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Soluciones Soluciones Document Transcript

  • 22 Heat Engines, Entropy, and the Second Law of Thermodynamics ANSWERS TO QUESTIONS CHAPTER OUTLINE 22.1 Heat Engines and the Second Law of Thermodynamics Heat Pumps and Refrigerators Reversible and Irreversible Processes The Carnot Engine Gasoline and Diesel Engines Entropy Entropy Changes in Irreversible Processes Entropy on a Microscopic Scale 22.4 22.5 22.6 22.7 22.8 Q22.1 First, the efficiency of the automobile engine cannot exceed the Carnot efficiency: it is limited by the temperature of burning fuel and the temperature of the environment into which the exhaust is dumped. Second, the engine block cannot be allowed to go over a certain temperature. Third, any practical engine has friction, incomplete burning of fuel, and limits set by timing and energy transfer by heat. *Q22.2 22.2 22.3 For any cyclic process the total input energy must be equal to the total output energy. This is a consequence of the first law of thermodynamics. It is satisfied by processes ii, iv, v, vi, vii but not by processes i, iii, viii. The second law says that a cyclic process that takes in energy by heat must put out some of the energy by heat. This is not satisfied for processes v, vii, and viii. Thus the answers are (i) b (ii) a (iii) b (iv) a (v) c (vi) a (vii) c (viii) d. Q22.3 A higher steam temperature means that more energy can be extracted from the steam. For a constant temperature heat sink at Tc, and steam at Th, the efficiency of the power plant goes as Th − Tc T = 1 − c and is maximized for a high Th. Th Th Q22.4 No. The first law of thermodynamics is a statement about energy conservation, while the second is a statement about stable thermal equilibrium. They are by no means mutually exclusive. For the particular case of a cycling heat engine, the first law implies Qh = Weng + Qc , and the second law implies Qc > 0. Q22.5 Take an automobile as an example. According to the first law or the idea of energy conservation, it must take in all the energy it puts out. Its energy source is chemical energy in gasoline. During the combustion process, some of that energy goes into moving the pistons and eventually into the mechanical motion of the car. The chemical energy turning into internal energy can be modeled as energy input by heat. The second law says that not all of the energy input can become output mechanical energy. Much of the input energy must and does become energy output by heat, which, through the cooling system, is dissipated into the atmosphere. Moreover, there are numerous places where friction, both mechanical and fluid, turns mechanical energy into heat. In even the most efficient internal combustion engine cars, less than 30% of the energy from the fuel actually goes into moving the car. The rest ends up as useless heat in the atmosphere. 571 13794_22_ch22_p571-600.indd 571 1/8/07 7:53:24 PM
  • 572 Chapter 22 *Q22.6 Answer (b). In the reversible adiabatic expansion OA, the gas does work against a piston, takes in no energy by heat, and so drops in internal energy and in temperature. In the free adiabatic expansion OB, there is no piston, no work output, constant internal energy, and constant temperature for the ideal gas. The points O and B are on a hyperbolic isotherm. The points O and A are on an adiabat, steeper than an isotherm by the factor γ. Q22.7 A slice of hot pizza cools off. Road friction brings a skidding car to a stop. A cup falls to the floor and shatters. Your cat dies. Any process is irreversible if it looks funny or frightening when shown in a videotape running backwards. The free flight of a projectile is nearly reversible. Q22.8 (a) When the two sides of the semiconductor are at different temperatures, an electric potential (voltage) is generated across the material, which can drive electric current through an external circuit. The two cups at 50°C contain the same amount of internal energy as the pair of hot and cold cups. But no energy flows by heat through the converter bridging between them and no voltage is generated across the semiconductors. (b) A heat engine must put out exhaust energy by heat. The cold cup provides a sink to absorb output or wasted energy by heat, which has nowhere to go between two cups of equally warm water. *Q22.9 (i) Answer (a). The air conditioner operating in a closed room takes in energy by electric transmission and turns it all into energy put out by heat. That is its whole net effect. (ii) Answer (b). The frozen stuff absorbs energy by heat from the air. But if you fill the ice trays with tap water and put them back into the freezer, the refrigerator will pump more heat into the air than it extracts from the water to make it freeze. *Q22.10 (i) Answer (d). (ii) Answer (d). The second law says that you must put in some work to pump heat from a lower-temperature to a higher-temperature location. But it can be very little work if the two temperatures are very nearly equal. Q22.11 One: Energy flows by heat from a hot bowl of chili into the cooler surrounding air. Heat lost by the hot stuff is equal to heat gained by the cold stuff, but the entropy decrease of the hot stuff is less than the entropy increase of the cold stuff. Two: As you inflate a soft car tire at a service station, air from a tank at high pressure expands to fill a larger volume. That air increases in entropy and the surrounding atmosphere undergoes no significant entropy change. Three: The brakes of your car get warm as you come to a stop. The shoes and drums increase in entropy and nothing loses energy by heat, so nothing decreases in entropy. Q22.12 (a) For an expanding ideal gas at constant temperature, the internal energy stays constant. The gas must absorb by heat the same amount of energy that it puts out by work. Then its ⎛V ⎞ ∆Q entropy change is ∆ S = = nR ln ⎜ 2 ⎟ T ⎝ V1 ⎠ (b) For a reversible adiabatic expansion ∆Q = 0, and ∆S = 0. An ideal gas undergoing an irreversible adiabatic expansion can have any positive value for ∆S up to the value given in part (a). *Q22.13 Answer (f). The whole Universe must have an entropy change of zero or more. The environment around the system comprises the rest of the Universe, and must have an entropy change of +8.0 Jր K, or more. 13794_22_ch22_p571-600.indd 572 1/8/07 7:53:25 PM
