics is obeyed.

19.20. Model: For a refrigerator QH = QC + Win, and the coefficient of performance and the Carnot coefficientof performance are

KQ

W= C

in

KT

T TCarnotC

H C

=−

Visualize: Please refer to Figure Ex19.20.Solve: (a) For refrigerator (a) Q Q WH C in (60 J 40 J 20 J)= + = + , 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

KT

T TCarnotC

H C

300 K

400 K 300 K=

−=

−= 3

For refrigerator (a),

KQ

WK= = = <C

inCarnot

40 J

20 J2

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

KQ

WK= = = >C

inCarnot

40 J

10 J4

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

K K= = <30 J

20 J Carnot1 5.

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 coldreservoirs.Solve: The efficiency of a Carnot engine is

ηCarnotC

H

= −1T

T⇒ = −

+( )0 60 1.

TC

427 273 K⇒ = = °TC 280 K 7 C

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

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

η = =+

=+

= =W

Q

W

Q Wout

H

out

C out

200 J

600 J 200 J0 25 25. %

(b) The thermal efficiency of a Carnot engine is ηCarnot C H= −1 T T . For this to be 25%,

0 25 1 504 8. .= −+( )

⇒ = = °TTC

C400 273 K K 232 C

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

η = =+

=+

= =W

Q

W

Q Wout

H

out

C out

10 J

15 J 10 J%0 40 40.

(b) The efficiency of a Carnot engine is ηCarnot C H= −1 T T . The minimum temperature in the hot reservoir is foundas follows:

0 40 1. = − ⇒ = = °293 K488 K 215 C

HHT

T

This is the minimum possible temperature. In a real engine, the hot-reservoir temperature would be higher than215°C because no real engine can match the Carnot efficiency.

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

ηCarnotC

H

300 K

500 K= − = − = =1 1 0 40 40

T

T. %

(b) An engine with power output of 1000 W does Wout = 1000 J of work during each ∆t = 1 s. A Carnot engine has aheat input that is

QW

inout

Carnot

=1000 J

0.40 J= =

η2500

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

19.25. Model: The efficiency of a Carnot engine depends only on the temperatures of the hot and coldreservoirs.Solve: The efficiency of a Carnot engine is ηCarnot C H= −1 T T . This can be increased by either increasing the hot-reservoir temperature or decreasing the cold-reservoir temperature. For a 40% Carnot efficiency and a cold-reservoir temperature of 7°C,

ηCarnotH

280 K= = −0 40 1.T

⇒ =TH 467 K

A higher efficiency of 60% can be obtained by raising the hot-reservoir temperature to ′TH . Thus,

ηCarnotH

280 K= = −′

0 60 1.T

⇒ ′ =TH 700 K

The reservoir temperature must be increased by 233 K to cause the efficiency to increase from 40% to 60%.

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

η ηmax CarnotC

H

K

K= = − = −

+( )+( )

=1 1273 20

273 6000 6644

T

T.

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

QW

Hout J

0.1993 J= = =

η1000

5020

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

KT

T TCarnotC

H C

K

K K=

−=

− +( )+( ) − − +( )

=20 273

20 273 20 2736 325.

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

KQ

WW

Q

K= ⇒ = = =C

inin

C J /s

6.325 J /s = 32 W

20032

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

Q Q WH C in 200 J /s 32 J /s 232 J /s= + = + =

19.28. Model: The coefficient of performance of a Carnot refrigerator depends only on the temperatures of thecold and the hot reservoirs.Solve: For a Carnot refrigerator,

KT

T T

Q

WCarnotC

H C

C

in

=−

= ⇒−

=4.2 K

293 K 4.2 K

1.0 J

inW⇒ =Win 68.8 J

19.29. Model: Assume that the refrigerator is an ideal or Carnot refrigerator.Solve: (a) For the refrigerator, the coefficient of performance is

KQ

WQ KW= ⇒ = = ( )( ) =C

inC in 10 J 50 J5 0.

The heat energy exhausted per cycle is

Q Q WH C in 50 J 10 J 60 J= + = + =

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

KT

T T

T

TTCarnot

C

H C

C

CC300 K

250 K 23 C=−

⇒ =−

⇒ = = − °5 0.

19.30. Model: The coefficient of performance of a Carnot refrigerator depends only on the temperatures of thecold and hot reservoirs.Solve: Using the definition of the coefficient of performance for a Carnot refrigerator,

KT

T TCarnotC

H C

=−

⇒ =− +( )− − +( )

=−

5 013 273

13 273.

K

K

260 K

260 KH HT T⇒ =TH 312 K

To increase the coefficient of performance to 10.0, the hot-reservoir temperature changes to ′TH . Thus,

KTCarnot

H

260 K

260 K= =

′ −10 0. ⇒ ′ =TH 286 K

Since ′TH is less than TH, the hot-reservoir temperature must be decreased by 26 K or 26°C.

19.31. Solve: The work-kinetic energy theorem is that the work done by the car engine is equal to the changein the kinetic energy. Thus,

W K mvout 1500 kg 15 m / s 168,750 J= = = ( )( ) =∆ 12

2 12

2

Using the definition of thermal efficiency,

ηη

= ⇒ = = = ×W

QQ

Wout

HH

out 168,750 J J

0 208 44 105

..

That is, the burning of gasoline transfers into the engine 8.44 × 105 J of heat energy.

19.32. Solve: The work done by the engine is equal to the change in the gravitational potential energy. Thus,Wout = ∆Ugrav = mgh = (2000 kg)(9.8 m/s2)(30 m) = 588,000 J

The efficiency of this engine is

η η= = −

= −+( )

+( )

0 40 0 40 1 0 40 1273 20

273 2000. . .Carnot

C

H

K

K

T

T= 0.3484

The amount of heat energy transferred is calculated as follows:

ηη

= ⇒ = = = ×W

QQ

Wout

HH

out 588,000 J

0.3484 J1 69 106.

19.33. Solve: The mass of the water is

100 10103

3

× × × =−−

L m

1 L

1000 kg

m0.100 kg

3

3

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

Q mc T

Q mL

Q mc T

1

25

3

3 33 10

= = ( )( )( ) =

= = ( ) ×( ) =

= = ( )( )( ) =

∆

∆

0.100 kg 4186 J / kg K 15 K 6279 J

0.100 kg J / kg 33,300 J

0.100 kg 2090 J / kg K 15 K 3135 J

f .

⇒ = + + =Q Q Q QC 42,714 J1 2 3

Using the performance coefficient,

KQ

W W= ⇒ = ⇒C

in in

42,714 J4 0. Win

42,714 J

4.010,679 J= =

The heat exhausted into the room is thus

Q Q WH C in442,714 J 10,679 J 5.34 10 J= + = + = ×

19.34. Solve: An adiabatic process has Q = 0 and thus, from the first law, Ws = –∆Eth. For any ideal gasprocess, ∆Eth = nCV∆T, so Ws = –nCV∆T. We can use the ideal gas law to find

TpV

nRT

pV

nR

pV pV

nR

p V p V

nR= ⇒ ∆ = ∆ = − = −( ) ( ) ( )f i f f i i

Consequently, the work is

W nC T nCp V p V

nR

C

Rp V p Vs V V

f f i i Vf f i i= − ∆ = − −

= − −( )

Because CP = CV + R, we can use the specific heat ratio γ to find

γγ

= = + = + ⇒ =−

C

C

C R

C

C R

C R

C

RP

V

V

V

V

V

V/

/

1 1

1

With this, the work done in an adiabatic process is

WC

Rp V p V p V p V p V p Vs

Vf f i i f f i i f f i i= − − = −

−− = − −( ) ( ) ( ) / ( )

