19.1. Model: The heat engine follows a closed cycle, starting and ending in the original state. The cycle consists 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
Wout =
1 2
(200 kPa )(100 × 10 −6 m 3 ) = 10 J
The heat energy transferred into the engine is QH = 120 J. Because Wout = QH − QC , the heat energy exhausted is
QC = QH − Wout = 120 J − 10 J = 110 J (b) The thermal efficiency of the engine is
η=
Wout 10 J = = 0.0833 120 J QH
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
Wout = ( 400 kPa − 100 kPa )(100 × 10 −6 m 3 ) = 30 J The heat energy leaving the engine is QC = 90 J + 25 J = 115 J . The heat input is calculated as follows:
Wout = QH − QC ⇒ QH = QC + Wout = 115 J + 30 J = 145 J (b) The thermal efficiency of the engine is
η=
Wout 30 J = = 0.207 145 J QH
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 is QC = 40 kJ . The thermal efficiency of the heat engine is
η = 1−
QC 40 kJ = 1− = 0.273 55 kJ QH
(b) The work done by the engine per cycle is
Wout = QH − QC = 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 Wout = QH − QC ,
QH = Wout + QC = 20 J + 30 J = 50 J Thus, the efficiency of the engine is
η = 1−
QC 30 J = 1− = 0.40 50 J QH
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: W 100 J η = out ⇒ 0.40 = ⇒ QH = 250 J QH QH (b) Because Wout = QH − QC , the heat exhausted is
QC = QH − Wout = 250 J − 100 J = 150 J
19.6. Solve: The coefficient of performance of the refrigerator is K=
QC QH − Win 50 J − 20 J = = = 1.50 20 J Win Win
19.7. Solve: (a) The heat extracted from the cold reservoir is calculated as follows: K=
QC Q ⇒ 4.0 = C ⇒ QC = 200 J 50 J Win
(b) The heat exhausted to the hot reservoir is QH = QC + Win = 200 J + 50 J = 250 J
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 1s kJ × = 12.5 s 40 cycles cycle
(b) The heat input per cycle is calculated as follows: W 12.5 kJ η = out ⇒ QH = = 62.5 kJ 0.20 QH The heat exhausted per cycle is
QC = QH − Win = 62.5 kJ − 12.5 kJ = 50 kJ
19.9. Solve: The amount of heat discharged per second is calculated as follows: η=
1 Wout Wout 1 = ⇒ QC = Wout − 1 = (900 MW ) − 1 = 1.913 × 10 9 W 0.32 QH QC + Wout η
That is, each second the electric power plant discharges 1.913 × 10 9 J of energy into the ocean. Since a typical American house needs 2.0 × 10 4 J of energy per second for heating, the number of houses that could be heated with the waste heat is 1.913 × 10 9 J 2.0 × 10 4 J = 96, 000 .
(
)(
)
19.10. Solution: The amount of heat removed from the water in cooling it down in 1 hour is QC = mwater cwater ∆T . The mass of the water is
mwater = ρ water Vwater = (1000 kg / m 3 )(1 L ) = (100 kg / m 3 )(10 −3 m 3 ) = 1.0 kg ⇒ QC = (1.0 kg)( 4190 J / kg K )(20°C − 5°C) = 6.285 × 10 4 J
The rate of heat removal from the refrigerator is
6.285 × 10 4 J = 17.46 J / s 3600 s The refrigerator does work W = 8.0 W = 8.0 J / s to remove this heat. Thus the performance coefficient of the refrigerator is 17.46 J / s K= = 2.18 8.0 J / s QC =
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 ∆Eth = Q − Ws and Ws = 0 J , Q increases for process A. Process B is isothermal, so T is constant and hence ∆Eth = 0 J. The work done Ws is positive because the gas expands. Because Q = Ws + ∆Eth , Q is positive for process B. Process C is adiabatic, so Q = 0 J. W s is positive because of the increase in volume. Since Q = 0 J = Ws + ∆Eth , ∆Eth is negative for process C. 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, Ws is negative. Because Q = Ws + ∆Eth , Q also decreases for process D. ∆Eth Ws Q + + A 0 + + B 0 − + C 0 − − − D
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 is isothermal, so ∆T = 0 and ∆Eth = 0 J. The gas is compressed, so Ws is negative. Q = Ws for an isothermal process, so Q 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 makes Q = Ws + ∆Eth positive. Ws Q ∆Eth + A – 0 B – 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 the triangle is
Wout =
1 2
(3 atm − 1 atm)(600 × 10 −6 m 3 − 200 × 10 −6 m 3 )
1.013 × 10 5 Pa −6 3 = 12 2 atm × ( 400 × 10 m ) = 40.5 J 1 atm
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 =
1 2
( pmax − 100 kPa )(800 cm 3 − 200 cm 3 ) ⇒ 600 ×( 10 −6) m 3 2 60 J
⇒ pmax = 3.0 × 10 5 Pa = 300 kPa
= pmax − 1.0 × 10 5 Pa
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
Wout =
1 2
(300 kPa − 100 kPa )(600 cm 3 − 300 cm 3 ) = 12 (200 × 10 3 Pa )(300 × 10 −6 m 3 ) = 30 J
The thermal efficiency is
η=
Wout 30 J = = 0.0952 225 J + 90 J QH
(b) Because Wout = QH − QC , the heat exhausted is QC = QH − Wout = 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
Wout =
1 2
(300 kPa − 100 kPa )(600 cm 3 − 200 cm 3 ) = 12 (200 × 10 3 Pa )(400 × 10 −6 m 3 ) = 40 J
The heat exhausted is QC = 180 J + 100 J = 280 J . The thermal efficiency of the engine is
η=
Wout Wout 40 J = = = 0.125 QH QC + Wout 280 J + 40 J
(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 ηB = 1 − rp( pmax/pmin. The specific heat ratio for a diatomic gas is 7 C R γ = P = 25 = 1.4 CV 2 R
1−γ ) γ
, where rp is the pressure ratio
Solving the above equation for rp, γ
−1.4
(1 − ηB ) = rp(1−γ )/γ ⇒ rp = (1 − ηB ) 1−γ = (1 − 0.60) 0.4
= 24.7
19.18. Model: The Brayton cycle involves two adiabatic processes and two isobaric processes. The adiabatic processes involve compression and expansion through the turbine. 1−γ / γ Solve: The thermal efficiency for the Brayton cycle is ηB = 1 − rp( ) , where γ = CP / CV and rp is the pressure ratio. For a diatomic gas γ = 1.4. For an adiabatic process,
p1V1γ = p2 V2γ ⇒ p2 / p1 = (V1 / V2 ) Because the volume is halved, V2 = 12 V1 and hence
rp = p2 / p1 = (2) = 21.4 = 2.639 γ
The efficiency is −0.4
ηB = 1 − (2.639) 1.4 = 0.242
γ
19.19. Model: The efficiency of a Carnot engine (ηCarnot) depends only on the temperatures of the hot and cold reservoirs. 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, QH = Wout + QC . For engine (a), QH = 50 J, QC = 20 J and Wout = 30 J, so the first law of thermodynamics is obeyed. For engine (b), QH = 10 J, QC = 7 J and Wout = 4 J, so the 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 T 300 K ηCarnot = 1 − C = 1 − = 0.50 600 K TH Engine (a) has
η = 1−
QC Wout 30 J = = = 0.60 50 J QH QH
This is larger than ηCarnot, thus violating the second law of thermodynamics. For engine (b),
η=
Wout 4J = = 0.40 < ηCarnot 10 J QH
so the second law is obeyed. Engine (c) has a thermal efficiency that is
η= so the second law of thermodynamics is obeyed.