  • Heat Engines, Entropy, and the Second Law of Thermodynamics *Q22.14 (i) (ii) 573 Consider the area that fits under each of the arrows, between its line segment and the horizontal axis. Count it as positive for arrows to the right, zero for vertical arrows, and negative for arrows tending left. Then E > F > G > H = D > A > B > C. The thin blue hyperbolic lines are isotherms. Each is a set of points representing states with the same internal energy for the ideal gas simple. An arrow tending farther from the origin than the BE hyperbola represents a process for which internal energy increases. So we have D = E > C > B = F > G > A = H. (iii) The arrows C and G are along an adiabat. Visualize or sketch in a set of these curves, uniformly steeper than the blue isotherms. The energy input by heat is determined by how far above the starting adiabat the process arrow ends. We have E > D > F > C = G > B > H > A. *Q22.15 Processes C and G are adiabatic. They can be carried out reversibly. Along these arrows entropy does not change. Visualize or sketch in a set of these adiabatic curves, uniformly steeper than the blue isotherms. The entropy change is determined by how far above the starting adiabat the process arrow ends. We have E > D > F > C = G > B > H > A. *Q22.16 (a) The reduced flow rate of ‘cooling water’ reduces the amount of heat exhaust Qc that the plant can put out each second. Even with constant efficiency, the rate at which the turbines can take in heat is reduced and so is the rate at which they can put out work to the generators. If anything, the efficiency will drop, because the smaller amount of water carrying the heat exhaust will tend to run hotter. The steam going through the turbines will undergo a smaller temperature change. Thus there are two reasons for the work output to drop. (b) The engineer’s version of events, as seen from inside the plant, is complete and correct. Hot steam pushes hard on the front of a turbine blade. Still-warm steam pushes less hard on the back of the blade, which turns in response to the pressure difference. Higher temperature at the heat exhaust port in the lake works its way back to a corresponding higher temperature of the steam leaving a turbine blade, a smaller temperature drop across the blade, and a lower work output. *Q22.17 Answer (d). Heat input will not necessarily produce an entropy increase, because a heat input could go on simultaneously with a larger work output, to carry the gas to a lower-temperature, lower-entropy final state. Work input will not necessarily produce an entropy increase, because work input could go on simultaneously with heat output to carry the gas to a lower-volume, lower-entropy final state. Either temperature increase at constant volume, or volume increase at constant temperature, or simultaneous increases in both temperature and volume, will necessarily end in a more disordered, higher-entropy final state. Q22.18 An analogy used by Carnot is instructive: A waterfall continuously converts mechanical energy into internal energy. It continuously creates entropy as the organized motion of the falling water turns into disorganized molecular motion. We humans put turbines into the waterfall, diverting some of the energy stream to our use. Water flows spontaneously from high to low elevation and energy spontaneously flows by heat from high to low temperature. Into the great flow of solar radiation from Sun to Earth, living things put themselves. They live on energy flow, more than just on energy. A basking snake diverts energy from a high-temperature source (the Sun) through itself temporarily, before the energy inevitably is radiated from the body of the snake to a lowtemperature sink (outer space). A tree builds organized cellulose molecules and we build libraries and babies who look like their grandmothers, all out of a thin diverted stream in the universal flow of energy crashing down to disorder. We do not violate the second law, for we build local reductions in the entropy of one thing within the inexorable increase in the total entropy of the Universe. Your roommate’s exercise puts energy into the room by heat. 13794_22_ch22_p571-600.indd 573 1/8/07 7:53:26 PM
  • 574 Chapter 22 Q22.19 Either statement can be considered an instructive analogy. We choose to take the first view. All processes require energy, either as energy content or as energy input. The kinetic energy which it possessed at its formation continues to make the Earth go around. Energy released by nuclear reactions in the core of the Sun drives weather on the Earth and essentially all processes in the biosphere. The energy intensity of sunlight controls how lush a forest or jungle can be and how warm a planet is. Continuous energy input is not required for the motion of the planet. Continuous energy input is required for life because energy tends to be continuously degraded, as heat flows into lower-temperature sinks. The continuously increasing entropy of the Universe is the index to energy-transfers completed. Arnold Sommerfeld suggested the idea for this question. Q22.20 Shaking opens up spaces between jellybeans. The smaller ones more often can fall down into spaces below them. The accumulation of larger candies on top and smaller ones on the bottom implies a small increase in order, a small decrease in one contribution to the total entropy, but the second law is not violated. The total entropy increases as the system warms up, its increase in internal energy coming from the work put into shaking the box and also from a bit of gravitational energy loss as the beans settle compactly together. SOLUTIONS TO PROBLEMS Section 22.1 Heat Engines and the Second Law of Thermodynamics Weng *P22.2 e= Qc = Qh − Weng = 360 J − 25.0 J = 335 J Qh = 25.0 J = 0.069 4 or 6.94% 360 J (a) (b) P22.1 The engine’s output work we identify with the kinetic energy of the bullet: Weng = K = e= 1 1 2 mv 2 = 0.002 4 kg ( 320 m s ) = 123 J 2 2 Weng Qh Weng 123 J = = 1.12 × 10 4 J e 0.011 Qh = Weng + Qc Qh = The energy exhaust is Qc = Qh − Weng = 1.12 × 10 4 J − 123 J = 1.10 × 10 4 J Q = mc∆T T ∆T = P22.3 (a) We have e = Weng Qh = Q 1.10 × 10 4 J kg°C = = 13.7°C mc 1.80 kg 448 J Qh − Qc Q = 1 − c = 0.250 Qh Qh with Qc = 8 000 J, we have Qh = 10.7 kJ (b) Weng = Qh − Qc = 2 667 J and from P = 13794_22_ch22_p571-600.indd 574 Weng ∆t , we have ∆t = Weng P = 2 667 J = 0.533 s 5 000 J s 1/8/07 7:53:27 PM