1

11

γγ

19.35. Visualize: The output from a perfectly reversible heat engine is used to drive a refrigerator.

Solve: Suppose the refrigerator is a perfectly reversible Carnot refrigerator. Then ′QH = QH and ′QC = QC. That is,a Carnot refrigerator exactly replaces all the heat that had been transferred by the Carnot engine, so there’s no nettransfer of heat. Now suppose the refrigerator is a superefficient refrigerator with K > KCarnot. Since K = QC/Win, asuperefficient refrigerator can pump more heat from the cold reservoir with the same work Win than can a Carnotrefrigerator. That is, ′QC > QC. Since ′QH = ′QC + Win, a larger amount of heat being pumped with the same Win

requires that ′QH > QH. Thus the net result of using a Carnot heat engine to drive a superefficient refrigerator is thatheat is transferred from the cold reservoir to the hot reservoir without outside assistance. But such a process isforbidden by the second law of thermodynamics, so we must conclude that no refrigerator can have a largercoefficient of performance than a Carnot refrigerator.

19.36. Solve: For any heat engine, η = 1 – QC/QH. For a Carnot heat engine, ηCarnot = 1 – TC/TH. Thus a propertyof the Carnot cycle is that QC/QH = TC/TH. Consequently, the coefficient of performance of a Carnot refrigerator is

KQ

W

Q

Q Q

Q Q

Q Q

T T

T T

T

T TCarnotC

in

C

H C

C H

C H

C H

C H

C

H C

= =−

=−

=−

=−

/

/

/

/1 1

19.37. Solve: We can also obtain the desired result using the η and K equations for the Carnot engine and theCarnot refrigerator:

KT

T T

T T

T T=

−=

− ( )C

H C

C H

C H1 η η= − ⇒ = −1 1

T

T

T

TC

H

C

H

Hence the equation for K becomes

K = −− −( ) = −1

1 1

1ηη

ηη

19.38. Visualize:

The figure shows two Carnot engines operating between temperatures TH and TC with intermediate temperature T3.Solve: A single Carnot engine operating between temperatures TH and TC does work

W QT

TQout Carnot H

C

HH= = −

η 1

For the two Carnot engines, the first operates between TH and T3 and does work

W QT

TQ1 H

3

HH= = −

η1 1

Similarly, the second heat engine, between T3 and TC, does work

W QT

TQ2 H

C

3H= ′ = −

′η2 1

For any heat engine η = 1 – QC/QH, and for a Carnot heat engine ηCarnot = 1 – TC/TH. Thus a property of the Carnotcycle is that QC/QH = TC/TH and thus QC = (TC/TH)QH. Applying this result to the second engine,

′ = =

=QT

TQ

T

T

T

TQ

T

TQH

3

CC

3

C

C

HH

3

HH

With this result for ′QH , the work done by the second engine is

WT

TQ

T

T

T

TQ

T

T

T

TQ2

C

3H

C

3 HH

H

C

HH= −

′ = −

= −

1 1 3 3

Consequently, the combined work of the two engines is

W WT

TQ

T

T

T

TQ

T

TQ W1

3

HH

H

C

HH

C

HH out+ = −

+ −

= −

=231 1

The combined work is exactly equal to the work of the single Carnot engine operating between TH and TC.

19.39. 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

η ηmax .= = − = −+( )

+( )=Carnot

C

H

K

K1 1

273 5

273 300 0825

T

T = 8.25%

19.40. Model: The ideal gas in the Carnot engine follows a closed cycle in four steps. During the isothermalexpansion at temperature TH, heat QH is transferred from the hot reservoir into the gas. During the isothermalcompression at TC, heat QC is removed from the gas. No heat is transferred during the remaining two adiabatic steps.Solve: The thermal efficiency of the Carnot engine is

ηCarnotC

H

out

H

= − =1T

T

W

Q⇒ − =1

323 K

573 K 1000 JoutW ⇒ =Wout 436 J

Using Q Q WH C out= + , we obtain

Q Q Q Wisothermal C H out 1000 J 436 J 564 J= = − = − =

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

ηCarnotC

H

= −1T

T⇒ = −

+0 25 1.

T

TC

C 80 K⇒ = = − °TC 240 K C33

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

19.42. Model: The efficiency of a Carnot engine depends only on the reservoir temperatures.Solve: The Carnot heat-engine efficiency is η = 1 – TC/TH. The efficiency can be increased either by decreasingTC or by increasing TH. First consider lowering the cold reservoir to TC – ∆T. This changes the efficiency to

η η η η+ ∆ = − − ∆ = − + ∆ = + ∆ ⇒ ∆ = ∆C

C

H

C

H H HC

H

1 1T T

T

T

T

T

T

T

T

T

T

Now raise the hot reservoir by the same ∆T to TH + ∆T. This changes the efficiency to

η η η

η η

+ ∆ = −+ ∆

= − + −+ ∆

= + −+ ∆

⇒ ∆ = −+ ∆

= ∆+ ∆

=+ ∆

∆ =+ ∆

× ∆

HC

H

C

H

C

H

C

H

C

H

C

H

HC

H

C

HC

H H

C

H H

C

HC

1 1T

T T

T

T

T

T

T

T T

T

T

T

T T

T

T

T

T TT

T

T T T

T

T T

T

T

T

T T( )

Now TC < TH and TH + ∆T > TH, so the ratio TC/( TH + ∆T) < 1. Consequently, ∆ηH < ∆ηC. In other words, loweringthe cold reservoir temperature has the larger effect.

19.43. Solve: If QC = 23 QH, then Wout = QH – QC = 1

3 QC. Thus the efficiency is

η = = =W

Q

Q

Qout

H

H

H

13 1

3

The efficiency of a Carnot engine is η = 1 – TC/TH. Thus

11

3

2

3− = ⇒ =T

T

T

TC

H

C

H

19.44. Visualize: Refer to Figure P19.44 in your textbook.Solve: (a) Q1 is given as 1000 J. Using the energy transfer equation for the heat engine,

Q Q W Q Q W Q Q WH C out out out= + ⇒ = + ⇒ = −1 2 2 1

The thermal efficiency of a Carnot engine is

η = − = − = =1 1 0 501

T

T

W

QC

H

out300 K

600 K.

⇒ = − = −( ) = ( ) −( ) =Q Q Q Q2 1 1 1 1 1 0 50η η 1000 J 500 J.

To determine Q3 and Q4, we turn our attention to the Carnot refrigerator, which is driven by the output of the heatengine with Win = Wout. The coefficient of performance is

KT

T T

Q

W

Q

W

Q

Q

Q K Q

=−

=−

= = = =

⇒ = = ( )( )( ) =

C

H C

C

in out

400 K

500 K 400 K

1000 J 2000 J

4 0

4 0 0 50

4 4

1

4 1

.

. .

ηη

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

Q W Q Q Q3 4 1 4 0 50= + = + = ( )( ) + =out 1000 J 2000 J 2500 Jη .

(b) From part (a) Q3 = 2500 J and Q1 = 1000 J , so Q Q3 1> .(c) Although Q1 = 1000 J and Q3 = 2500 J, the two devices together do not violate the second law of thermodynamics.This is because the hot and cold reservoirs are different for the heat engine and the refrigerator.