10 J = 0.333 < ηCarnot 30 J
19.20. Model: For a refrigerator QH = QC + Win, and the coefficient of performance and the Carnot coefficient of performance are K=
QC Win
K Carnot =
TC TH − TC
Visualize: Please refer to Figure Ex19.20. Solve: (a) For refrigerator (a) QH = QC + Win (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
K Carnot =
TC 300 K = =3 TH − TC 400 K − 300 K
For refrigerator (a),
K=
QC 40 J = = 2 < K Carnot Win 20 J
so the second law of thermodynamics is obeyed. For refrigerator (b),
K=
QC 40 J = = 4 > K Carnot Win 10 J
so the second law of thermodynamics is violated. For refrigerator (c),
K= so the second law is obeyed.
30 J = 1.5 < K Carnot 20 J
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
ηCarnot = 1 −
TC TC ⇒ 0.60 = 1 − ⇒ TC = 280 K = 7°C TH (427 + 273) K
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: Assume that the heat engine follows a closed cycle. Solve:
(a) The engine’s efficiency is
η=
Wout Wout 200 J = = = 0.25 = 25% QH QC + Wout 600 J + 200 J
(b) The thermal efficiency of a Carnot engine is ηCarnot = 1 − TC TH . For this to be 25%,
0.25 = 1 −
TC ⇒ TC = 504.8 K = 232°C (400 + 273) K
19.23. 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
η=
Wout Wout 10 J = = = 0.40 = 40% QH QC + Wout 15 J + 10 J
(b) The efficiency of a Carnot engine is ηCarnot = 1 − TC TH . The minimum temperature in the hot reservoir is found as follows: 293 K 0.40 = 1 − ⇒ TH = 488 K = 215°C TH 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.
19.24. Solve: (a) The efficiency of the Carnot engine is ηCarnot = 1 −
TC 300 K = 1− = 0.40 = 40% 500 K TH
(b) An engine with power output of 1000 W does Wout = 1000 J of work during each ∆t = 1 s. A Carnot engine has a heat input that is W 1000 J Qin = out = = 2500 J 0.40 ηCarnot during each ∆t = 1 s. The rate of heat input is 2500 J/s = 2500 W. (c) Wout = Qin − Qout , so the heat output during ∆t = 1 s is Qout = Qin − Wout = 1500 J . The rate of heat output is thus 1500 J/s = 1500 W.
19.25. Model: The efficiency of a Carnot engine depends only on the temperatures of the hot and cold reservoirs. Solve: The efficiency of a Carnot engine is ηCarnot = 1 − TC TH . This can be increased by either increasing the hotreservoir temperature or decreasing the cold-reservoir temperature. For a 40% Carnot efficiency and a coldreservoir temperature of 7°C, 280 K ηCarnot = 0.40 = 1 − ⇒ TH = 467 K TH A higher efficiency of 60% can be obtained by raising the hot-reservoir temperature to TH′ . Thus, 280 K ⇒ TH′ = 700 K TH′ The reservoir temperature must be increased by 233 K to cause the efficiency to increase from 40% to 60%.
ηCarnot = 0.60 = 1 −
19.26. Model: The maximum possible efficiency for a heat engine is provided by the Carnot engine. Solve:
The maximum efficiency is
ηmax = ηCarnot = 1 −
TC (273 + 20) K = 1− = 0.6644 TH (273 + 600) K
Because the heat engine is running at only 30% of the maximum efficiency, η = (0.30)ηmax = 0.1993 . The amount of heat that must be extracted is W 1000 J QH = out = = 5020 J 0.1993 η
19.27. 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
K Carnot =
TC (−20 + 273) K = 6.325 = TH − TC (20 + 273) K − ( −20 + 273) K
(b) The rate at which work is done on the refrigerator is found as follows:
K=
QC Q 200 J /s ⇒ Win = C = = 32 J /s = 32 W 6.325 Win K
(c) The heat exhausted to the hot side per second is
QH = QC + Win = 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 the cold and the hot reservoirs. Solve: For a Carnot refrigerator,
K Carnot =
TC Q 4.2 K 1.0 J = C ⇒ = ⇒ Win = 68.8 J 293 K − 4.2 K Win TH − TC Win
19.29. Model: Assume that the refrigerator is an ideal or Carnot refrigerator. Solve:
(a) For the refrigerator, the coefficient of performance is
K=
QC ⇒ QC = KWin = (5.0)(10 J ) = 50 J Win
The heat energy exhausted per cycle is
QH = QC + Win = 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 be obtained as follows:
K Carnot =
TC TC ⇒ 5.0 = ⇒ TC = 250 K = −23°C 300 K − TC TH − TC
19.30. Model: The coefficient of performance of a Carnot refrigerator depends only on the temperatures of the cold and hot reservoirs. Solve: Using the definition of the coefficient of performance for a Carnot refrigerator,
K Carnot =
TC 260 K (−13 + 273) K ⇒ 5.0 = = ⇒ TH = 312 K TH − TC TH − ( −13 + 273) K TH − 260 K
To increase the coefficient of performance to 10.0, the hot-reservoir temperature changes to TH′ . Thus,
K Carnot = 10.0 =
260 K ⇒ TH′ = 286 K TH′ − 260 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 change in the kinetic energy. Thus,
Wout = ∆K = 12 mv 2 =
1 2
(1500 kg)(15 m / s)2 = 168,750 J
Using the definition of thermal efficiency,
η=
Wout W 168,750 J ⇒ QH = out = = 8.44 × 10 5 J η 0.20 QH
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 T (273 + 20) K η = 0.40 ηCarnot = 0.401 − C = 0.401 − = 0.3484 TH (273 + 2000) K The amount of heat energy transferred is calculated as follows: W W 588,000 J η = out ⇒ QH = out = = 1.69 × 10 6 J 0.3484 QH η
19.33. Solve: The mass of the water is 10 −3 m 3 1000 kg × = 0.100 kg 1L m3 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 100 × 10 −3 L ×
Q1 = mc∆T = (0.100 kg)( 4186 J / kg K )(15 K ) = 6279 J
(
)
Q2 = mLf = (0.100 kg) 3.33 × 10 5 J / kg = 33,300 J Q3 = mc∆T = (0.100 kg)(2090 J / kg K )(15 K ) = 3135 J
⇒ QC = Q1 + Q2 + Q3 = 42,714 J Using the performance coefficient, Q 42,714 J 42,714 J K = C ⇒ 4.0 = ⇒ Win = = 10,679 J Win Win 4.0 The heat exhausted into the room is thus
QH = QC + Win = 42,714 J + 10,679 J = 5.34 × 10 4 J
19.34. Solve: An adiabatic process has Q = 0 and thus, from the first law, Ws = –∆Eth. For any ideal gas process, ∆Eth = nCV∆T, so Ws = –nCV∆T. We can use the ideal gas law to find
T=
p V − pi Vi pV ∆( pV ) ( pV ) f − ( pV ) i ⇒ ∆T = = = f f nR nR nR nR
Consequently, the work is