  • Heat Engines, Entropy, and the Second Law of Thermodynamics P22.4 (a) 575 The input energy each hour is J revolution ) ( 2 500 rev min ) 60 min = 1.18 × 10 9 J h 1h 1L ⎛ ⎞ = 29.4 L h implying fuel input (1.18 × 10 9 J h ) ⎝ 4.03 × 10 7 J ⎠ ( 7.89 × 10 (b) 3 Qh = Weng + Qc . For a continuous-transfer process we may divide by time to have Qh Weng Qc = + ∆t ∆t ∆t Useful power output = Weng ∆t = Qh Qc − ∆t ∆t ⎛ 7.89 × 10 3 J 4.58 × 10 3 J ⎞ 2 500 rev 1 min =⎜ − = 1.38 × 10 5 W ⎝ revolution ⎠ revolution ⎟ 1 min 60 s e P eng P ⎛ 1 hp ⎞ = 1.38 × 10 5 W ⎜ = 185 hp ⎝ 746 W ⎟ ⎠ 1.38 × 10 5 J s ⎛ 1 rev ⎞ = 527 N ⋅ m ⎝ ⎠ ( 2 500 rev 60 s ) ⎜ 2π rad ⎟ (c) P (d) Qc 4.58 × 10 J ⎛ 2 500 rev ⎞ = = 1.91 × 10 5 W ∆t revolution ⎝ 60 s ⎠ eng = τω ⇒ τ = eng ω = 3 P22.5 The heat to melt 15.0 g of Hg is Qc = mL f = (15 × 10 −3 kg ) (1.18 × 10 4 J kg ) = 177 J The energy absorbed to freeze 1.00 g of aluminum is Qh = mL f = (10 −3 kg ) ( 3.97 × 10 5 J/ kg ) = 397 J Weng = Qh − Qc = 220 J and the work output is e= The theoretical (Carnot) efficiency is Section 22.2 P22.6 Weng Qh = 220 J = 0.554, or 55.4% 397 J Th − Tc 933 K − 243.1 K = = 0.749 = 74.9% Th 933 K Heat Pumps and Refrigerators COP ( refrigerator ) = Qc W (a) If Qc = 120 J and COP = 5.00, then W = 24.0 J (b) Heat expelled = Heat removed + Work done. Qh = Qc + W = 120 J + 24 J = 144 J 13794_22_ch22_p571-600.indd 575 1/8/07 7:53:27 PM
  • 576 P22.7 Chapter 22 COP = 3.00 = Qc Q . Therefore, W = c . W 3.00 The heat removed each minute is QC = ( 0.030 0 kg ) ( 4 186 J kg °C ) ( 22.0°C ) + ( 0.030 0 kg ) ( 3.33 × 10 5 J kg ) t + ( 0.030 0 kg ) ( 2 090 J kg °C ) ( 20.0°C ) = 1.40 × 10 4 J min Qc = 233 J s t or, Thus, the work done per second is P = 233 J s = 77.8 W 3.00 (a) ⎛ 10.0 Btu ⎞ ⎛ 1055 J ⎞ ⎛ 1 h ⎞ ⎛ 1 W ⎞ = 2.93 ⎝ ⎠ ⎠⎝ ⎝ h ⋅ W ⎠ ⎝ 1 Btu ⎠ ⎜ 3 600 s ⎟ ⎜ 1 J s ⎟ (b) P22.8 The energy extracted by heat from the cold side divided by required work input is by ( COP )refrigerator definition the coefficient of performance for a refrigerator: (c) With EER 5, 5 Btu 10 000 Btu h : = P h⋅W Energy purchased is P = 10 000 Btu h = 2 000 W = 2.00 kW 5 Btu h ⋅ W P ∆t = ( 2.00 kW )(1 500 h ) = 3.00 × 10 3 kWh Cost = ( 3.00 × 10 3 kWh ) ( 0.100 $ kWh ) = $300 With EER 10, 10 Btu 10 000 Btu h = : h⋅W P Energy purchased is P = 10 000 Btu h = 1 000 W = 1.00 kW 10 Btu h ⋅ W P ∆t = (1.00 kW )(1 500 h ) = 1.50 × 10 3 kWh Cost = (1.50 × 10 3 kWh ) ( 0.100 $ kWh ) = $150 Thus, the cost for air conditioning is half as much for an air conditioner with EER 10 compared with an air conditioner with EER 5. Section 22.3 Reversible and Irreversible Processes Section 22.4 The Carnot Engine P22.9 Tc = 703 K Th = 2 143 K ∆T 1 440 = = 67.2% Th 2 143 (a) ec = (b) Qh = 1.40 × 10 5 J, Weng = 0.420 Qh P 13794_22_ch22_p571-600.indd 576 = Weng ∆t = 5.88 × 10 4 J = 58.8 kW 1s 1/8/07 7:53:28 PM
  • Heat Engines, Entropy, and the Second Law of Thermodynamics P22.10 When e = ec, 1− (a) Qh = (W Tc Weng = Th Qh eng ) ր ∆t ∆t 1 − (Tc րTh ) = Weng ր ∆t and (1.50 × 10 Qh ր ∆t 5 = 1− 577 Tc Th W ) ( 3 600 s ) 1 − 293ր773 Qh = 8.70 × 108 J = 870 MJ (b) ⎛ Weng ⎞ Qc = Qh − ⎜ ∆t = 8.70 × 108 − (1.50 × 10 5 ) ( 3 600 ) = 3.30 × 108 J = 330 MJ ⎝ ∆t ⎟ ⎠ *P22.11 We use amounts of energy to find the actual efficiency. Qh = Qc + Weng = 20 kJ + 1.5 kJ = 21.5 kJ e = WengրQh = 1.5 kJր21.5 kJ = 0.0698 We use temperatures to find the Carnot efficiency of a reversible engine eC = 1 − TcրTh = 1 – 373 Kր453 K = 0.177 The actual efficiency of 0.0698 is less than four-tenths of the Carnot efficiency of 0.177. n *P22.12 (a) (b) eC = 1 − TcրTh = 1 − 350ր500 = 0.300 In eC = 1 − TcրTh we differentiate to find deC րdTh = 0 – Tc(−1)Th–2 = TcրTh2 = 350ր5002 = 1.40 × 10 −3 This is the increase of efficiency per degree of increase in the temperature of the hot reservoir. (c) In eC = 1 − TcրTh we differentiate to find deC րdTc = 0 − 1րTh = −1ր500 = –2.00 × 10–3 Then deC ր(−dTc) = +2.00 × 10 −3 This is the increase of efficiency per degree of decrease in the temperature of the cold reservoir. Note that it is a better deal to cool the exhaust than to supercharge the firebox. Isothermal expansion at Th = 523 K Isothermal compression at P22.13 Tc = 323 K Gas absorbs 1 200 J during expansion. (a) ⎛T ⎞ 323 ⎞ Qc = Qh ⎜ c ⎟ = 1 200 J ⎛ = 741 J ⎝ 523 ⎠ ⎝ Th ⎠ (b) Weng = Qh − Qc = (1 200 − 741) J = 459 J ec ,s = 1 − Tc ( 273 + 20 ) K = 1− = 0.530 Th ( 273 + 350 ) K ec ,w = 1 − 283 = 0.546 623 Then the actual winter efficiency is 13794_22_ch22_p571-600.indd 577 The Carnot summer efficiency is And in winter, P22.14 0.546 ⎞ 0.320 ⎛ = 0.330 ⎝ 0.530 ⎠ or 33.0% 1/8/07 7:53:29 PM
  • 578 Chapter 22 γ P22.15 (a) ⎛ Pf V f ⎞ ⎛ PVi ⎞ i In an adiabatic process, Pf V fγ = PViγ . Also, ⎜ i ⎟ =⎜ T ⎟ Tf ⎠ ⎝ i ⎠ ⎝ ⎛ Pf ⎞ Dividing the second equation by the first yields T f = Ti ⎜ ⎟ ⎝ Pi ⎠ γ −1 2 5 Since γ = for Argon, = = 0.400 and we have γ 5 3 ⎛ 300 × 10 3 Pa ⎞ T f = (1 073 K ) ⎜ ⎝ 1.50 × 10 6 Pa ⎟ ⎠ (b) = 564 K − nCV ∆T = or t t ( −80.0 kg)(1 mol/0.0399 kg)( 3 )(8.314 J mol ⋅ K )( 564 − 1 073) K 2 = 60.0 s = 2.12 × 10 5 W = 212 kW Tc 564 K = 1− = 0.475 or 47.5% Th 1 073 K (c) eC = 1 − (a) emax = 1 − (b) P = Tc 278 = 1− = 5.12 × 10 −2 = 5.12% Th 293 Weng = 75.0 × 10 6 J s ∆t Weng = ( 75.0 × 10 6 J s ) ( 3 600 s h ) = 2.70 × 1011 J h Therefore, From e = *P22.17 (a) 0. 400 Weng = P (c) (γ −1) γ ∆Eint = nCV ∆T = Q − Weng = 0 − Weng, so Weng = − nCV ∆T , and the power output is P P22.16 γ Weng Qh we find Qh = Weng e = 2.70 × 1011 J h = 5.27 × 1012 J h = 5.27 TJ h 5.12 × 10 −2 As fossil-fuel prices rise, this way to use solar energy will become a good buy. e= Weng1 + Weng2 Q1h = e1Q1h + e2Q2 h Q1h Now Q2 h = Q1c = Q1h − Weng1 = Q1h − e1Q1h So e= e1Q1h + e2 (Q1h − e1Q1h ) = e1 + e2 − e1e2 Q1h continued on next page 13794_22_ch22_p571-600.indd 578 1/8/07 7:53:31 PM