19.45. Model: A heat pump is a refrigerator that is cooling the already cold outdoors and warming the indoorswith its exhaust heat.Solve: (a) The coefficient of performance for this heat pump is K Q W= =5 0. C in , where QC is the amount ofheat removed from the cold reservoir. QH is the amount of heat exhausted into the hot reservoir. QH = QC + Win,where Win is the amount of work done on the heat pump. We have

Q W Q W W WC in H in in in= ⇒ = + =5 0 5 0 6 0. . .

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

Q W WH in in15 kJ 6.015 kJ

6.02.5 kJ= = ⇒ = =

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

s200 hrs

3600 s

1 hr MJ× × × =1

40270

$$

The monthly heating cost in the house using a heat pump is2.5 kJ

s200 hrs

3600 s

1 hr

1$

40 MJ× × × = $45

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

ηmaxC

H

300 K

500 K= − = − =1 1 0 40

T

T.

The efficiency of the engine designed by the first student is

η1 0 44= = =W

Qout

H

110 J

250 J.

Because η1 > ηmax, 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 η2 =90 0 36 J 250 J max= <. η in agreement with the second law of thermodynamics. Applying the first law ofthermodynamics,

Q Q WH C out 250 J 170 J 90 J= + ⇒ = +

we see the first law is violated. This design will not work as claimed. The design by the third student satisfies thefirst law of thermodynamics because Q Q WH C out 250 J= + = . The thermal efficiency of this engine is η3 =0 36. < ηmax , 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 3 × 105 kg of coal is burnt. The volume of coal is

3 1024

5× × × = kg

1 hour

m

1500 kg hour 4800 m

33

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

η = = ×× × × ×

= ××

= =W

Qout

H5 9

J / s3 10 kg

1 hr

J

kg

1 hr

3600 s

J

2.333 10 J

7 50 1028 10

7 50 100 321 32 1

8

6

8. .. . %

Assess: An efficiency of 32.1% is typical of power plants.

19.48. Model: The power plant is treated as a heat engine.Solve: (a) The maximum possible thermal efficiency of the power plant is

ηmaxC

H

303 K

573 K= − = − = =1 1 0 471 47 1

T

T. . %

(b) The plant’s actual efficiency is

η = = ××

=W

Qout

H

J / s

J / s

700 10

2000 100 350 35 0

6

6 . . %

(c) Because QH = QC + Wout,

Q Q W QC H out C9 J / s J / s 1.3 10 J / s= − ⇒ = × − × = ×2 0 10 0 7 109 9. .

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

m = × × × × = ×−

1 2 1010

3 333 1083

4. . L

hour

1 hr

3600 s

m

1 L

1000 kg

m kg

3

3

The change in the temperature as QC J= ×1 3 109. of heat is transferred to m = ×3 333 104. kg of water iscalculated as follows:

Q mc TC = ∆ ⇒ × = ×( )( )1 3 109. J 3.333 10 kg 4186 J / kg K4 ∆T ⇒ = °∆T 9 C

The exit temperature is18 C 9 C 27 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

m = × × × × = ×−

1 0 1010

2 778 1083

4. . L

hr

1 hr

3600 s

m

1 L

1000 kg

m kg / s

3

3

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

Q mc TC kg / s 4186 J / kg K 27 C 16 C J / s= = ×( )( ) ° − °( ) = ×∆ 2 778 10 1 279 104 9. .

The amount of heat per second into the power plant is

Q W QH out C9 9 J / s 1.279 10 J / s 2.029 10 J / s= + = × + × = ×0 750 109.

Finally, the power plant’s thermal efficiency is

η = = ××

= =W

Qout

H

J / s

J / s

0 750 10

2 029 100 37 37

9

9

.

.. %

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

Wout

J

s

3600 s

1 hr

24 hr

1 day J= × × × = ×1 0 10 8 64 109 13. .

(b) The volume of water is V = =1 109 km m3 3.The amount of energy is

Q mc TH3

3 m1000 kg

m4190 J / kg K 1.0 K= = ( )

( )( )∆ 109 = ×4 19 1015. J

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

19.51. Visualize:

Solve: This problem does not have a unique answer. There are many three-step engines that meet these criteria.However, simple cycles consist of three straight-line segments and form a triangle in the pV diagram. The figureabove shows an example. The work done by the gas during one cycle is the area enclosed within the curve. Wehave

W p Vs cycle12( ) = ∆ ∆

We know that

∆p p p= − = − = =max min 3 atm atm 2 atm 202,600 Pa1

⇒ =( )

= ( ) = × =−∆∆

VW

p

2 24 00 10 4S cycle 3 340.5 J

202,600 Pa m 400 cm.

The maximum volume is

V V Vmax min3500 cm= + =∆

The figures below show possible pV diagram of this heat engine.

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

2 and C RP = 52 .

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

Q E nC T

TnC R

T T T

1 2

32

32

2 1 2

→ = = =

⇒ = =( )( ) =

( )( )( )=

⇒ − = ⇒ = + =

3750 J

3750 J 3750 J

1.0 mol

3750 J

1.0 mol 8.31 J / mol K301 K

300.8 K 300.8 K 300 K 601 K

th V

V

∆ ∆

∆

To find volume V2,

V VnRT

p2 11

15

3

3 0 108 31 10= = = ( )( )( )

×= × −1.0 mol 8.31 J / mol K 300 K

Pa m3

..

The pressure p2 can be obtained from the isochoric condition as follows:

p

T

p

Tp

T

Tp2

2

1

12

2

11

5 53 00 10 6 01 10= ⇒ = =

×( ) = ×601 K

300 K Pa Pa. .

With the above values of p2, V2 and T2, we can now obtain p3, V3 and T3. We have

V V p p

T

V

T

VT

V

VT T

3 22

3 25

3

3

2

23

3

22 2

2 1 662 10 6 01 10

2

= = × = = ×

= ⇒ = = =

−. . m Pa

1202 K

3

For the isobaric process 2 → 3,

Q nC T R T T

W p V V

E Q W

2 352 3 2

52

3 3 25 3

2 3

6 01 10 8 31 10

→

→−

→ →

= = ( )( ) −( ) = ( )( )( )( ) =

= −( ) = ×( ) ×( ) =

= − = − =

P

S 2 33

th S 2 3

1.0 mol 1.0 mol 8.31 J / mol K 601 K 12,480 J

Pa m 4990 J

12,480 J 4990 J 7490 J

∆

∆

. .

We are now able to obtain p4, V4 and T4. We have

V V p p

T

p

T

pT

p

pT

4 32

4 15

4

4

3

34

4

33

5

5

1 662 10 3 00 10

3 00 10

6 01 10

= = × = = ×

= ⇒ = = ××

( ) =

−. .

.

.

m Pa

Pa

Pa1202 K 600 K

3

For isochoric process 3 → 4,

Q nC T R T T

W E Q W

3 432 4 3

32

3 40

→

→ → →

= = ( )( ) −( ) = ( )( )( ) −( ) = −

= ⇒ = − = −V

S 3 4 th S 3 4

1.0 mol 1.0 mol 8.31 J / mol K 602 7500 J

J 7500 J

∆

∆

For isobaric process 4 → 1,

Q nC T

W p V V

E Q W

4 152

4 1 45 3 2

4 1

3 00 10 8 31 10 1 662 10

→

→− −

→ →

= = ( ) ( ) −( ) = −

= −( ) = ×( ) × × − ×( ) = −

= − = − − −( ) = −

P

S 4 13 3

th S 4 1

1.0 mol 8.31 J / mol K 300 K 600 K 6230 J

Pa m m 2490 J

6230 J 2490 J 3740 J

∆

∆

. . .