Ws = − nCV ∆T = − nCV
C pf Vf − pi Vi = − V ( pf Vf − pi Vi ) nR R
Because CP = CV + R, we can use the specific heat ratio γ to find
γ =
CP CV + R CV / R + 1 = = ⇒ CV CV CV / R
CV 1 = R γ −1
With this, the work done in an adiabatic process is
Ws = −
CV 1 ( pf Vf − pi Vi ) = − ( pf Vf − pi Vi ) = ( pf Vf − pi Vi ) / (1 − γ ) γ −1 R
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 net transfer of heat. Now suppose the refrigerator is a superefficient refrigerator with K > KCarnot. Since K = QC/Win, a superefficient refrigerator can pump more heat from the cold reservoir with the same work Win than can a Carnot refrigerator. 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 that heat is transferred from the cold reservoir to the hot reservoir without outside assistance. But such a process is forbidden by the second law of thermodynamics, so we must conclude that no refrigerator can have a larger coefficient 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 property of the Carnot cycle is that QC/QH = TC/TH. Consequently, the coefficient of performance of a Carnot refrigerator is K Carnot =
QC QC QC / QH T C / TH TC = = = = 1 − T C / TH Win QH − QC 1 − QC / QH TH − TC
19.37. Solve: We can also obtain the desired result using the η and K equations for the Carnot engine and the Carnot refrigerator:
K=
TC TC TH = TH − TC 1 − (TC TH )
η = 1−
Hence the equation for K becomes
K=
1−η 1−η = 1 − (1 − η) η
TC T ⇒ C = 1−η TH TH
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
T Wout = ηCarnot QH = 1 − C QH TH For the two Carnot engines, the first operates between TH and T3 and does work
T W1 = η1QH = 1 − 3 QH TH Similarly, the second heat engine, between T3 and TC, does work
T W2 = η2 QH′ = 1 − C QH′ T3 For any heat engine η = 1 – QC/QH, and for a Carnot heat engine ηCarnot = 1 – TC/TH. Thus a property of the Carnot cycle is that QC/QH = TC/TH and thus QC = (TC/TH)QH. Applying this result to the second engine,
QH′ =
T T3 T T QC = 3 C QH = 3 QH TC TC TH TH
With this result for QH′ , the work done by the second engine is
T T T T T W2 = 1 − C QH′ = 1 − C 3 QH = 3 − C QH T3 T3 TH TH TH Consequently, the combined work of the two engines is
T T T T W1 + W2 = 1 − 3 QH + 3 − C QH = 1 − C QH = Wout T T T T H H H H 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
ηmax = ηCarnot = 1 −
TC (273 + 5) K = 1− = 0.0825 = 8.25% TH (273 + 30) K
19.40. Model: The ideal gas in the Carnot engine follows a closed cycle in four steps. During the isothermal expansion at temperature TH, heat QH is transferred from the hot reservoir into the gas. During the isothermal compression 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 T W Wout 323 K ηCarnot = 1 − C = out ⇒ 1 − ⇒ Wout = 436 J = TH QH 573 K 1000 J Using QH = QC + Wout , we obtain
Qisothermal = QC = QH − Wout = 1000 J − 436 J = 564 J
19.41. Solve: Substituting into the formula for the efficiency of a Carnot engine, ηCarnot = 1 −
TC TC ⇒ 0.25 = 1 − ⇒ TC = 240 K = −33°C TH TC + 80 K
The hot-reservoir temperature is TH = TC + 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 decreasing TC or by increasing TH. First consider lowering the cold reservoir to TC – ∆T. This changes the efficiency to
η + ∆ηC = 1 −
TC − ∆T T ∆T ∆T = 1− C + =η+ TH TH TH TH
⇒ ∆ηC =
∆T TH
Now raise the hot reservoir by the same ∆T to TH + ∆T. This changes the efficiency to
η + ∆ηH = 1 − ⇒ ∆ηH =
TC T T TC T TC = 1− C + C − =η+ C − TH + ∆T TH TH TH + ∆T TH TH + ∆T
TC TC TC ∆T TC ∆T = − = TC × ∆ηC = TH TH + ∆T TH (TH + ∆T ) TH + ∆T TH TH + ∆T
Now TC < TH and TH + ∆T > TH, so the ratio TC/( TH + ∆T) < 1. Consequently, ∆ηH < ∆ηC. In other words, lowering the cold reservoir temperature has the larger effect.
19.43. Solve:
If QC =
2 3
QH, then Wout = QH – QC =
η=
1 3
QC. Thus the efficiency is
1 Wout Q 1 = 3 H = 3 QH QH
The efficiency of a Carnot engine is η = 1 – TC/TH. Thus T 1 1− C = ⇒ TH 3
TC 2 = TH 3
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,
QH = QC + Wout ⇒ Q1 = Q2 + Wout ⇒ Q2 = Q1 − Wout The thermal efficiency of a Carnot engine is
η = 1−
TC W 300 K = 1− = 0.50 = out 600 K TH Q1
⇒ Q2 = Q1 − ηQ1 = Q1 (1 − η) = (1000 J )(1 − 0.50) = 500 J To determine Q3 and Q4, we turn our attention to the Carnot refrigerator, which is driven by the output of the heat engine with Win = Wout. The coefficient of performance is
K=
TC Q Q Q 400 K = = 4.0 = C = 4 = 4 TH − TC 500 K − 400 K Win Wout ηQ1 ⇒ Q4 = KηQ1 = ( 4.0)(0.50)(1000 J ) = 2000 J
Using now the energy transfer equation Win + Q4 = Q3 , we have Q3 = Wout + Q4 = ηQ1 + Q4 = (0.50)(1000 J ) + 2000 J = 2500 J (b) From part (a) Q3 = 2500 J and Q1 = 1000 J , so Q3 > Q1 . (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 indoors with its exhaust heat. Solve: (a) The coefficient of performance for this heat pump is K = 5.0 = QC Win , where QC is the amount of heat 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 QC = 5.0Win ⇒ QH = 5.0Win + Win = 6.0Win If the heat pump is to deliver 15 kJ of heat per second to the house, then 15 kJ QH = 15 kJ = 6.0Win ⇒ Win = = 2.5 kJ 6.0 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 hrs × × = $270 s 1 hr 40 MJ The monthly heating cost in the house using a heat pump is 2.5 kJ 3600 s 1$ × 200 hrs × × = $45 s 1 hr 40 MJ
19.46. Solve: The maximum possible efficiency of the heat engine is ηmax = 1 −
TC 300 K = 1− = 0.40 500 K TH
The efficiency of the engine designed by the first student is W 110 J η1 = out = = 0.44 250 J QH 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 J 250 J = 0.36 < ηmax in agreement with the second law of thermodynamics. Applying the first law of thermodynamics,