  • Heat Engines, Entropy, and the Second Law of Thermodynamics (b) e = e1 + e2 − e1e2 = 1 − Ti T +1− c Th Ti = 2− 579 ⎛ T ⎞⎛ T ⎞ − ⎜1− i ⎟ ⎜1− c ⎟ ⎝ Th ⎠ ⎝ Ti ⎠ Ti Tc T T T T − −1+ i + c − c = 1− c Th Ti Th Ti Th Th The combination of reversible engines is itself a reversible engine so it has the Carnot efficiency. No improvement in net efficiency has resulted. (c) Weng1 + Weng2 With Weng2 = Weng1, e = 1− 2Weng1 Q1h = 2e1 ⎛ T ⎞ Tc = 2 ⎜1 − i ⎟ Th ⎝ Th ⎠ 0− Q1h = 2T Tc = 1− i Th Th 2Ti = Th + Tc Ti = (d) 1 (Th + Tc ) 2 Ti T = 1− c Th Ti e1 = e2 = 1 − Ti 2 = TcTh Ti = (ThTc ) 12 *P22.18 (a) “The actual efficiency is two thirds the Carnot efficiency” reads as an equation Weng Qh = Weng Qc + Weng 2 ⎛ T ⎞ 2 Th − Tc = ⎜1 − c ⎟ = . 3 ⎝ Th ⎠ 3 Th All the T ’s represent absolute temperatures. Then Qc + Weng Weng Qc = Weng = 1.5 Th Th − Tc 0.5 Th + Tc Th − Tc Qc Weng Qc ∆t = 1.5 Th 1.5 Th − Th + Tc −1= Th − Tc Th − Tc = Weng 0.5 Th + Tc 0.5 Th + 383 K = 1.40 MW ∆t Th − Tc Th − 383 K The dominating Th in the bottom of this fraction means that the exhaust power decreases as the f irebox temperature increases. (b) Qc 0.5 Th + 383 K 0.5(1073 K) + 383 K ( = 1.40 MW = 1.40 MW = 1.87 MW ∆t (1073 − 383) K Th − 383 K (c) We require Qc 1 0.5 Th + 383 K = 2 1.87 MW = 1.40 MW Th − 383 K ∆t 6 0.5 Th + 383 K = 0.666Th − 255 K (d) 13794_22_ch22_p571-600.indd 579 0.5 Th + 383 K = 0.666 Th − 383 K Th = 638 K/0.166 = 3.84 × 10 3 K The minimum possible heat exhaust power is approached as the firebox temperature goes to infinity, and it is |Qc|ր∆t = 1.40 MW( 0.5ր1) = 0.7 MW. The heat exhaust power cannot be as small as (1ր4)(1.87 MW) = 0.466 MW. So no answer exists. The energy exhaust cannot be that small. 1/8/07 7:53:32 PM
  • 580 Chapter 22 P22.19 P22.20 (a) Tc 270 = = 9.00 ∆T 30.0 ( COP )refrig = First, consider the adiabatic process D → A: γ γ γ PDVD = PAVA so 53 ⎛V ⎞ 10.0 L ⎞ PD = PA ⎜ A ⎟ = 1 400 kPa ⎛ = 712 kPa ⎝ 15.0 L ⎠ ⎝ VD ⎠ Also ⎛ nRTD ⎞ γ ⎛ nRTA ⎞ γ ⎜ V ⎟ VD = ⎜ V ⎟ VA ⎝ D ⎠ ⎝ A ⎠ or ⎛V ⎞ TD = TA ⎜ A ⎟ ⎝V ⎠ γ −1 D 10.0 ⎞ = 720 K ⎛ ⎝ 15.0 ⎠ 23 = 549 K Now, consider the isothermal process C → D: TC = TD = 549 K γ ⎛V ⎞ ⎡ ⎛ V ⎞ ⎤⎛V ⎞ P Vγ PC = PD ⎜ D ⎟ = ⎢ PA ⎜ A ⎟ ⎥ ⎜ D ⎟ = A γA−1 ⎝ VC ⎠ ⎢ ⎝ VD ⎠ ⎥ ⎝ VC ⎠ VCVD ⎣ ⎦ 1 400 kPa (10.0 L ) = 445 kPa 23 24.0 L (15.0 L ) 53 PC = γ Next, consider the adiabatic process B → C: PBVBγ = PCVC But, PC = γ ⎛V ⎞ PAVA from above. Also considering the isothermal process, PB = PA ⎜ A ⎟ γ −1 VCVD ⎝ VB ⎠ ⎛V ⎞ γ ⎛ P Vγ ⎞ γ VV 10.0 L ( 24.0 L ) Hence, PA ⎜ A ⎟ VB = ⎜ A γA−1 ⎟ VC which reduces to VB = A C = = 16.0 L VD 15.0 L ⎝ VB ⎠ ⎝ VCVD ⎠ ⎛V ⎞ 10.0 L ⎞ Finally, PB = PA ⎜ A ⎟ = 1 400 kPa ⎛ = 875 kPa ⎝ 16.0 L ⎠ ⎝ VB ⎠ State P (kPa) V (L) T (K) A 1 400 10.0 720 B 875 16.0 720 C 445 24.0 549 D 712 15.0 549 continued on next page 13794_22_ch22_p571-600.indd 580 1/8/07 7:53:33 PM
  • Heat Engines, Entropy, and the Second Law of Thermodynamics (b) For the isothermal process A → B: 581 ∆Eint = nCV ∆T = 0 ⎛V ⎞ ⎛ 16.0 ⎞ = +6.58 kJ so Q = −W = nRT ln ⎜ B ⎟ = 2.34 mol (8.314 J mol ⋅ K ) ( 720 K ) ln ⎝ 10.0 ⎠ ⎝ VA ⎠ For the adiabatic process B → C : Q= 0 3 ∆Eint = nCV (TC − TB ) = 2.34 mol ⎡ (8.314 J mol ⋅ K ) ⎤ ( 549 − 720 ) K = −4.98 kJ ⎢2 ⎥ ⎦ ⎣ and W = −Q + ∆Eint = 0 + ( −4.98 kJ ) = −4.98 kJ For the isothermal process C → D: ∆Eint = nCV ∆T = 0 ⎛V ⎞ 15.0 ⎞ and Q = −W = nRT ln ⎜ D ⎟ = 2.34 mol (8.314 J mol ⋅ K ) ( 549 K ) ln ⎛ = −5.02 kJ ⎝ 24.0 ⎠ ⎝ VC ⎠ Finally, for the adiabatic process D → A: Q= 0 3 ∆Eint = nCV (TA − TD ) = 2.34 mol ⎡ (8.314 J mol ⋅ K ) ⎤ ( 720 − 549 ) K = +4.98 kJ ⎢2 ⎥ ⎦ ⎣ and W = −Q + ∆Eint = 0 + 4.98 kJ = +4.98 kJ Process Q (kJ) W (kJ) ∆Eint (kJ) A→B +6.58 −6.58 0 B→C 0 −4.98 −4.98 C→D −5.02 +5.02 0 D→A 0 +4.98 +4.98 ABCDA +1.56 −1.56 0 The work done by the engine is the negative of the work input. The output work Weng is given by the work column in the table with all signs reversed. (c) e= Weng Qh ec = 1 − P22.21 (a) = −WABCD 1.56 kJ = = 0.237 or 23.7% QA→ B 6.58 kJ Tc 549 = 1− = 0.237 or 23.7% Th 720 For a complete cycle, ∆Eint = 0 and ⎡ (Qh ) ⎤ W = Qh − Qc = Qc ⎢ − 1⎥ ⎢ Qc ⎥ ⎣ ⎦ The text shows that for a Carnot cycle (and only for a reversible cycle) Qh Th = Qc Tc Therefore, ⎡ T − Tc ⎤ W = Qc ⎢ h ⎥ ⎣ Tc ⎦ (b) We have the definition of the coefficient of performance for a refrigerator, Using the result from part (a), this becomes 13794_22_ch22_p571-600.indd 581 COP = Tc Th − Tc COP = Qc W 1/8/07 7:53:33 PM
  • 582 Chapter 22 Qc + W T 295 = h = = 11.8 W ∆T 25 P22.22 ( COP )heat pump = P22.23 ( COP )Carnot refrig = Q 4.00 Tc = = 0.013 8 = c ∆T 289 W ∴ W = 72.2 J per 1 J energy removed by heat. P22.24 COP = 0.100 COPCarnot cycle or ⎛Q ⎞ Qh ⎛ 1 ⎞ = 0.100 ⎜ h ⎟ = 0.100 ⎜ ⎝ Carnot efficiency ⎟ ⎠ W ⎝ W ⎠ Carnot cycle Qh ⎛ Th ⎞ 293 K ⎛ ⎞ = 0.100 ⎜ ⎟ = 0.100 ⎝ 293 K − 268 K ⎠ = 1.17 W ⎝ Th − Tc ⎠ FIG. P22.24 Thus, 1.17 joules of energy enter the room by heat for each joule of work done. t *P22.25 Qc Qc ր ∆t Tc = COPC ( refrigerator ) = = W Th − Tc W ր ∆t 0.150 W 260 K = W ր∆t 40.0 K W ⎛ 40.0 K ⎞ = 0.150 W ⎜ = 23.1 mW ⎝ 260 K ⎟ ⎠ ∆t P P22.26 = e= W = 0.350 Qh W = 0.350Qh Qh = W + Qc Qc = 0.650Qh COP ( refrigerator ) = Qc 0.650Qh = = 1.86 W 0.350Qh Section 22.5 P22.27 (a) Gasoline and Diesel Engines PViγ = Pf V fγ i γ 1.40 ⎛V ⎞ ⎛ 50.0 cm 3 ⎞ Pf = Pi ⎜ i ⎟ = ( 3.00 × 10 6 Pa ) ⎜ = 244 kPa ⎝ 300 cm 3 ⎟ ⎠ ⎝ Vf ⎠ Vi (b) W = ∫ PdV Vi V P = Pi ⎛ i ⎞ ⎝V⎠ γ Integrating, ⎡ ⎛ V ⎞ γ −1 ⎤ ⎛ 1 ⎞ W =⎜ PVi ⎢1 − ⎜ i ⎟ ⎥ i ⎝ γ − 1⎟ ⎠ ⎢ ⎝ Vf ⎠ ⎥ ⎦ ⎣ ⎡ ⎛ 50.0 cm 3 ⎞ 0.400 ⎤ = ( 2.50 ) ( 3.00 × 10 6 Pa ) ( 5.00 × 10 −5 m 3 ) ⎢1 − ⎜ ⎥ 3 ⎟ ⎢ ⎥ ⎣ ⎝ 300 cm ⎠ ⎦ = 192 J 13794_22_ch22_p571-600.indd 582 1/8/07 7:53:35 PM