WS (J) Q (J) ∆Eth (J)

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

η = =+

=+

= =→ →

W

Q

W

Q Qout

H

out 2500 J

3750 J 12,480 J1 2 2 3

0 154 15 4. . %

Assess: For a closed cycle, as expected, (Ws)net = Qnet and (∆Eth)net = 0 J

19.53. Model: The heat engine follows a closed cycle. For a diatomic gas, C RV = 52 and C RP = 7

2 .Visualize: Please refer to Figure P19.53.Solve: (a) Since T1 = 293 K, the number of moles of the gas is

np V

RT= =

× ×( ) ×( )( )( )

= ×−

−1 1

1

5 60 5 1 013 10 10 10. . Pa m

8.31 J / mol K 293 K2.08 10 mol

3

4

At point 2, V V p p2 1 2 14 3= = and . The temperature is calculated as follows:

p V

T

p V

TT

p

p

V

VT1 1

1

2 2

22

2

1

2

11 3 4 293 3516= ⇒ = = ( )( )( ) = K K

At point 3, V V V p p3 2 1 3 14= = = and . The temperature is calculated as before:

Tp

p

V

VT3

3

1

3

11 1 4 293 1172= = ( )( )( ) = K K

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

Ws3 3 3 3

3

atm cm cm atm atm cm cm

m atm Pa

1 atm J

= ( ) −( ) + −( ) −( )= ×( )( ) ×

=−

0 5 40 10 1 5 0 5 40 10

30 10 11 013 10

3 04

12

65

. . .

..

The change in the thermal energy is

∆ ∆E nC Tth V mol 8.31 J / mol K 3516 K 293 K 13.93 J= = ×( ) ( ) −( ) =−2 08 10 4 52.

The heat is Q W E= + =s th 16.97 J∆ . For process 2 → 3, the work done is Ws = 0 J and

Q E nC T n R T T= = = ( ) −( )= ×( ) ( ) −( ) = −−

∆ ∆th V

mol 8.31 J / mol K 1172 K 3516 K J

52 3 2

4 522 08 10 10 13. .

For process 3 → 1,

W

E nC T

s3 3 3

th V

atm 10 cm 40 cm Pa m 1.52 J

mol 8.31 J / mol K 293 K 1172 K J

= ( ) −( ) = × ×( ) − ×( ) = −

= = ×( ) ( ) −( ) = −

−

−

0 5 0 5 1 013 10 30 10

2 08 10 3 80

5 6

4 52

. . .

. .∆ ∆

The heat is Q E W= + = −∆ th s 5.32 J .

Ws (J) Q (J) ∆Eth

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

η = = = =W

Qnet

H

1.52 J

16.97 J0 0896 8 96. . %

(c) The power output of the engine is

500 revolutions

min

1 min

60 s revolution1.52 J / s 12.7 Wnet× × = × =W 500

60Assess: For a closed cycle, as expected, (Ws)net = Qnet and (∆Eth)net = 0 J.

19.54. Model: For the closed cycle, process 1 → 2 is isothermal, process 2 → 3 is isobaric, and process 3 → 1is 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 WS and Qfor each of the three processes that make up this cycle. Using the ideal-gas equation the number of moles of the gas is

np V

RT= =

×( ) ×( )( )( )

=−

1 1

1

5 61 013 10 600 10

3000 0244

..

Pa m

8.31 J / mol K K mol

3

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

CR

R C C R RV P V =−

= = + =1

4 5γ

At point 2, process 1 → 2 is isothermal, so we can find the pressure p2 as follows:

p V p V pV

Vp p p1 1 2 2 2

1

21

4

4 1 1

6 00 10

2 00 103= ⇒ = = ×

×= = = ×

−

−

.

.

m

m3 atm 3.039 10 Pa

3

35

At point 3, process 2 → 3 is isobaric, so we can find the temperature T3 as follows:

V

T

V

TT

V

VT T T2

2

3

33

3

22

4

4 2 2

6 00 10

2 00 103= ⇒ = = ×

×= =

−

−

.

.

m

m900 K

3

3

Point p V T

1 1.0 atm = 1.013 × 105 Pa 6.00 × 10−4 m3 300 K

2 3.0 atm = 3.039 × 105 Pa 2.00 × 10−4 m3 300 K

3 3.0 atm = 3.039 × 105 Pa 6.00 × 10−4 m3 900 K

Process 1 → 2 is isothermal:

W p V V VS 66.8 J( ) = ( ) = −12 1 1 2 1ln Q W12 12

= ( ) = −S 66.8 J

Process 2 → 3 is isobaric:

W p V p V VS 121.6 J( ) = = −( ) =23 2 2 3 2∆ Q nC T nC T T23 3 2= = −( ) =P P 608.3 J∆

Process 3 → 1 is isochoric:

WS( ) =31

0 J Q nC T nC T T31 1 3= = −( ) = −V V 486.7 J∆

We find that

WS cycle66.8 J 121.6 J 0 J 54.8 J( ) = − + + = Qcycle 66.8 J 608.3 J 486.7 J 54.8 J= − + − =

These are equal, as they should be. Knowing that the work done is Wout = (WS)cycle = 54.8 J/cycle, an engine operating at20 cycles/s has power output

Pout

J

cycle

20 cycle

s

J

s1096 W 1.096 kW= × = = =54 8

1096.

(b) Only Q23 is positive, so Qin = Q23 = 608 J. Thus, the thermal efficiency is

η = = = =W

Qout

in

54.8 J

608.3 J0 090 9 0. . %

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

2 and C RP = 52 for a monatomic gas.

Visualize: Please refer to Figure P19.55.Solve: (a) Using the ideal-gas equation, the number of moles is

np V

RT= =

×( ) ×( )=

−1 1

1

5 61 0 10 100 100 00401

..

Pa m

(8.31 J / mol K)(300 K) mol

3

We can use the adiabat 2 → 3 to calculate p2 as follows:

p V p V p pV

Vp

V

V3 3 2 2 2 33

21

3

2

5

5 3

61 0 10 1 981 10γ γγ γ

= ⇒ =

=

= ×( )

= ×. ./

Pa600 cm

100 cmPa

3

3

T2 can be determined from the ideal-gas equation and is

Tp V

nR22 2

6 61 981 10 100 10

8 31 0 00401= =

×( ) ×( )=

−.

( . )( . )

Pa m

J / mol K mol5943 K

3

At point 3, p3 = p1 and

T

V

T

VT

V

VT3

3

1

13

3

11

6

6

600 10

100 10300= ⇒ = = ×

×( ) =

−

−

m

m K 1800 K

3

3

Now we can calculate Ws, Q, and ∆Eth for the three processes involved in the cycle. For process 1 → 2, Ws 1→2 = 0 Jand

∆E Q nC T T n R T Tth V= = −( ) = ( ) −( )→1 2 2 132 2 1 = 282.2 J

For process 2 → 3, Q2→3 = 0 J and

∆E nC T T n R T Tth V= −( ) = ( ) −( )3 232 3 2 = –207.2 J

Because, ∆Eth = W + Q, W = −Ws, so Ws 2→3 = −∆Eth = +207.2 J for process 2 → 3. For process 3 → 1,

∆E nC T T n R T Tth V= −( ) = ( ) −( ) =1 332 1 3 −75.0 J

Q nC T T n R T T3 1 1 352 1 3→ = −( ) = ( ) −( ) = −P 125.0 J

The work done Ws 3→1 is the area under the p-versus-V graph. We have

Ws 3 13 6 3 Pa m 600 10 m J→

− −= ×( ) × − ×( ) = −100 10 100 10 50 03 6 .