QH = QC + Wout ⇒ 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 the first law of thermodynamics because QH = QC + Wout = 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 × 10 5 kg m3 × × 24 hour = 4800 m 3 1 hour 1500 kg The height of the room will be 48 m. (b) The thermal efficiency of the power plant is
η=
Wout 7.50 × 10 8 J / s 7.50 × 10 8 J = = = 0.321 = 32.1% 3 × 10 5 kg 28 × 10 6 J 1 hr 2.333 × 10 9 J QH × × 1 hr kg 3600 s
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
ηmax = 1 −
TC 303 K = 1− = 0.471 = 47.1% 573 K TH
(b) The plant’s actual efficiency is
η=
Wout 700 × 10 6 J / s = 0.350 = 35.0% 2000 × 10 6 J / s QH
(c) Because QH = QC + Wout,
QC = QH − Wout ⇒ QC = 2.0 × 10 9 J / s − 0.7 × 10 9 J / s = 1.3 × 10 9 J / s The mass of water that flows per second through the condenser is
m = 1.2 × 10 8
L 1 hr 10 −3 m 3 1000 kg × × × = 3.333 × 10 4 kg hour 3600 s 1L m3
The change in the temperature as QC = 1.3 × 10 9 J of heat is transferred to m = 3.333 × 10 4 kg of water is calculated as follows:
QC = mc∆T ⇒ 1.3 × 10 9 J = (3.333 × 10 4 kg)( 4186 J / kg K )∆T ⇒ ∆T = 9°C The exit temperature is 18°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 × 10 8
L 1 hr 10 −3 m 3 1000 kg × × × = 2.778 × 10 4 kg / s hr 3600 s 1L m3
The amount of heat transferred per second to the cooling water is thus
QC = mc∆T = (2.778 × 10 4 kg / s)( 4186 J / kg K )(27°C − 16°C) = 1.279 × 10 9 J / s The amount of heat per second into the power plant is
QH = Wout + QC = 0.750 × 10 9 J / s + 1.279 × 10 9 J / s = 2.029 × 10 9 J / s Finally, the power plant’s thermal efficiency is W 0.750 × 10 9 J / s η = out = = 0.37 = 37% 2.029 × 10 9 J / s QH
19.50. Solve: (a) The energy supplied in one day is Wout = 1.0 × 10 9
J 3600 s 24 hr × × = 8.64 × 1013 J s 1 hr 1 day
(b) The volume of water is V = 1 km 3 = 10 9 m 3 . The amount of energy is 1000 kg QH = mc∆T = 10 9 m 3 (4190 J / kg K)(1.0 K) = 4.19 × 1015 J m3 (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 surface 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:
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 figure above shows an example. The work done by the gas during one cycle is the area enclosed within the curve. We have
(Ws )cycle = 12 ∆p∆V We know that
∆p = pmax − pmin = 3 atm − 1 atm = 2 atm = 202,600 Pa ⇒ ∆V =
2(WS ) cycle ∆p
=
2( 40.5 J ) = 4.00 × 10 −4 m 3 = 400 cm 3 202,600 Pa
The maximum volume is
Vmax = Vmin + ∆V = 500 cm 3 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 isochoric
and process 2 → 3 and process 4 → 1 being isobaric. For a monatomic gas, CV = 23 R and CP = 25 R . Visualize: Please refer to Figure P19.52. Solve: (a) The first law of thermodynamics is Q = ∆Eth + WS . For the isochoric process 1 → 2, WS 1 → 2 = 0 J. Thus,
Q1→2 = 3750 J = ∆Eth = nCV ∆T ⇒ ∆T =
3750 J 3750 J 3750 J = = = 301 K nCV (1.0 mol)( 23 R) (1.0 mol)( 23 )(8.31 J / mol K) ⇒ T2 − T1 = 300.8 K ⇒ T2 = 300.8 K + 300 K = 601 K
To find volume V2,
V2 = V1 =
nRT1 (1.0 mol)(8.31 J / mol K )(300 K ) = = 8.31 × 10 −3 m 3 3.0 × 10 5 Pa p1
The pressure p2 can be obtained from the isochoric condition as follows:
p2 p T 601 K = 1 ⇒ p2 = 2 p1 = 3.00 × 10 5 Pa = 6.01 × 10 5 Pa 300 K T2 T1 T1
(
)
With the above values of p2, V2 and T2, we can now obtain p3, V3 and T3. We have
V3 = 2V2 = 1.662 × 10 −2 m 3
p3 = p2 = 6.01 × 10 5 Pa
T3 T2 V = ⇒ T3 = 3 T2 = 2T2 = 1202 K V3 V2 V2 For the isobaric process 2 → 3,
Q2→3 = nCP ∆T = (1.0 mol)( 25 R)(T3 − T2 ) = (1.0 mol)( 25 )(8.31 J / mol K )(601 K ) = 12,480 J
(
)
WS 2→3 = p3 (V3 − V2 ) = 6.01 × 10 5 Pa (8.31 × 10 −3 m 3 ) = 4990 J ∆Eth = Q2→3 − WS 2→3 = 12,480 J − 4990 J = 7490 J
We are now able to obtain p4, V4 and T4. We have
V4 = V3 = 1.662 × 10 −2 m 3
p4 = p1 = 3.00 × 10 5 Pa
3.00 × 10 5 Pa T T4 p = 3 ⇒ T4 = 4 T3 = (1202 K ) = 600 K p4 p3 p3 6.01 × 10 5 Pa For isochoric process 3 → 4,
Q3→4 = nCV ∆T = (1.0 mol)( 23 R)(T4 − T3 ) = (1.0 mol)( 23 )(8.31 J / mol K )( −602) = −7500 J WS 3→4 = 0 J ⇒ ∆Eth = Q3→4 − WS 3→4 = −7500 J For isobaric process 4 → 1,
Q4→1 = nCP ∆T = (1.0 mol) 25 (8.31 J / mol K )(300 K − 600 K ) = −6230 J
(
)
WS 4→1 = p4 (V1 − V4 ) = 3.00 × 10 5 Pa × (8.31 × 10 −3 m 3 − 1.662 × 10 −2 m 3 ) = −2490 J ∆Eth = Q4→1 − WS 4→1 = −6230 J − ( −2490 J ) = −3740 J WS (J)
Q (J)
∆Eth (J)
1→2
0
3750
2→3
4990
12,480
7490
3→4
0
–7500
–7500
4→1
–2490
–6230
–3740
2500
2500
0
Net
3750
(b) The thermal efficiency of this heat engine is
η=
Wout Wout 2500 J = = = 0.154 = 15.4% QH Q1→2 + Q2→3 3750 J + 12,480 J
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, CV = 25 R and CP = 27 R . Visualize: Please refer to Figure P19.53. Solve: (a) Since T1 = 293 K, the number of moles of the gas is
n=
(
)
0.5 × 1.013 × 10 5 Pa (10 × 10 −6 m 3 ) p1V1 = = 2.08 × 10 −4 mol RT1 (8.31 J / mol K)(293 K)
At point 2, V2 = 4V1 and p2 = 3 p1 . The temperature is calculated as follows:
p1V1 p2 V2 p V = ⇒ T2 = 2 2 T1 = (3)( 4)(293 K ) = 3516 K T1 T2 p1 V1 At point 3, V3 = V2 = 4V1 and p3 = p1 . The temperature is calculated as before:
T3 =
p3 V3 T1 = (1)( 4)(293 K ) = 1172 K p1 V1
For process 1 → 2, the work done is the area under the p-versus-V curve. That is,
Ws = (0.5 atm)( 40 cm 3 − 10 cm 3 ) +
1 2
(1.5 atm − 0.5 atm)(40 cm 3 − 10 cm 3 )
1.013 × 10 5 Pa = (30 × 10 −6 m 3 )(1 atm) = 3.04 J 1 atm The change in the thermal energy is
∆Eth = nCV ∆T = (2.08 × 10 −4 mol) 25 (8.31 J / mol K )(3516 K − 293 K ) = 13.93 J
The heat is Q = Ws + ∆Eth = 16.97 J . For process 2 → 3, the work done is Ws = 0 J and
Q = ∆Eth = nCV ∆T = n( 25 R)(T3 − T2 ) For process 3 → 1,
= (2.08 × 10 −4 mol) 25 (8.31 J / mol K )(1172 K − 3516 K ) = −10.13 J
( mol)
)
Ws = (0.5 atm)(10 cm 3 − 40 cm 3 ) = 0.5 × 1.013 × 10 5 Pa ( −30 × 10 −6 m 3 ) = −1.52 J ∆Eth = nCV ∆T = (2.08 × 10
−4
5 2
(8.31 J / mol K)(293 K − 1172 K) = −3.80 J
The heat is Q = ∆Eth + Ws = −5.32 J . Ws (J)
Q (J)
∆Eth
1→2 2→3
3.04 0
16.97 −10.13
13.93 −10.13
3→1 Net
−1.52 1.52
−5.32 1.52