  • Heat Engines, Entropy, and the Second Law of Thermodynamics P22.28 583 (a), (b) The quantity of gas is n= 3 −6 3 PAVA (100 × 10 Pa ) ( 500 × 10 m ) = = 0.020 5 mol RTA (8.314 J mol ⋅ K ) ( 293 K ) Eint, A = 5 5 5 nRTA = PAVA = (100 × 10 3 Pa ) ( 500 × 10 −6 m 3 ) = 125 J 2 2 2 γ ⎛V ⎞ 1.40 In process AB, PB = PA ⎜ A ⎟ = (100 × 10 3 Pa ) (8.00 ) = 1.84 × 10 6 Pa ⎝V ⎠ B TB = 6 −6 3 PBVB (1.84 × 10 Pa ) ( 500 × 10 m 8.00 ) = = 673 K nR ( 0.020 5 mol ) (8.314 J mol ⋅ K ) Eint, B = so 5 5 nRTB = ( 0.020 5 mol ) (8.314 J mol ⋅ K ) ( 673 K ) = 287 J 2 2 ∆Eint, AB = 287 J − 125 J = 162 J = Q − Wout = 0 − Wout WAB = −162 J Process BC takes us to: PC = nRTC ( 0.020 5 mol ) (8.314 J mol ⋅ K ) (1 023 K ) = = 2.79 × 10 6 Pa VC 62.5 × 10 −6 m 3 Eint, C = 5 5 nRTC = ( 0.020 5 mol ) (8.314 J mol ⋅ K ) (1 023 K ) = 436 J 2 2 Eint, BC = 436 J − 287 J = 149 J = Q − Wout = Q − 0 QBC = 149 J In process CD: γ 1.40 ⎛V ⎞ 1 ⎞ PD = PC ⎜ C ⎟ = ( 2.79 × 10 6 Pa ) ⎛ = 1.52 × 10 5 Pa ⎝ 8.00 ⎠ ⎝ VD ⎠ TD = 5 −6 3 PDVD (1.52 × 10 Pa ) ( 500 × 10 m ) = = 445 K nR ( 0.020 5 mol ) (8.314 J mol ⋅ K ) Eint, D = 5 5 nRTD = ( 0.020 5 mol ) (8.314 J mol ⋅ K ) ( 445 K ) = 190 J 2 2 ∆Eint, CD = 190 J − 436 J = −246 J = Q − Wout = 0 − Wout WCD = 246 J and ∆Eint, DA = Eint, A − Eint, D = 125 J − 190 J = −65.0 J = Q − Wout = Q − 0 QDA = −65.0 J For the entire cycle, ∆Eint, net = 162 J + 149 − 246 − 65.0 = 0 . The net work is Weng = −162 J + 0 + 246 J + 0 = 84.3 J Qnet = 0 + 149 J + 0 − 65.0 J = 84.3 J continued on next page 13794_22_ch22_p571-600.indd 583 1/8/07 7:53:36 PM
  • 584 Chapter 22 The tables look like: State T (K) P (kPa) V (cm3) Eint (J) A 293 100 500 125 B 673 1 840 62.5 287 C 1 023 2 790 62.5 436 D 445 152 500 190 A 293 100 500 125 Process Q (J) output W (J) ∆Eint (J) AB 0 −162 162 BC 149 0 149 CD 0 246 −246 DA −65.0 0 84.3 84.3 ABCDA −65.0 0 (c) The input energy is Qh = 149 J , the waste is Qc = 65.0 J , and Weng = 84.3 J . (d) The efficiency is: e = (e) f Let f represent the angular speed of the crankshaft. Then is the frequency at which we 2 obtain work in the amount of 84.3 Jրcycle: Weng Qh = 84.3 J = 0.565 149 J f 1 000 J s = ⎛ ⎞ (84.3 J cycle ) ⎝ 2⎠ f = P22.29 2 000 J s = 23.7 rev s = 1.42 × 10 3 rev min 84.3 J cycle Compression ratio = 6.00, γ = 1.40 (a) ⎛V ⎞ Efficiency of an Otto-engine e = 1 − ⎜ 2 ⎟ ⎝V ⎠ γ −1 1 1 ⎞ e = 1− ⎛ ⎝ 6.00 ⎠ (b) Section 22.6 P22.30 0.400 = 51.2% If actual efficiency e′ = 15.0% losses in system are e − e′ = 36.2% Entropy For a freezing process, ∆S = 5 ∆Q − ( 0.500 kg ) ( 3.33 × 10 J kg ) = = −610 J K T 273 K *P22.31 The process of raising the temperature of the sample in this way is reversible, because an infinitesimal change would make δ negative, and energy would flow out instead of in. Then we may find the entropy change of the sample as Tf Tf Ti Ti ∆S = ∫ dS = ∫ 13794_22_ch22_p571-600.indd 584 dQ T f mcdT = = mc ln T T ∫Ti T Tf Ti = mc ⎡ ln T f − ln Ti ⎤ = mc ln(T f րTi ) ⎣ ⎦ 1/8/07 7:53:37 PM
  • Heat Engines, Entropy, and the Second Law of Thermodynamics *P22.32 (a) (b) 585 The process is isobaric because it takes place under constant atmospheric pressure. As described by Newton’s third law, the stewing syrup must exert the same force on the air as the air exerts on it. The heating process is not adiabatic (because energy goes in by heat), isothermal (T goes up), isovolumetric (it likely expands a bit), cyclic (it is different at the end), or isentropic (entropy increases). It could be made as nearly reversible as you wish, by not using a kitchen stove but a heater kept always just incrementally higher in temperature than the syrup. The process would then also be eternal, and impractical for food production. The final temperature is 100 − 0°C ⎞ 220°F = 212°F + 8°F = 100°C + 8°F ⎛ = 104°C ⎝ 212 − 32°F ⎠ For the mixture, Q = m1c1∆T + m2 c2 ∆T = ( 900 g 1 cal g ⋅ °C + 930 g 0.299 cal g ⋅ °C ) (104.4°C − 23°C ) = 9.59 × 10 4 cal = 4.02 × 10 5 J (c) Consider the reversible heating process described in part (a): f ∆S = dQ ∫i T = f ∫ ( m1c1 + m2 c2 ) dT T i = ( m1c1 + m2 c2 ) ln Tf Ti 4.186 J ⎞ ⎛ 1°C ⎞ ⎛ 273 + 104 ⎞ = [ 900 (1) + 930 ( 0.299 )] ( cal °C ) ⎛ ln ⎝ 1 cal ⎠ ⎝ 1 K ⎠ ⎝ 273 + 23 ⎠ = ( 4 930 J K ) 0.243 = 1.20 × 10 3 J K f P22.33 ∆S = dQ ∫i T = Tf ∫ Ti ⎛ Tf ⎞ mcdT = mc ln ⎜ ⎟ T ⎝ Ti ⎠ 353 ⎞ ∆S = 250 g (1.00 cal g ⋅ °C ) ln ⎛ = 46.6 cal K = 195 J K ⎝ 293 ⎠ Section 22.7 Entropy Changes in Irreversible Processes Q2 Q1 ⎛ 1 000 1 000 ⎞ − = − J K = 3.27 J K ⎝ ⎠ T2 T1 ⎜ 290 5 700 ⎟ P22.34 ∆S = P22.35 The car ends up in the same thermodynamic state as it started, so it undergoes zero changes in entropy. The original kinetic energy of the car is transferred by heat to the surrounding air, adding to the internal energy of the air. Its change in entropy is ∆S = 13794_22_ch22_p571-600.indd 585 1 2 mv 2 750 ( 20.0 ) = T 293 2 J K = 1.02 kJ K 1/8/07 7:53:38 PM
  • 586 Chapter 22 *P22.36 Define T1 = Temp Cream = 5.00°C = 278 K. Define T2 = Temp Coffee = 60.0°C = 333 K (20.0 g)T1 + (200 g)T2 = 55.0°C = 328 K 220 g The final temperature of the mixture is: Tf = The entropy change due to this mixing is ∆S = ( 20.0 g ) ∫ Tf T1 T f c dT cV dT + ( 200 g ) ∫ V T2 T T ⎛ Tf ⎞ ⎛ Tf ⎞ 328 ⎞ 328 ⎞ ∆S = (84.0 J K ) ln ⎜ ⎟ + (840 J K ) ln ⎜ ⎟ = (84.0 K J ) ln ⎛ + (840 J K ) ln ⎛ ⎝ 333 ⎠ ⎝ 278 ⎠ ⎝ T1 ⎠ ⎝ T2 ⎠ ∆S = +1.18 J K P22.37 Sitting here writing, I convert chemical energy, in ordered molecules in food, into internal energy that leaves my body by heat into the room-temperature surroundings. My rate of energy output is equal to my metabolic rate, 2 500 kcal d = 2 500 × 10 3 cal ⎛ 4.186 J ⎞ = 120 W 86 400 s ⎝ 1 cal ⎠ My body is in steady state, changing little in entropy, as the environment increases in entropy at the rate ∆S Q T Q ∆t 120 W = = = = 0.4 W K ~ 1 W K ∆t ∆t T 293 K When using powerful appliances or an automobile, my personal contribution to entropy production is much greater than the above estimate, based only on metabolism. P22.38 ciron = 448 J kg ⋅ °C ; cwater = 4 186 J kg ⋅ °C ) ( ( Qcold = −Qhot: 4.00 kg ( 4 186 J kg ⋅ °C ) T f − 10.0°C = − (1.00 kg ) ( 448 J kg ⋅ °C ) T f − 900°C which yields ) T f = 33.2°C = 306.2 K 306.2 K ∆S = ∫ 283 K 306.2 K cwater mwater dT c m dT + ∫ iron iron T T 1173 K ⎛ 306.2 ⎞ 306.2 ⎞ + c m ln ∆S = cwater mwater ln ⎛ ⎝ 283 ⎠ iron iron ⎜ 1173 ⎟ ⎠ ⎝ ∆S = ( 4 186 J kg ⋅ K ) ( 4.00 kg ) ( 0.078 8 ) + ( 448 J kg ⋅ K ) (1.00 kg ) ( −1.34 ) ∆S = 718 J K P22.39 ⎛ Vf ⎞ ∆S = nR ln ⎜ ⎟ = R ln 2 = 5.76 J K ⎝ Vi ⎠ There is no change in temperature for an ideal gas. FIG. P22.39 13794_22_ch22_p571-600.indd 586 1/8/07 7:53:38 PM