Ws (J) Q (J) ∆Eth (J)

1 → 2 0 282.2 282.22 → 3 207.2 0 −207.23 → 1 −50.0 −125.0 −75.0

Net 157.2 157.2 0

(b) The thermal efficiency of the engine is

η = = = =W

Qout

H

157.2 J

282.2 J0 522 52 2. . %

Assess: As expected for a closed cycle, (Ws)net = Qnet and (∆Eth)net = 0 J.

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 C RV = 5

2 and γ = 75 .

Visualize: Please refer to Figure P19.56.Solve: (a) The pressure p2 lies on the adiabat from 2 → 3. We can find the pressure as follows:

p V p V p pV

V2 2 3 3 2 33

2

5

7 5

51 0 10 6 964 10γ γγ

= ⇒ =

= ×( )

= ×. ./

Pa4000 cm

1000 cm Pa

3

3

The temperature T2 can be obtained from the ideal-gas equation relating points 1 and 2:

p V

T

p V

TT T

p

p

V

V1 1

1

2 2

22 1

2

1

2

1

5

5

6 964 10

4 0 101= ⇒ = = ( ) ×

×

( ) =300 K

Pa

Pa522.3 K

.

.

(b) The number of moles of the gas is

Rp V

RT= =

×( ) ×( )=

−1 1

1

5 34 0 10 1 0 100 1604

. ..

Pa m

(8.31 J / mol K)(300 K) mol

3

For isochoric process 1 → 2 , Ws = 0 J and

Q E nC T n R T= = = ( ) =∆ ∆ ∆th V J52 741 1.

For adiabatic process 2 → 3, Q = 0 J and

∆ ∆E nC T n R T Tth V 741.1 J= = ( ) −( ) = −52 3 2

Using the first law of thermodynamics, ∆Eth = ∆Ws + Q, which means Ws = −∆Eth = +741.1 J. Ws can also bedetermined from

Wp V p V nR T T

s

J / K 300 K 522.3 K741.1 J= −

−=

−( )−

= ( ) −( )−( ) =3 3 2 2 3 2

43

251 1γ γ

For isothermal process 3 → 1, ∆Eth = 0 J and

W nRTV

Vs 554.5 J= = −11

3

ln

Using the first law of thermodynamics, ∆E W Qth s= − + , Q W= = −s 554.5 J .

∆Eth (J) Ws (J) Q (J)

1 → 2 741.1 0 741.12 → 3 −741.1 741.1 03 → 1 0 −554.5 −554.5

Net 0 186.6 186.6

(c) The work per cycle is 186.6 J and the thermal efficiency is

η = = = =W

Qs

H

186.6 J

741.1 J0 252 25 2. . %

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

2 and γ = 75 .

Visualize: Please refer to Figure P19.57.Solve: (a) The number of moles of gas is

np V

RT= =

×( ) ×( )=

−2 2

2

5 64 0 10 1000 101203

..

Pa m

(8.31 J / mol K)(400 K)0 mol

3

Using p V p V1 1 2 2γ γ= , and reading V1 from the graph,

p pV

V1 22

1

5 14

7 5 44 0 10 5 743 10=

= ×( )( ) = ×γ

. ./

Pa Pa

With p1, V1, and nR having been determined, we can find T1 using the ideal-gas equation:

Tp V

nR11 1

4 65 743 10 4000 10

8 31 0 1203= =

×( ) ×( )=

−.

( . )( . )

m

J / mol K mol229.7 K

3

(b) For adiabatic process 1 → 2, Q = 0 J and

Wp V p V

s

3 3 Pa m Pa m425.7 K= −

−=

×( ) ×( ) − ×( ) ×( )−

= −− −

2 2 1 1

5 3 4 3

1

4 0 10 1 00 10 5 743 10 4 00 10

1 1 4γ. . . .

.

Because ∆E W Qth s= − + , ∆E Wth s 425.7 K= − = . For isothermal process 2 → 3, ∆Eth J= 0 . From Equation 17.15,

W nRTV

Vs 554.5 J= =23

2

ln

From the first law of thermodynamics, Q W= =s 554.5 J for process 2 → 3. For isochoric process 3 → 1, Ws = 0 J and

Q E nC T n R T T= = = ( ) −( )∆ ∆th V52 1 3 = −425.7 J

∆Eth (J) Ws (J) Q (J)

1 → 2 425.7 −425.7 02 → 3 0 554.5 554.53 → 1 −425.7 0 −425.7

Net 0 128.8 128.8

(c) The work per cycle is 128.8 J and the thermal efficiency of the engine is

η = = = =W

Qs

H

128.8 J

554.5 J0 232 23 2. . %

Assess: As expected, for a closed cycle W Qs net net( ) = and ∆Eth net0( ) = J.

19.58. Model: In the Brayton cycle, process 1 → 2 and process 3 → 4 are adiabatic, and process 2 → 3 andprocess 4 → 2 are isobaric. We will assume the gas to be diatomic, with C RP and = =7

2 1 40γ . .Visualize: Please refer to Figure P19.58.Solve: The number of moles of gas is

np V

RT= =

×( )( )=1 1

1

51 013 101 27

..

Pa 2.0 m

(8.31 J / mol K)(300 K)8 mol

3

We can find the volume at 2 from the adiabatic equation p V p V1 1 2 2γ γ= ,

V Vp

p2 11

2

1 11 41

10=

= ( ) =

γ

2.0 m 0.3861 m3 3.

The temperature T2 is determined from the ideal-gas equation as follows:

p V

T

p V

TT T

p

p

V

V1 1

1

2 2

22 1

2

1

2

1

10= ⇒ = = ( )

=300 K atm

1 atm

0.3861 m

2.0 m579.2 K

2

3

To find T3 we use the heat input:

Q nC T n R T T

T T T TH P J 675.33 J / K

846.1 K 579.2 K 846.1 K 1425.3 K

= × = = = ( )⇒ = ⇒ = + = + =

2 0 106 72

72

3 2

. ∆ ∆ ∆∆ ∆

We are now able to find V3 using the ideal-gas law. We have

V

T

V

TV V

T

T3

3

2

23 2

3

2

= ⇒ = = ( ) =0.3861 m

1425.3 K

579.2 K0.9501 m3 3

The volume at 4 is found by again using the adiabatic equation p V p V3 3 4 4γ γ= ,

V Vp

p4 33

4

1

11 410=

= ( )( ) =γ

0.9501 m 4.921 m3 3.

T4 is now obtained from

Tp V

nR44 4

51 013 10 4 921= =

×( )( )=

. . Pa m

(81.27 mol)(8.31 J / mol K)738.2 K

3

To find the engine’s efficiency, we must first find the total work done per cycle which is

W W W W Wnet = + + +→ → → →1 2 2 3 3 4 4 1

The four contributions are

Wp V p V nR T T

W p V

WnR T T

1 22 2 1 1 2 1 5

2 3 26 5

3 44 3

1 1

8 31

1 1 44 714 10

1 013 10 5 713 10

1

→

→

→

= −−

=−( )

−= ( )( ) −( )

−= − ×

= = ×( ) −( ) = ×

=−( )

−= ( )

γ γ

γ

81.27 mol J / mol K 579.2 K 300 K J

Pa 0.9501 m 0.3861 m J

81.27 mol

3 3

.

..

. .∆

88 31

1 1 41 160 10

1 013 10 2 959 10

6

4 1 45 5

.

..

. .

J / mol K 738.2 K 1425.3 K J

Pa 2.0 m 4.921 m J3 3

( ) −( )−

= + ×

= = ×( ) −( ) = − ×→W p V∆

Thus, Wnet J= ×9 64 105. . The thermal efficiency is

η η= ⇒ = ××

= =W

Qnet

H

J

J

9 64 10

2 0 100 482 48 2

5

6

.