−3.80 0
(b) The efficiency of the engine is
η=
Wnet 1.52 J = = 0.0896 = 8.96% 16.97 J QH
(c) The power output of the engine is Wnet revolutions 1 min 500 500 × × = × 1.52 J / s = 12.7 W min 60 s revolution 60 Assess: 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 → 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 WS 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
n=
(
)
1.013 × 10 5 Pa (600 × 10 −6 m 3 ) p1V1 = 0.0244 mol = RT1 (8.31 J / mol K)(300 K)
We are given that γ = 1.25, which means this is not a monatomic or a diatomic gas. The specific heats are R CV = = 4R CP = CV + R = 5 R 1−γ At point 2, process 1 → 2 is isothermal, so we can find the pressure p2 as follows:
p1V1 = p2 V2 ⇒ p2 =
V1 6.00 × 10 −4 m 3 p1 = p1 = 3 p1 = 3 atm = 3.039 × 10 5 Pa V2 2.00 × 10 −4 m 3
At point 3, process 2 → 3 is isobaric, so we can find the temperature T3 as follows: V V2 V3 6.00 × 10 −4 m 3 = ⇒ T3 = 3 T2 = T2 = 3T2 = 900 K 2.00 × 10 −4 m 3 T2 T3 V2
Point
p
V
T
1 2
1.0 atm = 1.013 × 105 Pa 3.0 atm = 3.039 × 105 Pa
6.00 × 10−4 m3 2.00 × 10−4 m3
300 K 300 K
3
3.0 atm = 3.039 × 105 Pa
6.00 × 10−4 m3
900 K
Process 1 → 2 is isothermal:
(WS )12 = p1V1 ln(V2 V1 ) = −66.8 J
Process 2 → 3 is isobaric:
(WS )23 = p2 ∆V = p2 (V3 − V2 ) = 121.6 J
Process 3 → 1 is isochoric:
(WS )31 = 0 J
Q12 = (WS )12 = −66.8 J Q23 = nCP ∆T = nCP (T3 − T2 ) = 608.3 J
Q31 = nCV ∆T = nCV (T1 − T3 ) = −486.7 J
We find that
(WS )cycle = −66.8 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 at 20 cycles/s has power output
Pout =
54.8 J 20 cycle J × = 1096 = 1096 W = 1.096 kW cycle s s
(b) Only Q23 is positive, so Qin = Q23 = 608 J. Thus, the thermal efficiency is W 54.8 J η = out = = 0.090 = 9.0% 608.3 J Qin
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. CV = 23 R and CP = 25 R for a monatomic gas. Visualize: Please refer to Figure P19.55. Solve: (a) Using the ideal-gas equation, the number of moles is
n=
(
)
1.0 × 10 5 Pa (100 × 10 −6 m 3 ) p1V1 = = 0.00401 mol RT1 (8.31 J / mol K)(300 K)
We can use the adiabat 2 → 3 to calculate p2 as follows: γ
γ
V V 600 cm 3 p3V3γ = p2 V2γ ⇒ p2 = p3 3 = p1 3 = 1.0 × 10 5 Pa 100 cm 3 V2 V2
(
)
5/ 3
= 1.981 × 10 6 Pa
T2 can be determined from the ideal-gas equation and is
T2 =
6 −6 3 p2 V2 (1.981 × 10 Pa )(100 × 10 m ) = 5943 K = (8.31 J / mol K )(0.00401 mol) nR
At point 3, p3 = p1 and
T3 T1 V 600 × 10 −6 m 3 = ⇒ T3 = 3 T1 = (300 K) = 1800 K 100 × 10 −6 m 3 V3 V1 V1 Now we can calculate Ws, Q, and ∆Eth for the three processes involved in the cycle. For process 1 → 2, Ws 1→2 = 0 J and
∆Eth = Q1→2 = nCV (T2 − T1 ) = n( 23 R)(T2 − T1 ) = 282.2 J For process 2 → 3, Q2→3 = 0 J and
∆Eth = nCV (T3 − T2 ) = n( 23 R)(T3 − T2 ) = –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,
∆Eth = nCV (T1 − T3 ) = n( 23 R)(T1 − T3 ) = −75.0 J Q3→1 = nCP (T1 − T3 ) = n( 25 R)(T1 − T3 ) = −125.0 J The work done Ws 3→1 is the area under the p-versus-V graph. We have
Ws 3→1 = (100 × 10 3 Pa )(100 × 10 −6 m 3 − 600 × 10 −6 m 3 ) = −50.0 J 1→2 2→3 3→1 Net (b) The thermal efficiency of the engine is
η=
Ws (J)
Q (J)
∆Eth (J)
0 207.2 −50.0 157.2
282.2 0 −125.0 157.2
282.2 −207.2 −75.0 0
Wout 157.2 J = = 0.522 = 52.2% 282.2 J QH
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 CV = 25 R 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: γ
V 4000 cm 3 p2 V2γ = p3V3γ ⇒ p2 = p3 3 = 1.0 × 10 5 Pa 1000 cm 3 V2
(
)
7/ 5
= 6.964 × 10 5 Pa
The temperature T2 can be obtained from the ideal-gas equation relating points 1 and 2:
6.964 × 10 5 Pa p1V1 p2 V2 p V = ⇒ T2 = T1 2 2 = (300 K ) (1) = 522.3 K T1 T2 p1 V1 4.0 × 10 5 Pa (b) The number of moles of the gas is
R=
(
)
4.0 × 10 5 Pa (1.0 × 10 −3 m 3 ) p1V1 = 0.1604 mol = (8.31 J / mol K)(300 K) RT1
For isochoric process 1 → 2 , Ws = 0 J and
Q = ∆Eth = nCV ∆T = n( 25 R)∆T = 741.1 J For adiabatic process 2 → 3, Q = 0 J and
∆Eth = nCV ∆T = n( 25 R)(T3 − T2 ) = −741.1 J Using the first law of thermodynamics, ∆Eth = ∆Ws + Q, which means Ws = −∆Eth = +741.1 J. Ws can also be determined from
Ws =
p3V3 − p2 V2 nR(T3 − T2 ) = = 1−γ 1−γ
( 43
J / K )(300 K − 522.3 K ) = 741.1 J (− 25 )
For isothermal process 3 → 1, ∆Eth = 0 J and
Ws = nRT1 ln
V1 = −554.5 J V3
Using the first law of thermodynamics, ∆Eth = − Ws + Q , Q = Ws = −554.5 J . 1→2 2→3 3→1 Net
∆Eth (J)
Ws (J)
Q (J)
741.1 −741.1 0 0
0 741.1 −554.5 186.6
741.1 0 −554.5 186.6
(c) The work per cycle is 186.6 J and the thermal efficiency is
η=
Ws 186.6 J = = 0.252 = 25.2% QH 741.1 J
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 CV = 25 R and γ = 75 . Visualize: Please refer to Figure P19.57. Solve: (a) The number of moles of gas is
n=
(
)
4.0 × 10 5 Pa (1000 × 10 −6 m 3 ) p2 V2 = 0.1203 mol = (8.31 J / mol K)(400 K) RT2
Using p1V1γ = p2 V2γ , and reading V1 from the graph, γ
V p1 = p2 2 = 4.0 × 10 5 Pa V1
(
)( )
1 7/ 5 4
= 5.743 × 10 4 Pa
With p1, V1, and nR having been determined, we can find T1 using the ideal-gas equation:
T1 =
4 −6 3 p1V1 (5.743 × 10 )( 4000 × 10 m ) = 229.7 K = (8.31 J / mol K )(0.1203 mol) nR
(b) For adiabatic process 1 → 2, Q = 0 J and
Ws =
(
)
4.0 × 10 5 Pa (1.00 × 10 −3 m 3 ) − (5.743 × 10 4 Pa )( 4.00 × 10 −3 m 3 ) p2 V2 − p1V1 = −425.7 K = 1−γ 1 − 1.4
Because ∆Eth = − Ws + Q , ∆Eth = − Ws = 425.7 K . For isothermal process 2 → 3, ∆Eth = 0 J . From Equation 17.15,
Ws = nRT2 ln
V3 = 554.5 J V2
From the first law of thermodynamics, Q = Ws = 554.5 J for process 2 → 3. For isochoric process 3 → 1, Ws = 0 J and
Q = ∆Eth = nCV ∆T = n( 25 R)(T1 − T3 ) = −425.7 J ∆Eth (J)
Ws (J) 1→2 −425.7 425.7 2→3 0 554.5 3→1 −425.7 0 Net 0 128.8 (c) The work per cycle is 128.8 J and the thermal efficiency of the engine is
η=
Ws 128.8 J = = 0.232 = 23.2% QH 554.5 J
Assess: As expected, for a closed cycle (Ws ) net = Qnet and ( ∆Eth ) net = 0 J.