  • Heat Engines, Entropy, and the Second Law of Thermodynamics P22.40 587 ⎛ Vf ⎞ ∆S = nR ln ⎜ ⎟ = ( 0.044 0 ) ( 2 ) R ln 2 ⎝ Vi ⎠ ∆S = 0.088 0 (8.314 ) ln 2 = 0.507 J K FIG. P22.40 Section 22.8 Entropy on a Microscopic Scale P22.42 (a) A 12 can only be obtained one way, as 6 + 6 (b) P22.41 A 7 can be obtained six ways: 6 + 1, 5 + 2, 4 + 3, 3 + 4, 2 + 5, 1 + 6 (a) The table is shown below. On the basis of the table, the most probable recorded result of a toss is 2 heads and 2 tails . (b) The most ordered state is the least likely macrostate. Thus, on the basis of the table this is either all heads or all tails . (c) The most disordered is the most likely macrostate. Thus, this is 2 heads and 2 tails . Result All heads 3H, 1T 2H, 2T 1H, 3T All tails Total 1 4 6 4 1 (a) Result All red 2R, 1G 1R, 2G All green Possible Combinations RRR RRG, RGR, GRR RGG, GRG, GGR GGG Total 1 3 3 1 (b) P22.43 Possible Combinations HHHH THHH, HTHH, HHTH, HHHT TTHH, THTH, THHT, HTTH, HTHT, HHTT HTTT, THTT, TTHT, TTTH TTTT Result All red 4R, 1G 3R, 2G 2R, 3G 1R, 4G All green 13794_22_ch22_p571-600.indd 587 Possible Combinations RRRRR RRRRG, RRRGR, RRGRR, RGRRR, GRRRR RRRGG, RRGRG, RGRRG, GRRRG, RRGGR, RGRGR, GRRGR, RGGRR, GRGRR, GGRRR GGGRR, GGRGR, GRGGR, RGGGR, GGRRG, GRGRG, RGGRG, GRRGG, RGRGG, RRGGG RGGGG, GRGGG, GGRGG, GGGRG, GGGGR GGGGG Total 1 5 10 10 5 1 1/8/07 7:53:40 PM
  • 588 Chapter 22 Additional Problems P22.44 The conversion of gravitational potential energy into kinetic energy as the water falls is reversible. But the subsequent conversion into internal energy is not. We imagine arriving at the same final state by adding energy by heat, in amount mgy, to the water from a stove at a temperature infinitesimally above 20.0°C. Then, 3 3 2 dQ Q mgy 5 000 m (1 000 kg m ) ( 9.80 m s ) ( 50.0 m ) = = = = 8.36 × 10 6 J K ∫T T T 293 K T 300 K *P22.45 For the Carnot engine, ec = 1 − c = 1 − = 0.600 Th 750 K ∆S = Weng Also, ec = so Qh = and Qc = Qh − Weng = 250 J − 150 J = 100 J (a) Qh = Weng eS = Qh Weng ec = 150 J = 250 J 0.600 FIG. P22.45 150 J = 214 J 0.700 Qc = Qh − Weng = 214 J − 150 J = 64.3 J (b) Qh,net = 214 J − 250 J = −35.7 J Qc,net = 64.3 J − 100 J = −35.7 J The net flow of energy by heat from the cold to the hot reservoir without work input, is impossible. For engine S: Qc = Qh − Weng = so (c) Weng = and (d) Weng eS − Weng Qc 100 J = 1 = 233 J − 1 0.700 − 1 FIG. P22.45(b) 1 eS Qh = Qc + Weng = 233 J + 100 J = 333 J Qh,net = 333 J − 250 J = 83.3 J Wnet = 233 J − 150 J = 83.3 J Qc,net = 0 The output of 83.3 J of energy from the heat engine by work in a cyclic process without any exhaust by heat is impossible. FIG. P22.45(d) continued on next page 13794_22_ch22_p571-600.indd 588 1/8/07 7:53:42 PM
  • Heat Engines, Entropy, and the Second Law of Thermodynamics Both engines operate in cycles, so ∆SS = ∆SCarnot = 0 For the reservoirs, (e) ∆Sh = − 589 Qh Q and ∆Sc = + c Th Tc Thus, ∆Stotal = ∆SS + ∆SCarnot + ∆Sh + ∆Sc = 0 + 0 − 83.3 J 0 + = −0.111 J K 750 K 300 K A decrease in total entropy is impossible. *P22.46 (a) Let state i represent the gas before its compression and state f afterwards, V f = diatomic ideal gas, Cv = 5 7 R, C p = R, 2 2 γ = and Cp CV Vi . For a 8 = 1.40. Next, PViγ = Pf V fγ i γ ⎛V ⎞ Pf = Pi ⎜ i ⎟ = Pi 81.40 = 18.4 Pi ⎝ Vf ⎠ PiVi = nRTi Pf V f = so T f = 2.30Ti 18.4 PVi i = 2.30 PVi = 2.30 nRTi = nRT f i 8 ( ) 5 5 5 ∆Eint = nCV ∆T = n R T f − Ti = nR1.30Ti = 1.30 PiVi 2 2 2 5 = 1.30 (1.013 × 10 5 N m 2 ) 0.12 × 10 −3 m 3 = 39.4 J 2 Since the process is adiabatic, (b) Q=0 ∆Eint = Q + W gives W = 39.4 J and 1 1 2 MR 2 = 5.1 kg ( 0.085 m ) = 0.018 4 kg ⋅ m 2 2 2 We want the flywheel to do work 39.4 J, so the work on the flywheel should be −39.4 J: The moment of inertia of the wheel is I = K rot i + W = K rot f 1 2 Iω i − 39.4 J = 0 2 ⎛ 2 ( 39.4 J ) ⎞ ωi = ⎜ ⎝ 0.018 4 kg ⋅ m 2 ⎟ ⎠ (c) 12 = 65.4 rad s Now we want W = 0.05K rot i 1 39.4 J = 0.05 0.018 4 kg ⋅ m 2ω i2 2 ⎛ 2 ( 789 J ) ⎞ ω =⎜ ⎝ 0.018 4 kg ⋅ m 2 ⎟ ⎠ 13794_22_ch22_p571-600.indd 589 12 = 293 rad s 1/8/07 7:53:43 PM
  • 590 P22.47 Chapter 22 (a) H ET so if all the electric energy is converted into internal energy, the steady-state ∆t condition of the house is described by H ET = Q . P electric = Therefore, P (b) For a heat pump, Q = 5 000 W ∆t T 295 K = h = = 10.92 27 K ∆T electric ( COP )Carnot = Qh Q ∆t = h W W ∆t Therefore, to bring 5 000 W of energy into the house only requires input power Pheat pump = W = Qh ∆t = 5 000 W = 763 W ∆t COP 6.56 −1 000 J ∆Shot = 600 K +750 J ∆Scold = 350 K Actual COP = 0.6 (10.92 ) = 6.55 = P22.48 (a) ∆SU = ∆Shot + ∆Scold = 0.476 J K (b) ec = 1 − T1 = 0.417 T2 Weng = ec Qh = 0.417 (1 000 J ) = 417 J (c) Wnet = 417 J − 250 J = 167 J T1 ∆SU = 350 K ( 0.476 J K ) = 167 J ⎛V ⎞ Q = nRT ln ⎜ 2 ⎟ ⎝ V1 ⎠ Q1 = nR ( 3Ti ) ln 2 and (a) For an isothermal process, Therefore, P22.49 1 Q3 = nR (Ti ) ln ⎛ ⎞ ⎝ 2⎠ 3 nR (Ti − 3Ti ) 2 3 = nR ( 3Ti − Ti ) 2 For the constant volume processes, Q2 = ∆Eint, 2 = and Q4 = ∆Eint, 4 The net energy by heat transferred is then Q = Q1 + Q2 + Q3 + Q4 or (b) FIG. P22.49 Q = 2nRTi ln 2 A positive value for heat represents energy transferred into the system. Therefore, Qh = Q1 + Q4 = 3nRTi (1 + ln 2 ) Since the change in temperature for the complete cycle is zero, ∆Eint = 0 and Weng = Q Therefore, the efficiency is 13794_22_ch22_p571-600.indd 590 ec = Weng Qh = Q 2 ln 2 = = 0.273 Qh 3 (1 + ln 2 ) 1/8/07 7:53:44 PM
  • Heat Engines, Entropy, and the Second Law of Thermodynamics P22.50 (a) 591 5 35.0°F = ( 35.0 − 32.0 ) °C = (1.67 + 273.15) K = 274.82 K 9 5 98.6°F = ( 98.6 − 32.0 ) °C = ( 37.0 + 273.15) K = 310.15 K 9 310.15 ∆Sice water = dQ dT ⎛ 310.15 ⎞ ∫ T = ( 453.6 g) (1.00 cal g ⋅ K ) × 274∫.82 T = 453.6 ln ⎝ 274.82 ⎠ = 54.86 cal K ∆Sbody = − Q ( 310.15 − 274.82 ) = − ( 453.6 ) (1.00 ) = −51.67 cal K 310.15 Tbody ∆Ssystem = 54.86 − 51.67 = 3.19 cal K (b) ( 453.6 ) (1) (TF − 274.82 ) = ( 70.0 × 10 3 ) (1) ( 310.15 − TF ) Thus, ( 70.0 + 0.453 6) × 103 TF = ⎡( 70.0 ) ( 310.15) + ( 0.453 6) ( 274.82)⎤ × 103 ⎣ ⎦ and TF = 309.92 K = 36.77°C = 98.19°F 309.92 ⎞ ∆Sice water = 453.6 ln ⎛ = 54.52 cal K ′ ⎝ 274.82 ⎠ 310.15 ⎞ ∆Sbody = − ( 70.0 × 10 3 ) ln ⎛ = −51.93 cal K ′ ⎝ 309.92 ⎠ ∆Ssys = 54.52 − 51.93 = 2.59 cal K ′ This is significantly less than the estimate in part (a). e P22.51 ec = 1 − Tc Weng Weng / ∆t : = = Th Qh Qh / ∆t Qh = Weng + Qc : Qc = mc∆T : Qh P = P Th = ∆t (1 − Tc Th ) Th − Tc Weng Qc Q = h − ∆t ∆t ∆t Qc P Th − P = P Tc = Th − Tc ∆t Th − Tc Qc ⎛ ∆m ⎞ P Tc c∆T = =⎜ ⎝ ∆t ⎟ ⎠ Th − Tc ∆t ∆m P Tc = ∆t (Th − Tc ) c∆T (1.00 × 109 W ) (300 K ) = 5.97 × 10 4 kg s ∆m = ∆t 200 K ( 4 186 J kg ⋅ °C ) ( 6.00°C ) 13794_22_ch22_p571-600.indd 591 1/8/07 7:53:46 PM