.. . %

As a comparison, we can calculate η from Equation 19.21 which is

η γγ

= −( )

= −( )

=−( )11

11

1048 21 0 4

1 4

rp

.

.

. %

where r p pp max min= = 10 .

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

2 and CP R.= 52

Visualize: Please refer to Figure P19.59.Solve: (a) At point 1: The pressure p1 = = ×1.0 atm 1.013 10 Pa5 and the volume V1

61000 10 1 0= × = ×− m3 .10 3− m3 . The number of moles is

n = =0.120 g

4 g / mol0.030 mol

Using the ideal-gas law,

Tp V

nR11 1

5 31 013 10 1 0 10= =

×( ) ×( )( )( )

=−. . Pa m

0.030 mol 8.31 J / mol K406 K

3

At point 2: The pressure p2 = = ×5.0 atm 5.06 10 Pa5 and V231 0 10= × −. m3. The temperature is

Tp V

nR22 2

5 35 06 10 1 0 10= =

×( ) ×( )( )( )

=−. . Pa m

0.030 mol 8.31 J / mol K2030 K

3

At point 3: The pressure is p3 = = ×1.0 atm 1.013 10 Pa5 and the temperature is T T3 2= = 2030 K . The volume is

V Vp

p3 22

3

3 31 0 105

5 0 10= = ×( ) = ×− −. . m

atm

1 atm m3 3

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

Q nC T R1 232→ = = ( )( ) −( )V 0.030 mol 2030 K 406 K∆ = 607 J

For isothermal process 2 → 3, ∆Eth 2 3 J→ = 0 and

Q W nRTV

V2 3 2 3 23

2

3

3

5 0 10

1 0 10815→ →

−

−= = = ( )( )( ) ××

=ln ln.

.0.030 mol 8.31 J / mol K 2030 K

m

m J

3

3

For isobaric process 3 → 1,

W p V3 1 35 3 31 013 10 1 0 10 5 0 10→

− −= = ×( ) × − ×( ) = −∆ . . . Pa m m 405 J3 3

Q nC T3 152→ = = ( )( )( ) −( )P 0.030 mol 8.31 J / mol K 406 K 2030 K∆ = −1012 J

The total work done is W W W Wnet = + +→ → →1 2 2 3 3 1 = 410 J. The total heat input is Q Q QH 1422 J= + =→ →1 2 2 3 . Theefficiency of the engine is

η = = =W

Qnet

H

410 J

1422 J28 8. %

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

ηmaxmin

max

406 K

2030 K= − = − =1 1 80

T

T%

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 amonatomic gas C RV = 3

2 and C RP = 52 .

Visualize: Please refer to Figure P19.60.Solve: (a) The three temperatures are

Tp V

nR11 1

5 34 0 10= =

×( )( )( )( )

=. Pa 0.025 m

2.0 mol 8.31 J / mol K601.7 K

Tp V

nR22 2

56 0 10= =

×( )( )( )( )

=. Pa 0.050 m

2.0 mol 8.31 J / mol K1805.1 K

3

Tp V

nR33 3

54 0 10= =

×( )( )( )( )

=. Pa 0.050 m

2.0 mol 8.31 J / mol K1203.4 K

3

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

W

E nC T R T T

s3 3 3 3

th V

Pa Pa 0.050 m 0.025 m Pa 0.050 m 0.025 m

J

2.0 mol

2.0 mol 8.31 J / mol K 1805.1 K 601.7 K J

= × − ×( ) −( ) + ×( ) −( )= ×

= = ( )( ) −( )= ( )( )( ) −( ) = ×

12

5 5 5

4

32 2 1

32

4

6 0 10 4 0 10 4 0 10

1 25 10

3 00 10

. . .

.

.

∆ ∆

The heat input is Q W E= + = ×s th J∆ 4 25 104. . For isochoric process 2 → 3, Ws = 0 J and

Q E nC T= = = ( ) ( ) −( ) = − ×∆ ∆th V 2.0 mol 8.31 J / mol K 1203.4 K 1805.1 K J32

41 50 10.

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

Ws3 3 Pa 0.025 m 0.050 m J= ×( ) −( ) = − ×4 0 10 1 0 105 4. .

∆ ∆E nC T n R T Tth V= = ( ) −( )32 1 3 = ( ) ( ) −( ) = − ×2.0 mol 8.31 J / mol K 601.7 K 1203.4 K J3

241 5 10.

The heat input is Q W E= + = − ×S th J∆ 2 50 104. .

∆Eth (J) Ws (J) Q (J)

1 → 2 3.0 × 104 1.25 × 104 4.25 × 104

2 → 3 −1.5 × 104 0 −1.50 × 104

3 → 1 −1.5 × 104 −1.0 ×104 −2.50 × 104

Net 0 2.5 × 103 2.5 × 103

(c) The thermal efficiency is

η = = ××

=W

Qnet

H

J

J

2 5 10

4 25 105 88

3

4

.

.. %

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, C R C RV P, , and = 1.4. = = =5

272

75γ

Visualize: Please refer to Figure P19.61.Solve: (a) We can find the temperature T2 from the ideal-gas equation as follows:

Tp V

nR22 2

5 34 0 10 1 0 10= =

×( ) ×( )( )( )

=−. . Pa m

0.020 mol 8.31 J / mol K2407 K

3

We can use the equation p V p V2 2 1 1γ γ= to find V1,

V Vp

p1 22

1

1

35

5

1 1 4

31 0 104 0 10

1 0 102 692 10=

= ×( ) ××

= ×− −

/ / .

..

..

γ

m Pa

Pa m3 3

The ideal-gas equation can now be used to find T1,

Tp V

nR11 1

5 31 0 10 2 692 10= =

×( ) ×( )( )( )

=−. . Pa m

0.020 mol 8.31 J / mol K1620 K

3

At point 3, V V3 1= so we have

Tp V

nR33 3

5 34 10 2 692 10

8 31= =

×( ) ×( )( )( )

=− Pa m

0.020 mol J / mol K6479 K

3.

.

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

Wp V p V nR T T

s

0.020 mol 8.31 J / mol K 2407 K 1620 K J= −

−=

−( )−

= ( )( ) −( )−( )

= −2 2 1 1 2 1

1 1 1 1 4327 0

γ γ ..

For isobaric process 2 → 3,

Q nC T n R T= = ( )( )P∆ ∆72 = ( ) ( ) −( )0.020 mol 8.31 J / mol K 6479 K 2407 K7

2 = 2369 J

∆ ∆ ∆E nC T n R Tth V 1692 J= = ( ) =52

The work done is the area under the p-versus-V graph. Hence,

Ws3 3 Pa m m J= ×( ) × − ×( ) =− −4 0 10 2 692 10 1 0 10 6775 3 3. . .