Q (J) 0 554.5 −425.7 128.8
19.58. Model: In the Brayton cycle, process 1 → 2 and process 3 → 4 are adiabatic, and process 2 → 3 and process 4 → 2 are isobaric. We will assume the gas to be diatomic, with CP = 27 R and γ = 1.40. Visualize: Please refer to Figure P19.58. Solve: The number of moles of gas is
n=
(
)
1.013 × 10 5 Pa (2.0 m 3 ) p1V1 = 81.27 mol = (8.31 J / mol K)(300 K) RT1
We can find the volume at 2 from the adiabatic equation p1V1γ = p2 V2γ , 1
1
p γ 1 1.4 V2 = V1 1 = (2.0 m 3 ) = 0.3861 m 3 10 p2 The temperature T2 is determined from the ideal-gas equation as follows:
p1V1 p2 V2 p V 10 atm 0.3861 m 2 = 579.2 K = ⇒ T2 = T1 2 2 = (300 K ) 1 atm 2.0 m 3 T1 T2 p1 V1 To find T3 we use the heat input:
QH = 2.0 × 10 6 J = nCP ∆T = 27 n R∆T =
7 2
(675.33 J / K)∆T
⇒ ∆T = 846.1 K ⇒ T3 = T2 + ∆T = 579.2 K + 846.1 K = 1425.3 K We are now able to find V3 using the ideal-gas law. We have
V3 V2 T 1425.3 K = ⇒ V3 = V2 3 = (0.3861 m 3 ) = 0.9501 m 3 579.2 K T3 T2 T2 The volume at 4 is found by again using the adiabatic equation p3V3γ = p4 V4γ , 1
1 p γ V4 = V3 3 = (0.9501 m 3 )(10) 1.4 = 4.921 m 3 p4
T4 is now obtained from
T4 =
(
)
1.013 × 10 5 Pa ( 4.921 m 3 ) p4 V4 = 738.2 K = (81.27 mol)(8.31 J / mol K) nR
To find the engine’s efficiency, we must first find the total work done per cycle which is
Wnet = W1→2 + W2→3 + W3→4 + W4→1 The four contributions are
W1→2 =
p2 V2 − p1V1 nR(T2 − T1 ) (81.27 mol)(8.31 J / mol K )(579.2 K − 300 K ) = = −4.714 × 10 5 J = 1−γ 1−γ 1 − 1.4
W2→3 = p2 ∆V = (1.013 × 10 6 Pa )(0.9501 m 3 − 0.3861 m 3 ) = 5.713 × 10 5 J W3→4 =
nR(T4 − T3 ) 1−γ
(
=
(81.27 mol)(8.31 J / mol K)(738.2 K − 1425.3 K) 1 − 1.4
)
W4→1 = p4 ∆V = 1.013 × 10 5 Pa (2.0 m 3 − 4.921 m 3 ) = −2.959 × 10 5 J
= +1.160 × 10 6 J
Thus, Wnet = 9.64 × 10 5 J . The thermal efficiency is
η=
Wnet 9.64 × 10 5 J ⇒η = = 0.482 = 48.2% 2.0 × 10 6 J QH
As a comparison, we can calculate η from Equation 19.21 which is
η = 1−
1
( ) rp
where rp = pmax pmin = 10 .
( ) γ −1 γ
= 1−
1
(10) 1.4
0.4
= 48.2%
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, CV = 23 R and CP = 25 R. Visualize: Please refer to Figure P19.59. Solve: (a) At point 1: The pressure p1 = 1.0 atm = 1.013 × 10 5 Pa and the volume V1 = 1000 × 10 −6 m 3 = 1.0 × 10 −3 m 3 . The number of moles is 0.120 g n= = 0.030 mol 4 g / mol Using the ideal-gas law,
T1 =
(
)
1.013 × 10 5 Pa (1.0 × 10 −3 m 3 ) p1V1 = 406 K = nR (0.030 mol)(8.31 J / mol K)
At point 2: The pressure p2 = 5.0 atm = 5.06 × 10 5 Pa and V2 = 1.0 × 10 −3 m 3 . The temperature is
(
)
5.06 × 10 5 Pa (1.0 × 10 −3 m 3 ) p2 V2 = 2030 K = nR (0.030 mol)(8.31 J / mol K) At point 3: The pressure is p3 = 1.0 atm = 1.013 × 10 5 Pa and the temperature is T3 = T2 = 2030 K . The volume is T2 =
V3 = V2
p2 5 atm = (1.0 × 10 −3 m 3 ) = 5.0 × 10 −3 m 3 1 atm p3
(b) For isochoric process 1 → 2, W1→2 = 0 J and
Q1→2 = nCV ∆T = (0.030 mol)( 23 R)(2030 K − 406 K ) = 607 J For isothermal process 2 → 3, ∆Eth 2→3 = 0 J and
Q2→3 = W2→3 = nRT2 ln For isobaric process 3 → 1,
5.0 × 10 −3 m 3 V3 = (0.030 mol)(8.31 J / mol K )(2030 K ) ln = 815 J 1.0 × 10 −3 m 3 V2
(
)
W3→1 = p3 ∆V = 1.013 × 10 5 Pa (1.0 × 10 −3 m 3 − 5.0 × 10 −3 m 3 ) = −405 J
Q3→1 = nCP ∆T = (0.030 mol)( 25 )(8.31 J / mol K )( 406 K − 2030 K ) = −1012 J The total work done is Wnet = W1→2 + W2→3 + W3→1 = 410 J. The total heat input is QH = Q1→2 + Q2→3 = 1422 J . The efficiency of the engine is W 410 J η = net = = 28.8% 1422 J QH (c) The maximum possible efficiency of a heat engine that operates between Tmax and Tmin is
ηmax = 1 −
Tmin 406 K = 1− = 80% 2030 K Tmax
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 CV = 23 R and CP = 25 R . Visualize: Please refer to Figure P19.60. Solve: (a) The three temperatures are
(
)
(
)
(
)
T1 =
4.0 × 10 5 Pa (0.025 m 3 ) p1V1 = 601.7 K = nR (2.0 mol)(8.31 J / mol K)
T2 =
6.0 × 10 5 Pa (0.050 m 3 ) p2 V2 = 1805.1 K = nR (2.0 mol)(8.31 J / mol K)
T3 =
4.0 × 10 5 Pa (0.050 m 3 ) p3V3 = 1203.4 K = nR (2.0 mol)(8.31 J / mol K)
(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
Ws =
1 2
(6.0 × 10
5
)
(
)
Pa − 4.0 × 10 5 Pa (0.050 m 3 − 0.025 m 3 ) + 4.0 × 10 5 Pa (0.050 m 3 − 0.025 m 3 )
= 1.25 × 10 J 4
∆Eth = nCV ∆T = (2.0 mol)( 23 R)(T2 − T1 ) = (2.0 mol)( 23 )(8.31 J / mol K )(1805.1 K − 601.7 K ) = 3.00 × 10 4 J The heat input is Q = Ws + ∆Eth = 4.25 × 10 4 J . For isochoric process 2 → 3, Ws = 0 J and
Q = ∆Eth = nCV ∆T = (2.0 mol) 23 (8.31 J / mol K )(1203.4 K − 1805.1 K ) = −1.50 × 10 4 J For isobaric process 3 → 1, the work done is the area under the p-versus-V curve. Hence,
(
)
Ws = 4.0 × 10 5 Pa (0.025 m 3 − 0.050 m 3 ) = −1.0 × 10 4 J
∆Eth = nCV ∆T = n( 23 R)(T1 − T3 ) = (2.0 mol) 23 (8.31 J / mol K )(601.7 K − 1203.4 K ) = −1.5 × 10 4 J The heat input is Q = WS + ∆Eth = −2.50 × 10 4 J .