  • 592 P22.52 Chapter 22 ec = 1 − Tc Weng Weng / ∆t = = Th Qh Qh / ∆t Qh P P Th = = ∆t 1 − (Tc / Th ) Th − Tc Qc ⎛ Qh ⎞ P Tc = −P = Th − Tc ∆t ⎜ ∆t ⎟ ⎝ ⎠ Qc = mc∆T , where c is the specific heat of water. Therefore, Qc ⎛ ∆m ⎞ P Tc =⎜ ⎟ c∆T = Th − Tc ∆t ⎝ ∆t ⎠ and ∆m = ∆t PT c (Th − Tc ) c∆T We test for dimensional correctness by identifying the units of the right-hand side: W ⋅ °C ( J s ) kg = = kg s , as on the left hand side. Think of yourself as a power-company °C ( J kg ⋅ °C ) °C J engineer arranging to have enough cooling water to carry off your thermal pollution. If the plant power P increases, the required flow rate increases in direct proportion. If environmental regulations require a smaller temperature change ∆T, then the required flow rate increases again, now in inverse proportion. Next note that Th is in the bottom of the fraction. This means that if you can run the reactor core or firebox hotter, the required coolant flow rate decreases! If the turbines take in steam at higher temperature, they can be made more efficient to reduce waste heat output. 13794_22_ch22_p571-600.indd 592 1/8/07 7:53:47 PM
  • Heat Engines, Entropy, and the Second Law of Thermodynamics P22.53 593 Like a refrigerator, an air conditioner has as its purpose the removal of energy by heat from the cold reservoir. Tc 280 K = = 14.0 Th − Tc 20 K Its ideal COP is COPCarnot = (a) 0.400 (14.0 ) = 5.60 = Its actual COP is 5.60 Qh Q Q − 5.60 c = c ∆t ∆t ∆t 5.60 (10.0 kW ) = 6.60 Weng Qc Qc ∆t = Qh − Qc Qh ∆t − Qc ∆t = Qc ∆t and Qc = 8.48 kW ∆t Qh Qc − = 10.0 kW − 8.48 kW = 1.52 kW ∆t ∆t (b) Qh = Weng + Qc : (c) The air conditioner operates in a cycle, so the entropy of the working fluid does not change. The hot reservoir increases in entropy by ∆t Qh (10.0 × 10 3 J s ) ( 3 600 s ) = = 1.20 × 10 5 J K 300 K Th The cold room decreases in entropy by ∆S = − Qc (8.48 × 103 J s ) (3 600 s ) = −1.09 × 105 J K =− Tc 280 K The net entropy change is positive, as it must be: +1.20 × 10 5 J K − 1.09 × 10 5 J K = 1.09 × 10 4 J K Tc 280 K = = 11.2 Th − Tc 25 K The new ideal COP is COPCarnot = We suppose the actual COP is (d) 0.400 (11.2 ) = 4.48 As a fraction of the original 5.60, this is drop by 20.0% . 13794_22_ch22_p571-600.indd 593 4.48 = 0.800, so the fractional change is to 5.60 1/8/07 7:53:47 PM
  • 594 Chapter 22 *P22.54 (a) For the isothermal process AB, the work on the gas is ⎛V ⎞ WAB = − PAVA ln ⎜ B ⎟ ⎝ VA ⎠ 50.0 ⎞ WAB = −5 (1.013 × 10 5 Pa ) (10.0 × 10 −3 m 3 ) ln ⎛ ⎝ 10.0 ⎠ WAB = −8.15 × 10 3 J where we have used 1.00 atm = 1.013 × 10 5 Pa and 1.00 L = 1.00 × 10 −3 m 3 FIG. P22.54 WBC = − PB ∆V = − (1.013 × 10 5 Pa ) ⎡(10.0 − 50.0 ) × 10 −3 ⎤ m 3 = +4.05 × 10 3 J ⎣ ⎦ WCA = 0 and Weng = −WAB − WBC = 4.10 × 10 3 J = 4.10 kJ Since AB is an isothermal process, ∆Eint, AB = 0 and (b) QAB = −WAB = 8.15 × 10 3 J For an ideal monatomic gas, CV = 3R 5R and CP = 2 2 )( ( ) 1.013 × 10 5 50.0 × 10 −3 PBVB 5.06 × 10 3 TB = TA = = = R nR R PCVC (1.013 × 10 ) (10.0 × 10 = nR R 5 Also, TC = −3 ) = 1.01 × 10 3 R ⎛ 3 ⎞ ⎛ 5.06 × 10 3 − 1.01 × 10 3 ⎞ QCA = nCV ∆T = 1.00 ⎜ R⎟ ⎜ ⎟ R ⎝ 2 ⎠⎝ ⎠ = 6.08 kJ so the total energy absorbed by heat is QAB + QCA = 8.15 kJ + 6.08 kJ = 14.2 kJ (c) 5 5 QBC = nCP ∆T = ( nR∆T ) = PB ∆VBC 2 2 5 (1.013 × 105 ) ⎡(10.0 − 50.0 ) × 10 −3 ⎤ = −1.01 × 10 4 J = −10.1 kJ ⎣ ⎦ 2 Weng Weng 4.10 × 10 3 J e= = = = 0.288 or 28.8% Qh QAB + QCA 1.42 × 10 4 J QBC = (d) (e) A Carnot engine operating between Thot = TA = 5060րR and Tcold = TC = 1010րR has efficiency 1 − Tc րTh = 1 − 1ր5 = 80.0%. The three-process engine considered in this problem has much lower efficiency. 13794_22_ch22_p571-600.indd 594 1/8/07 7:53:48 PM
  • Heat Engines, Entropy, and the Second Law of Thermodynamics PVi = nRTi i and n = 1.00 mol At point B, 3PVi = nRTB i so TB = 3Ti At point C, (3Pi ) ( 2Vi ) = nRTC and TC = 6Ti At point D, Pi ( 2Vi ) = nRTD so 595 TD = 2Ti *P22.55 At point A, The heat for each step in the cycle is found using CV = and CP = 5R : 2 3R 2 QAB = nCV ( 3Ti − Ti ) = 3nRTi FIG. P22.55 QBC = nCP ( 6Ti − 3Ti ) = 7.50 nRTi QCD = nCV ( 2Ti − 6Ti ) = −6nRTi QDA = nCP (Ti − 2Ti ) = −2.50 nRTi Qentering = Qh = QAB + QBC = 10.5nRTi (a) Therefore, (b) Qleaving = Qc = QCD + QDA = 8.50 nRTi (c) Actual efficiency, e= (d) Carnot efficiency, ec = 1 − Qh − Qc = 0.190 Qh Tc T = 1 − i = 0.833 Th 6Ti The Carnot efficiency is much higher. f P22.56 ∆S = f f dQ nCP dT −1 ∫i T = ∫i T = nCP ∫i T dT = nCP ln T Tf Ti ⎛ Tf ⎞ = nCP ln T f − ln Ti = nCP ln ⎜ ⎟ ⎝ Ti ⎠ ( ) ⎛ PV f nR ⎞ = nCP ln 3 ∆S = nCP ln ⎜ ⎠ ⎝ nR PVi ⎟ 13794_22_ch22_p571-600.indd 595 1/8/07 7:53:49 PM
  • 596 P22.57 Chapter 22 (a) The ideal gas at constant temperature keeps constant internal energy. As it puts out energy by work in expanding it must take in an equal amount of energy by heat. Thus its entropy increases. Let Pi , Vi, Ti represent the state of the gas before the isothermal expansion. Let PC, VC, Ti represent the state after this process, so that PiVi = PCVC. Let Pi, 3Vi, Tf represent the state after the adiabatic compression. Then γ PCVC = Pi ( 3Vi ) Substituting PC = gives γ PViVC −1 = Pi ( 3γ Viγ ) i Then γ VC −1 = 3γ Viγ −1 and γ PVi i VC VC = 3γ Vi (γ −1) The work output in the isothermal expansion is C C ⎛V ⎞ W = ∫ PdV = nRTi ∫ V −1dV = nRTi ln ⎜ C ⎟ = nRTi ln 3γ ⎝ Vi ⎠ i i This is also the input heat, so the entropy change is ( ∆S = (γ − 1) CV and CP = γR γ −1 R γ −1 The pair of processes considered here carries the gas from the initial state in Problem 56 to the final state there. Entropy is a function of state. Entropy change does not depend on path. Therefore the entropy change in Problem 56 equals ∆Sisothermal + ∆Sadiabatic in this problem. Since ∆Sadiabatic = 0, the answers to Problems 56 and 57(a) must be the same. Vf W= ∫ PdV = nRT Vi 13794_22_ch22_p571-600.indd 596 = R , CV = ∆S = nCP ln 3 Then the result is (b) i CP = γ CV = CV + R we have *P22.58 (a) ) = nRT ⎛ γ γ− 1⎞ ln 3 ⎜ ⎟ ⎝ ⎠ ⎛ γ ⎞ Q = nR ⎜ ln 3 T ⎝ γ − 1⎟ ⎠ Since (b) (γ −1) 2Vi ∫ Vi ⎛ 2V ⎞ dV = (1.00 ) RT ln ⎜ i ⎟ = RT ln 2 V ⎝ Vi ⎠ While it lasts, this process does convert all of the energy input into work output. But the gas sample is in a different state at the end than it was at the beginning. The process cannot be done over unless the gas is recompressed by a work input. To be practical, a heat engine must operate in a cycle. The second law refers to a heat engine operating in a cycle, so this process is consistent with the second law of thermodynamics. 1/8/07 7:53:50 PM