For isochoric process 3 → 1, Ws = 0 J and

∆ ∆E Q nC Tth V 0.020 mol 8.31 J / mol K 1620 K 6479 K J= = = ( )( )( ) −( ) = −52 2019

∆Eth (J) Ws (J) Q (J)

1 → 2 327 −327 02 → 3 1692 677 23693 → 1 −2019 0 −2019

Net 0 350 350

(c) The engine’s thermal efficiency is

η = = =350 J

2369 J0 148 14 8. . %

19.62. Model: The closed cycle of the heat engine involves the following processes: one isothermal compression,one isobaric expansion, and one isochoric cooling.Visualize:

Solve: The number of moles of helium is

n = =2.0 g

4.0 g / mol0.50 mol

The total work done by the engine per cycle is W W W Wnet = + +→ → →1 2 2 3 3 1. For the isothermal process,

W nRTV

V1 2 12

1→ = ln = ( )( )( ) ( ) = −0.50 mol 8.31 J / mol K 273 K Jln . .0 5 786 2

For the isobaric process,

W p V pV

p V nRT2 3 2 11

1 1 122→ = ( ) = ( )

= =∆ = ( )( )( ) =0.50 mol 8.31 J / mol K 273 K 1134.3 J

For the isochoric process, W3→1 = 0 J. Thus, Wnet = 348.1 J.For calculating heat transfers, note that T T2 1 273= = K and, since process 3 1→ is isochoric, T3 =

( )2 5461 atm / 1 atm K.T = For the isothermal process, Q W1 2 1 2→ →= = −786.2 J because ∆Eth 1 2→ = 0 J. For theisobaric process,

Q nC T n R T T2 352 3 2

52→ = = ( ) −( ) = ( )( )( ) =P 0.50 mol 8.31 J / mol K 273 K 2835.8 J∆

For the isochoric process,

Q nC T n R T T3 132 1 3

32 1701 5→ = = ( ) −( ) = ( )( )( )( ) = −v 0.50 mol 8.31 J / mol K –273 K J∆ .

From Q1→2, Q2→3, and Q3→1, we see that QH = 2835.8 J. Thus, the thermal efficiency is

η = = =348.1 J

2835.8 J0 123 12 3. . %

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

2 .

Visualize:

Solve: Using the ideal-gas law,

pnRT

V11

1

54 986 10= = ( )( )( )×

= ×−

0.20 mol 8.31 J / mol K 600 K

2.0 10 m Pa3 3 .

At point 2, because of the isothermal conditions, T T2 1 600= = K and

p pV

V2 11

2

53

354 986 10

2 0 10

4 0 102 493 10= = ×( ) ×

×

= ×−

−..

.. Pa

m

m Pa

3

3

At point 3, because it is an isochoric process, V V3 2 4000= = cm and3

p pT

T3 23

2

5 52 493 10 1 247 10= = ×( ) = ×. . Pa

300 K

600 K Pa

Likewise at point 4, T T4 3 300= = K and

p pV

V4 33

4

53

51 247 104 0 10

2 493 10= = ×( ) ××

= ×−

−..

. Pa m

2.0 10 m Pa

3

3 3

Let us now calculate Wnet = W1→2 + W2→3 + W3→4 + W4→1. For the isothermal processes,

W nRTV

V1 2 12

1

2→ = = ( )( )( ) ( ) =ln ln0.20 mol 8.31 J / mol K 600 K 691.2 J

W nRTV

V3 4 34

3

12→ = = ( )( )( ) ( ) = −ln ln0.20 mol 8.31 J / mol K 300 K 345.6 J

For the isochoric processes, W2→3 = W4→1 = 0 J. Thus, Wnet = 345.6 J.Because Q = Ws + ∆Eth,

Q W E1 2 1 2 1 2→ → →= + ( )∆ th = 691.2 J + 0 J = 691.2 J

For the first isochoric process,

Q nC T R T T2 332 3 2

320 20

→ = = ( )( ) −( )= ( ) ( ) −( ) = −

V 0.20 mol

mol 8.31 J / mol K 300 K 600 K 747.9 K

∆

.

For the second isothermal process

Q W E3 4 3 4 3 4→ → →= + ( )∆ th = −345.6 J + 0 J = −345.6 J

For the second isochoric process,

Q nC T n R T T4 132 1 4

32

→ = = ( ) −( )= ( )( )( ) −( ) =

V

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

∆

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

η = = = =W

Qnet

H

345.6 J

1439.1 J0 240 24 0. . %

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

Solve: (a) Except in an adiabatic process, heat must be transferred into the gas to raise its temperature. Thus heatis 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 hotreservoir, so the heat transferred is QH. Similarly, heat energy is transferred out of the gas during processes 2 → 1and 1 → 4. This requires that the reservoir temperature be T < 300 K. This is the cold reservoir, and the energytransferred during these two processes is QC.(c) The heat energies were calculated in Example 19.2, but now they have the opposite signs.

QH = Q43 + Q32 = 7.09 × 105 J + 15.19 × 105 J = 22.28 × 105 J

QC = Q21 + Q14 = 21.27 × 105 J + 5.06 × 105 J = 26.33 × 105 J

(d) For a counterclockwise cycle in the pV diagram, the work is Win. Its value is the area inside the curve, which is Win =(∆p)(∆V) = (2 × 101,300 Pa)(2 m3) = 4.05 × 105 J. Note that Win = QC – QH, as expected from energy conservation.(e) Since QC > QH, more heat is exhausted to the cold reservoir than is extracted from the hot reservoir. In thisdevice, work is used to transfer energy “downhill,” from hot to cold. The exhaust energy is QC = QH + Win > QH.This is the energy-transfer diagram of Figure 19.19.(f) No. A refrigerator uses work input to transfer heat energy from the cold reservoir to the hot reservoir. Thisdevice 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 coldreservoir at 0°C, what temperature (in °C) must the hot reservoir be?(b)

0 80 10

273

273

2730 20. .= −

° +( )+( ) ⇒

+= ⇒ = °

C 2731092 C

H HHT T

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 4 0 4. = ⇒ =Q W Q WC in C in . This means

QH = QC + Win = 4Win + Win = 5Win ⇒ = = =WQ

inH

5

100 J

520 J

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

19.67. Solve: (a) A heat engine operates at 20% efficiency and produces 20 J of work in each cycle. What isthe net heat extracted from the hot reservoir and the net heat exhausted in each cycle?(b) We have 0 20 1. = − Q QC H . Using the first law of thermodynamics,

W Q Q Q Qout H C C H20 J 20 J= − = ⇒ = −Substituting into the definition of efficiency,

0 20 120

1 120 20

. = − − = − + =Q

Q Q QH

H H H

J J J ⇒ = =QH

J J

20

0 20100

.

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

19.68. Solve: (a)

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

1

21 0 10 2 05pmax . .− ×( )( ) = × Pa m 4.0 10 J3 5 ⇒ = × =pmax .5 0 105 Pa 500 kPa

19.69. Model: The heat engine follows a closed cycle, starting and ending in the original state.Visualize: Please refer to Figure CP19.69. The figure indicates the following seven steps. First, the pin is insertedwhen the heat engine has the initial conditions. Second, heat is turned on and the pressure increases at constant volumefrom 1 atm to 3 atm. Third, the pin is removed. The flame continues to heat the gas and the volume increases atconstant pressure from 50 cm3 to 100 cm3. Fourth, the pin is inserted and some of the weights are removed. Fifth, thecontainer is placed on ice and the gas cools at constant volume to a pressure of 1 atm. Sixth, with the container still onice, the pin is removed. The gas continues to cool at constant pressure to a volume of 50 cm3. Seventh, with no ice orflame, the pin is inserted back in and the weights returned bringing the engine back to the initial conditions and readyto start over.Solve: (a)

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

Wout = (∆p)(∆V) = (2 × 101,300 Pa)(50 × 10–6 m3) = 10.13 J

(c) Heat energy is input during processes 1 → 2 and 2 → 3, so QH = Q12 + Q23. This is a diatomic gas, with CV = 32 R

and CP = 52 R. The number of moles of gas is

np V

RT= = × =

−1 1

1

6101 300

8 310 00208

( , )

( ..