∆Eth (J) 1→2 2→3 3→1 Net
Ws (J)
3.0 × 10 −1.5 × 104 −1.5 × 104 0 4
Q (J)
1.25 × 10 0 −1.0 ×104 2.5 × 103
4
(c) The thermal efficiency is
η=
Wnet 2.5 × 10 3 J = = 5.88% 4.25 × 10 4 J QH
4.25 × 104 −1.50 × 104 −2.50 × 104 2.5 × 103
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, CV = 25 R, CP = 27 R, and γ = 75 = 1.4. Visualize: Please refer to Figure P19.61. Solve: (a) We can find the temperature T2 from the ideal-gas equation as follows:
T2 =
(
)
4.0 × 10 5 Pa (1.0 × 10 −3 m 3 ) p2 V2 = 2407 K = nR (0.020 mol)(8.31 J / mol K)
We can use the equation p2 V2γ = p1V1γ to find V1,
p V1 = V2 2 p1
1/ γ
4.0 × 10 5 Pa = (1.0 × 10 −3 m 3 ) 1.0 × 10 5 Pa
The ideal-gas equation can now be used to find T1,
(
1/1.4
= 2.692 × 10 −3 m 3
)
T1 =
1.0 × 10 5 Pa (2.692 × 10 −3 m 3 ) p1V1 = 1620 K = nR (0.020 mol)(8.31 J / mol K)
T3 =
4 × 10 5 Pa (2.692 × 10 −3 m 3 ) p3V3 = 6479 K = nR (0.020 mol)(8.31 J / mol K)
At point 3, V3 = V1 so we have
(
)
(b) For adiabatic process 1 → 2, Q = 0 J, ∆Eth = − Ws , and
Ws =
p2 V2 − p1V1 nR(T2 − T1 ) (0.020 mol)(8.31 J / mol K )(2407 K − 1620 K ) = −327.0 J = = 1−γ 1−γ (1 − 1.4)
For isobaric process 2 → 3,
Q = nCP ∆T = n( 27 R)( ∆T ) = (0.020 mol) 27 (8.31 J / mol K )(6479 K − 2407 K ) = 2369 J
∆Eth = nCV ∆T = n( 25 R)∆T = 1692 J The work done is the area under the p-versus-V graph. Hence,
(
)
Ws = 4.0 × 10 5 Pa (2.692 × 10 −3 m 3 − 1.0 × 10 −3 m 3 ) = 677 J For isochoric process 3 → 1, Ws = 0 J and
∆Eth = Q = nCV ∆T = (0.020 mol)( 25 )(8.31 J / mol K )(1620 K − 6479 K ) = −2019 J ∆Eth (J) 1→2 2→3 3→1 Net
327 1692 −2019 0
Ws (J) −327 677 0 350
(c) The engine’s thermal efficiency is
η=
350 J = 0.148 = 14.8% 2369 J
Q (J) 0 2369 −2019 350
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 = 0.50 mol 4.0 g / mol
The total work done by the engine per cycle is Wnet = W1→2 + W2→3 + W3→1 . For the isothermal process,
W1→2 = nRT1 ln
V2 = (0.50 mol)(8.31 J / mol K )(273 K ) ln(0.5) = −786.2 J V1
For the isobaric process,
V W2→3 = p2 ( ∆V ) = (2 p1 ) 1 = p1V1 = nRT1 = (0.50 mol)(8.31 J / mol K )(273 K ) = 1134.3 J 2 For the isochoric process, W3→1 = 0 J. Thus, Wnet = 348.1 J. For calculating heat transfers, note that T2 = T1 = 273 K and, since process 3 → 1 is isochoric, T3 = (2 atm / 1 atm ) T1 = 546 K. For the isothermal process, Q1→2 = W1→2 = −786.2 J because ∆Eth 1→2 = 0 J. For the isobaric process,
Q2→3 = nCP ∆T = n( 25 R)(T3 − T2 ) =
5 2
(0.50 mol)(8.31 J / mol K)(273 K) = 2835.8 J
For the isochoric process,
Q3→1 = nCv ∆T = n( 23 R)(T1 − T3 ) = ( 23 )(0.50 mol)(8.31 J / mol K )( –273 K ) = −1701.5 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 = 0.123 = 12.3% 2835.8 J
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 CV = 23 R . Visualize:
Solve:
Using the ideal-gas law,
p1 =
nRT1 (0.20 mol)(8.31 J / mol K )(600 K ) = = 4.986 × 10 5 Pa 2.0 × 10 −3 m 3 V1
At point 2, because of the isothermal conditions, T2 = T1 = 600 K and
2.0 × 10 −3 m 3 V1 5 = 4.986 × 10 5 Pa = 2.493 × 10 Pa 4.0 × 10 −3 m 3 V2
(
p2 = p1
)
At point 3, because it is an isochoric process, V3 = V2 = 4000 cm 3 and
T3 300 K = 1.247 × 10 5 Pa = 2.493 × 10 5 Pa 600 K T2
(
p3 = p2
)
Likewise at point 4, T4 = T3 = 300 K and
4.0 × 10 −3 m 3 V3 5 = 1.247 × 10 5 Pa = 2.493 × 10 Pa 2.0 × 10 −3 m 3 V4
(
p4 = p3
)
Let us now calculate Wnet = W1→2 + W2→3 + W3→4 + W4→1. For the isothermal processes,
W1→2 = nRT1 ln W3→4 = nRT3 ln
V2 = (0.20 mol)(8.31 J / mol K )(600 K ) ln(2) = 691.2 J V1
V4 = (0.20 mol)(8.31 J / mol K )(300 K ) ln( 12 ) = −345.6 J V3
For the isochoric processes, W2→3 = W4→1 = 0 J. Thus, Wnet = 345.6 J. Because Q = Ws + ∆Eth,
Q1→2 = W1→2 + ( ∆Eth )1→2 = 691.2 J + 0 J = 691.2 J
For the first isochoric process,
Q2→3 = nCV ∆T = (0.20 mol)( 23 R)(T3 − T2 ) = (0.20 mol) 23 (8.31 J / mol K )(300 K − 600 K ) = −747.9 K For the second isothermal process
Q3→4 = W3→4 + ( ∆Eth )3→4 = −345.6 J + 0 J = −345.6 J For the second isochoric process,
Q4→1 = nCV ∆T = n( 23 R)(T1 − T4 ) = (0.20 mol)( 23 )(8.31 J / mol K )(600 K − 300 K ) = 747.9 K
Thus, QH = Q1→2 + Q4→1 = 1439.1 J . The thermal efficiency of the engine is
η=
Wnet 345.6 J = = 0.240 = 24.0% 1439.1 J QH
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 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 QH. 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 transferred 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 this device, 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. 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) 273 (0°C + 273) 0.80 = 1 − ⇒ = 0.20 ⇒ TH = 1092°C T + T 273 (H ) H + 273
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 = QC Win ⇒ QC = 4Win . This means QH = QC + Win = 4Win + Win = 5Win ⇒ Win = Hence, QC = QH − Win = 100 J − 20 J = 80 J .