  • Heat Engines, Entropy, and the Second Law of Thermodynamics P22.59 The heat transfer over the paths CD and BA is zero since they are adiabatic. Over path BC: Qc = QDA Adiabatic Processes QDA = nCV (TA − TD ) < 0 Therefore, P QBC = nCP (TC − TB ) > 0 Over path DA: 597 and B C D Qh = QBC A The efficiency is then e = 1− Qc (T − TA ) CV = 1− D Qh (TC − TB ) CP Vi V FIG. P22.59 1 ⎡ T − TA ⎤ e = 1− ⎢ D γ ⎣ TC − TB ⎥ ⎦ P22.60 3Vi Simply evaluate the maximum (Carnot) efficiency. eC = ∆T 4.00 K = = 0.014 4 Th 277 K The proposal does not merit serious consideration. Operating between these temperatures, this device could not attain so high an efficiency. *P22.61 (a) 20.0°C Tf ⎤ ⎡ Tf = 1.00 kg ( 4.19 kJ kg ⋅ K ) ⎢ ln + ln ⎥ T1 T2 ⎦ ⎣ ⎛ 293 293 ⎞ ⋅ = ( 4.19 kJ K ) ln ⎜ ⎝ 283 303 ⎟ ⎠ Tf (c) ∆S = +4.88 J K (d) 13794_22_ch22_p571-600.indd 597 ∆S = mc ln Yes, the mixing is irreversible. Entropy has increased. s T1 + mc ln Tf (b) T2 1/8/07 7:53:51 PM
  • 598 P22.62 Chapter 22 (a) Use the equation of state for an ideal gas nRT P 1.00 (8.314 ) ( 600 ) VA = = 1.97 × 10 −3 m 3 9 25.0 (1.013 × 10 5 ) V= VC = 1.00 (8.314 ) ( 400 ) = 32.8 × 10 −3 m 3 8 1.013 × 10 5 FIG. P22.62 Since AB is isothermal, PAVA = PBVB and since BC is adiabatic, γ PBVBγ = PCVC Combining these expressions, ⎡⎛ P ⎞ V γ ⎤ VB = ⎢⎜ C ⎟ C ⎥ ⎣⎝ PA ⎠ VA ⎦ 1 (γ −1) ⎡⎛ 1.00 ⎞ ( 32.8 × 10 −3 m 3 )1.40 ⎤ ⎥ =⎢ −3 3 ⎥ ⎢⎝ 25.0 ⎠ 1.97 × 10 m ⎦ ⎣ (1 0.400 ) ⎡ 25.0 ⎞ (1.97 × 10 −3 m 3 )1.40 ⎤ ⎥ = ⎢⎛ −3 3 ⎢⎝ 1.00 ⎠ 32.8 × 10 m ⎥ ⎣ ⎦ (1 0.400 ) VB = 11.9 × 10 −3 m 3 1 (γ −1) Similarly, ⎡⎛ P ⎞ V γ ⎤ VD = ⎢⎜ A ⎟ A ⎥ ⎣⎝ PC ⎠ VC ⎦ or VD = 5.44 × 10 −3 m 3 Since AB is isothermal, PAVA = PBVB and ⎛V ⎞ ⎛ 1.97 × 10 −3 m 3 ⎞ PB = PA ⎜ A ⎟ = 25.0 atm ⎜ = 4.14 atm ⎝ 11.9 × 10 −3 m 3 ⎟ ⎠ ⎝ VB ⎠ Also, CD is an isothermal and ⎛V ⎞ ⎛ 32.8 × 10 −3 m 3 ⎞ PD = PC ⎜ C ⎟ = 1.00 atm ⎜ = 6.03 atm ⎝ 5.44 × 10 −3 m 3 ⎟ ⎠ ⎝ VD ⎠ Solving part (c) before part (b): ec = 1 − Tc 400 K = 1− = 0.333 Th 600 K (c) For this Carnot cycle, (b) Energy is added by heat to the gas during the process AB. For the isothermal process, ∆Eint = 0. ⎛V ⎞ and the first law gives QAB = −WAB = nRTh ln ⎜ B ⎟ ⎝ VA ⎠ ⎛ 11.9 ⎞ or Qh = QAB = 1.00 mol (8.314 J mol ⋅ K ) ( 600 K ) ln ⎜ ⎝ 1.97 ⎟ ⎠ = 8.97 kJ Then, from e= Weng Qh the net work done per cycle is Weng = ec Qh = 0.333 (8.97 kJ ) = 2.99 kJ 13794_22_ch22_p571-600.indd 598 1/8/07 7:53:52 PM
  • Heat Engines, Entropy, and the Second Law of Thermodynamics 599 ANSWERS TO EVEN PROBLEMS P22.2 13.7°C P22.4 (a) 29.4 Lրh P22.6 (a) 24.0 J P22.8 (a) 2.93 (b) coefficient of performance for a refrigerator (c) The cost for air conditioning is half as much for an air conditioner with EER 10 compared with an air conditioner with EER 5. P22.10 (a) 870 MJ P22.12 (a) 0.300 P22.14 33.0% P22.16 (a) 5.12% (b) 5.27 TJրh a good buy. P22.18 (a) |Qc|ր∆t = 700 kW(Th + 766 K)ր(Th − 383 K) The exhaust power decreases as the firebox temperature increases. (b) 1.87 MW (c) 3.84 × 103 K (d) No answer exists. The energy exhaust cannot be that small. P22.20 (a) State A B C D (b) (c) 527 N . m (b) 185 hp (d)1.91 × 105 W (b) 144 J (b) 330 MJ (b) 1.40 × 10−3 (c) 2.00 × 10−3 Process A→B B→C C→D D→A ABCDA P (kPa) 1 400 875 445 712 (c) As fossil-fuel prices rise, this way to use solar energy will become V (L) T (K) 10.0 720 16.0 720 24.0 549 15.0 549 Q (kJ) 6.58 0 −5.02 0 1.56 W (kJ) −6.58 −4.98 5.02 4.98 −1.56 ∆Eint (kJ) 0 −4.98 0 4.98 0 (c) 23.7%; see the solution P22.22 11.8 P22.24 1.17 J P22.26 1.86 P22.28 (a), (b) see the solution (e) 1.42 × 103 revրmin P22.30 −610 JրK 13794_22_ch22_p571-600.indd 599 (c) Qh = 149 J; Qc = 65.0 J; Weng = 84.3 J (d) 56.5% 1/8/07 7:53:53 PM
  • 600 Chapter 22 P22.32 (a) The process is isobaric because it takes place under constant atmospheric pressure. The heating process is not adiabatic (because energy goes in by heat), isothermal (T goes up), isovolumetric (it likely expands a bit), cyclic (it is different at the end), or isentropic (entropy increases). It could be made as nearly reversible as you wish, by not using a kitchen stove but a heater kept always just incrementally higher in temperature than the syrup. (b) 402 kJ (c) 1.20 kJրK P22.34 3.27 JրK P22.36 +1.18 JրK P22.38 718 JրK P22.40 0.507 JրK P22.42 (a) 2 heads and 2 tails P22.44 8.36 MJրK P22.46 (a) 39.4 J P22.48 (a) 0.476 JրK (b) 417 J P22.50 (a) 3.19 calրK (b) 98.19ºF, 2.59 calրK This is significantly less than the estimate in part (a). P22.52 (Th − Tc ) c∆T P22.54 (a) 4.10 kJ (b) 14.2 kJ (c) 10.1 kJ (d) 28.8% (e) The three-process engine considered in this problem has much lower efficiency than the Carnot efficiency. P22.56 nCpln3 P22.58 (a) see the solution (b) While it lasts, this process does convert all of the energy input into work output. But the gas sample is in a different state at the end than it was at the beginning. The process cannot be done over unless the gas is recompressed by a work input. To be practical, a heat engine must operate in a cycle. The second law refers to a heat engine operating in a cycle, so this process is consistent with the second law of thermodynamics. P22.60 The proposal does not merit serious consideration. Operating between these temperatures, this device could not attain so high an efficiency. P22.62 (a) (b) All heads or all tails (b) 65.4 radրs = 625 revրmin (c) 293 radրs = 2 790 revրmin (c) Wnet = T1 ∆ SU = 167 J PT c P, atm A V, L 25.0 1.97 B 4.14 11.9 C 1.00 32.8 D 6.03 (b) 2.99kJ 13794_22_ch22_p571-600.indd 600 (c) 2 heads and 2 tails 5.44 (c) 33.3% 1/8/07 7:53:53 PM