Pa)(50 10 m

J / mol K)(293 K) mol

3

Process 1 → 2 is isochoric, so T2 = (p2/p1)T1 = 3T1 = 879 K. Process 2 → 3 is isobaric, so T3 = (V3/V2)T2 = 2T2 =1758 K. Thus

Q nC T nR T T1252 2 1

52 0 00208= ∆ = − =V mol)(8.31 J / mol K)(586 K) = 25.32 J( ) ( .

Similarly,

Q nC T nR T T2372 3 2

72 0 00208= ∆ = − =P mol)(8.31 J / mol K)(879 K) = 53.18 J( ) ( .

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

η = = = =W

Qout

H

J

78.50 J

10 130 129 12 9

.. . %

19.70. Model: System 1 undergoes an isochoric process and system 2 undergoes an isobaric process.Visualize: Please refer to Figure CP19.70.Solve: (a) Heat will flow from system 1 to system 2 because system 1 is hotter. Because there is no heat inputfrom (or loss to) the outside world, we have Q1 + Q2 = 0 J. Heat Q1, which is negative, will change the temperatureof system 1. Heat Q2 will both change the temperature of system 2 and do work by lifting the piston. But theseconsequences of heat flow don’t change the fact that Q1 + Q2 = 0 J. System 1 undergoes constant volume coolingfrom T1i = 600 K to Tf. System 2, whose pressure is controlled by the weight of the piston, undergoes constantpressure heating from T2i = 300 K to Tf. Thus,

Q Q n C T T n C T T1 2 1 20+ = = −( ) + −( ) J V f 1i P f 2i = ( ) −( ) + ( ) −( )n R T T n R T T132 2

52f 1i f 2i

Solving this equation for Tf gives

Tn T n T

n nf1i 2i 0.060 mol 600 K 0.030 mol 300 K

3 0.060 mol 0.030 mol464 K= +

+= ( )( ) + ( )( )

( ) + ( )=3 5

3 5

3 5

51 2

1 2

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

Q n C T T n R T T2 2 252= −( ) = ( ) −( ) =P f 2i f 2i 102.0 J

(c) The change of thermal energy in system 2 is

∆ ∆E n C T n R T T Qth V f 2i 61.2 J= = ( ) −( ) = =2 232

35 2

According to the first law of thermodynamics, Q2 = Ws + ∆Eth. Thus, the work done by system 2 is Ws = Q − ∆Eth =102.0 J − 61.2 J = 40.8 J. The work is done to lift the weight w mg= by a height ∆y. So,

W w y mg ys = =∆ ∆ ⇒ = =( )( ) =∆y

W

mgs

2

40.8 J

2.0 kg 9.8 m / s2.08 m

(d) The fraction of heat converted to work is

W

Qs 40.8 J

102.0 J2

0 40 40= = =. %

19.71. Model: Process 1 → 2 and process 3 → 4 are adiabatic, and process 2 → 3 and process 4 → 1 areisochoric.Visualize: Please refer to Figure CP19.71.Solve: (a) For adiabatic process 1 → 2, Q12 = 0 J and

Wp V p V nR T T

122 2 1 1 2 1

1 1= −

−=

−( )−γ γ

For isochoric process 2 → 3, W23 = 0 J and Q nC T T23 3 2= −( )V . For adiabatic process 3 → 4, Q34 = 0 J and

Wp V p V nR T T

344 4 3 3 4 3

1 1= −

−=

−( )−γ γ

For isochoric process 4 → 1, W41 = 0 J and Q nC T T41 1 4= −( )V . The work done per cycle is

W W W W WnR T T nR T T

net J J= + + + =−( )

−+ +

−( )−

+12 23 34 412 1 4 3

10

10

γ γ=

−− + −( )nR

T T T T1 2 1 4 3γ

(b) The thermal efficiency of the heat engine is

η = = − = − = −−( )−( )

W

Q

Q

Q

Q

Q

nC T T

nC T Tout

H

C

H

V

V

1 1 141

23

4 1

3 2

The last step follows from the fact that T3 > T2 and T4 > T1. We will now simplify this expression further as follows:

pV pVV nRTV nRT V nRT V T TV

VT rγ γ γ γ γ

γγ= = ⇒ = ⇒ =

=− − − −−

−1 11 1

12 2

12 1

1

2

1

11

Similarly, T T r3 41= −γ . The equation for thermal efficiency now becomes

η γ γ= − −−− −1 4 1

41

11

T T

T r T r= − −1

11rγ

(c)

19.72. Model: For the Diesel cycle, process 1 → 2 is an adiabatic compression, process 2 → 3 is an isobaricexpansion, 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

rV

V

V

V= = = =max

min

1

2

105021

cm

50 cm

3

3

The number of moles of gas is

np V

RT= =

×( ) ×( )( ) ( )[ ]

−1 1

1

6

8 31

1.013 10 Pa 1050 10 m

J / mol K 25 + 273 K= 0.0430 mol

5 3

.

For an adiabatic process,

p V p V1 1 2 2γ γ= ⇒ =

= = × = = ×pV

Vp r p2

1

21 1

1 4021γ

γ . 1 atm 71.0 atm 7.19 10 Pa6

T V T V TV

VT r T1 1

12 2

12

1

2

1

11

10 4121 298γ γ

γγ− −

−−= ⇒ =

= = × =. K 1007 K

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

Q nC T TQ

nC= ⇒ =P

P

∆ ∆

The gas has a specific heat ratio γ = 1.40 = = =7 5 52

72/ , . thus and V PC R C R Knowing CP, we can calculate first

∆T = 800 K and then T3 = T2 + ∆T = 1807 K. Finally, for an isobaric process we have

V

T

V

TV

T

TV2

2

3

33

3

22

6 650 10 89 7 10= ⇒ = = ×( ) = ×− −1807 K

1007 K m m3 3.

Process 3 → 4 is an adiabatic expansion to V4 = V1. Thus,

p V p V pV

Vp

T V T V TV

VT

3 3 4 4 43

43

6

6

1 4

6

3 31

4 41

43

4

1

3

6

89 7 107 19 10

89 7 10

γ γγ

γ γγ

= ⇒ =

= ××

×( ) = × =

= ⇒ =

= ×

−

−

− −− −

..

.

. m

1050 10 m Pa 2.30 10 Pa 2.27 atm

m

3

35

3

10501050 10 m1807 K = 675 K3×

( )−6

0 4.

Point p V T

1 1.00 atm = 1.013 × 105 Pa 1050 × 10−6 m3 298 K = 25°C

2 71.0 atm = 7.19 × 106 Pa 50.0 × 10−6 m3 1007 K = 734°C

3 71.0 atm = 7.19 × 106 Pa 89.7 × 10−6 m3 1807 K = 1534°C

4 2.27 atm = 2.30 × 105 Pa 1050 × 10−6 m3 675 K = 402°C

(b) For adiabatic process 1 → 2,

Wp V p V

s 633 J( ) = −−

= −12

2 2 1 1

1 γ

For isobaric process 2 → 3,

W p V p V Vs 285 J( ) = = −( ) =23 2 2 3 2∆

For adiabatic process 3 → 4,

Wp V p V

s 1009 J( ) = −−

=34

4 4 3 3

1 γ

For isochoric process 4 → 1, Ws( ) =41

0 J. Thus,

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

(c) The efficiency is

η = = =W

Qout

H

661 J

1000 J66 1. %

(d) The power output of one cylinder is

661 J

cycle

2400 cycle

min

min

60 sec

J

s kW× × = =1

26 440 26 4, .

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