QH 100 J = = 20 J 5 5
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 0.20 = 1 − QC QH . Using the first law of thermodynamics, Wout = QH − QC = 20 J ⇒ QC = QH − 20 J Substituting into the definition of efficiency, Q − 20 J 20 J 20 J 20 J 0.20 = 1 − H = 1−1+ = ⇒ QH = = 100 J QH QH QH 0.20 The heat exhausted is QC = QH − 20 J = 100 J − 20 J = 80 J .
19.68.
Solve:
(a)
In this heat engine, 400 kJ of work is done each cycle. What is the maximum pressure? (b) 1 pmax − 1.0 × 10 5 Pa 2.0 m 3 = 4.0 × 10 5 J ⇒ pmax = 5.0 × 10 5 Pa = 500 kPa 2
(
)(
)
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 inserted when the heat engine has the initial conditions. Second, heat is turned on and the pressure increases at constant volume from 1 atm to 3 atm. Third, the pin is removed. The flame continues to heat the gas and the volume increases at constant pressure from 50 cm3 to 100 cm3. 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 50 cm3. 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: 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 = and CP = 25 R. The number of moles of gas is
n=
3 2
R
p1V1 (101, 300 Pa)(50 × 10 −6 m 3 ) = = 0.00208 mol (8.31 J / mol K)(293 K) RT1
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
Q12 = nCV ∆T = 25 nR(T2 − T1 ) = 25 (0.00208 mol)(8.31 J / mol K)(586 K) = 25.32 J Similarly,
Q23 = nCP ∆T = 27 nR(T3 − T2 ) = 27 (0.00208 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 W 10.13 J η = out = = 0.129 = 12.9% 78.50 J QH
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 input from (or loss to) the outside world, we have Q1 + Q2 = 0 J. Heat Q1, which is negative, will change the temperature of system 1. Heat Q2 will both change the temperature of system 2 and do work by lifting the piston. But these consequences of heat flow don’t change the fact that Q1 + Q2 = 0 J. System 1 undergoes constant volume cooling from T1i = 600 K to Tf. System 2, whose pressure is controlled by the weight of the piston, undergoes constant pressure heating from T2i = 300 K to Tf. Thus,
Q1 + Q2 = 0 J = n1CV (Tf − T1i ) + n2 CP (Tf − T2i ) = n1 ( 23 R)(Tf − T1i ) + n2 ( 25 R)(Tf − T2i )
Solving this equation for Tf gives
Tf =
3n1T1i + 5n2 T2i 3(0.060 mol)(600 K ) + 5(0.030 mol)(300 K ) = 464 K = 3(0.060 mol) + 5(0.030 mol) 3n1 + 5n2
(b) Knowing Tf, we can compute the heat transferred from system 1 to system 2:
Q2 = n2 CP (Tf − T2i ) = n2 ( 25 R)(Tf − T2i ) = 102.0 J (c) The change of thermal energy in system 2 is
∆Eth = n2 CV ∆T = n2 ( 23 R)(Tf − T2i ) = 35 Q2 = 61.2 J 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 40.8 J = 2.08 m Ws = w∆y = mg∆y ⇒ ∆y = s = mg (2.0 kg) 9.8 m / s 2
(
(d) The fraction of heat converted to work is
Ws 40.8 J = = 0.40 = 40% Q2 102.0 J
)
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, Q12 = 0 J and
W12 =
p2 V2 − p1V1 nR(T2 − T1 ) = 1−γ 1−γ
For isochoric process 2 → 3, W23 = 0 J and Q23 = nCV (T3 − T2 ) . For adiabatic process 3 → 4, Q34 = 0 J and
W34 =
p4 V4 − p3V3 nR(T4 − T3 ) = 1−γ 1−γ
For isochoric process 4 → 1, W41 = 0 J and Q41 = nCV (T1 − T4 ) . The work done per cycle is
Wnet = W12 + W23 + W34 + W41 =
nR(T2 − T1 ) 1−γ
+0 J+
nR(T4 − T3 ) 1−γ
+0 J =
nR (T2 − T1 + T4 − T3 ) 1−γ
(b) The thermal efficiency of the heat engine is
η=
Q nCV (T4 − T1 ) Q Wout = 1 − C = 1 − 41 = 1 − QH QH Q23 nCV (T3 − T2 )
The last step follows from the fact that T3 > T2 and T4 > T1. We will now simplify this expression further as follows:
V pV γ = pVV γ −1 = nRTV γ −1 ⇒ nRT1V1γ −1 = nRT2 V2γ −1 ⇒ T2 = T1 1 V2 Similarly, T3 = T4 r γ −1 . The equation for thermal efficiency now becomes
η = 1−
(c)
T4 − T1 1 = 1 − γ −1 T4 r γ −1 − T1r γ −1 r
γ −1
= T1r γ −1
19.72. Model: For the Diesel cycle, process 1 → 2 is an adiabatic compression, process 2 → 3 is an isobaric expansion, 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
Vmax V1 1050 cm 3 = = = 21 50 cm 3 Vmin V2
r= The number of moles of gas is
n=
(
)
1.013 × 10 5 Pa (1050 × 10 −6 m 3 ) p1V1 = 0.0430 mol = RT1 (8.31 J / mol K)[(25 + 273) K]
For an adiabatic process, γ
V p1V1γ = p2 V2γ ⇒ p2 = 1 p1 = r γ p1 = 211.40 × 1 atm = 71.0 atm = 7.19 × 10 6 Pa V2 V T1V1γ −1 = T2 V2γ −1 ⇒ T2 = 1 V2
γ −1
T1 = r γ −1T1 = 210.41 × 298 K = 1007 K
Process 2 → 3 is an isobaric heating with Q = 1000 J. Constant pressure heating obeys
Q = nCP ∆T ⇒ ∆T =
Q nCP
The gas has a specific heat ratio γ = 1.40 = 7 / 5, thus CV = 25 R and CP = 27 R. Knowing CP, we can calculate first ∆T = 800 K and then T3 = T2 + ∆T = 1807 K. Finally, for an isobaric process we have
T V2 V3 1807 K = ⇒ V3 = 3 V2 = (50 × 10 −6 m 3 ) = 89.7 × 10 −6 m 3 1007 K T2 T3 T2 Process 3 → 4 is an adiabatic expansion to V4 = V1. Thus, γ
V 89.7 × 10 −6 m 3 p3V3γ = p4 V4γ ⇒ p4 = 3 p3 = 1050 × 10 −6 m 3 V4 γ −1 3 3
TV
γ −1 4 4
=TV
V ⇒ T4 = 3 V4
Point
p
γ −1
1.4
(7.19 × 10
6
89.7 × 10 −6 m 3 T3 = 1050 × 10 −6 m 3
Pa ) = 2.30 × 10 5 Pa = 2.27 atm 0.4
(1807 K) = 675 K
V
T −6
1
1.00 atm = 1.013 × 10 Pa
1050 × 10 m
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 × 10 Pa
5
5
−6
3
1050 × 10 m
3
(b) For adiabatic process 1 → 2,
(Ws )12 =
p2 V2 − p1V1 = −633 J 1−γ
For isobaric process 2 → 3,
(Ws )23 = p2 ∆V = p2 (V3 − V2 ) = 285 J For adiabatic process 3 → 4,
(Ws )34 =
p4 V4 − p3V3 = 1009 J 1−γ
For isochoric process 4 → 1, (Ws ) 41 = 0 J. Thus,
(Ws )cycle = Wout = (Ws )12 + (Ws )23 + (Ws )34 + (Ws )41 = 661 J
675 K = 402°C
(c) The efficiency is
η=
Wout 661 J = = 66.1% 1000 J QH
(d) The power output of one cylinder is
661 J 2400 cycle 1 min J × × = 26, 440 = 26.4 kW cycle min 60 sec s For an 8-cylinder engine the power will be 211 kW or 283 horsepower.