PSE9e ISM chapter22 final tủ tài liệu training

22 Heat Engines, Entropy, and the Second Law of Thermodynamics CHAPTER OUTLINE 22.1 Heat Engines and the Second Law of Thermodynamics 22.2 Heat Pumps and Refrigerators 22.3 Reversible and Irreversible Processes 22.4 The Carnot Engine 22.5 Gasoline and Diesel Engines 22.6 Entropy 22.7 Changes in Entropy for Thermodynamic Systems 22.8 Entropy and the Second Law * An asterisk indicates a question or problem new to this edition ANSWERS TO OBJECTIVE QUESTIONS OQ22.1 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 OQ22.2 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, lowerentropy 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 higherentropy final state 1147 © 2014 Cengage Learning All Rights Reserved May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part 1148 OQ22.3 Heat Engines, Entropy, and the Second Law of Thermodynamics Answer (c) The coefficient of performance of this refrigerator is COP = Qc W = 115 kJ = 6.39 18.0 kJ OQ22.4 Answer (c) Choice (c) is a statement of the first law of thermodynamics, not the second law Choices (a), (b), (d), and (e) are alternative statements of the second law, (a) being the Kelvin-Planck formulation, (b) the Carnot statement, (d) the Clausius statement, and (e) summarizes the primary consequence of all these various statements OQ22.5 (i) Answer (b) (ii) Answer (a) (iii) Answer (b) (iv) Answer (a) (v) Answer (c) (vi) Answer (a) 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), and (vi) but not by processes (i) and (iii) 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 process (v) OQ22.6 Answer (a) The air conditioner operates on a cyclic process so the change in the internal energy of the refrigerant is zero Then, the conservation of energy gives the thermal energy exhausted to the room as Qh = Qc + Weng, where Qc is the thermal energy the air conditioner removes from the room and Weng is the work done to operate the device Since Weng > 0, the air conditioner is returning more thermal energy to the room than it is removing, so the average temperature in the room will increase OQ22.7 Answer (c) The maximum theoretical efficiency (the Carnot efficiency) of a device operating between absolute temperatures Tc < Th is ec = − Tc/Th For the given steam turbine, this is ec = − 3.0 × 102 K = 0.33 or 33% 450 K OQ22.8 Answer (d) 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 J/K, or more OQ22.9 Answer: E > D > C > B > A Recall that for and ideal gas, PV = nRT, and Cv = 3R/2 and Cp = 5R/2 Process A: isobaric, volume V goes to 0.5V, so temperature T goes to 0.5T, dQ = nCpdT, so dS = nCpdT/T; therefore ΔS = − nR ln 2 © 2014 Cengage Learning All Rights Reserved May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part Chapter 22 1149 Process B: isothermal, volume V goes to 0.5V, so temperature T is constant and pressure P goes to 2P, dQ = pdV = nRT(dV/V), so dS = nR(dV/V); therefore ΔS = –nR ln Process C: adiabatic, Q = 0; therefore ΔS = Process D: isovolumetric, pressure P goes to 2P, so temperature T goes to 2T, dQ = nCvdT, so dS = n Cv dT/T; therefore ΔS = nR ln 2 Process E: isobaric, volume V goes to 2V; therefore ΔS = OQ22.10 nR ln 2 Answer (b) From conservation of energy, the energy input to the engine must be Qh = Weng + Qc = 15.0 kJ + 37.0 kJ = 52.0 kJ so the efficiency is e= OQ22.11 Weng Qc = 15.0 kJ = 0.288 or 28.8% 52.0 kJ 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 Point A is at a lower temperature than O and point C is at an even lower temperature The only point that could possibly have the same temperature as O is point B ANSWERS TO CONCEPTUAL QUESTIONS CQ22.1 (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 © 2014 Cengage Learning All Rights Reserved May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part 1150 Heat Engines, Entropy, and the Second Law of Thermodynamics 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 CQ22.2 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 CQ22.3 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 > CQ22.4 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 potential 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 internal energy 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 internal energy in the atmosphere CQ22.5 Either statement can be considered an instructive analogy We choose to take the first view All processes require energy, either as energy content or as energy input The kinetic energy that it possessed at its formation continues to make the Earth go around Energy released by nuclear reactions in the core of the Sun drives weather on the Earth and essentially all processes in the biosphere © 2014 Cengage Learning All Rights Reserved May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part Chapter 22 1151 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 CQ22.6 (a) 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 (b) The free flight of a projectile is nearly reversible CQ22.7 (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 CQ22.8 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 = − c and is maximized for a high Th Th Th CQ22.9 (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 entropy ⎛V ⎞ ΔQ change is ΔS = = nR ln ⎜ ⎟ T ⎝ V1 ⎠ (b) For a reversible adiabatic expansion, ΔQ = and ΔS = An ideal gas undergoing an irreversible adiabatic expansion can have any positive value for ΔS up to the value given in part (a) CQ22.10 No Your roommate creates “order” locally, but as she works, she transfers energy by heat to the room, causing the net entropy to increase An analogy used by Carnot is instructive: A waterfall continuously converts mechanical energy into internal energy It © 2014 Cengage Learning All Rights Reserved May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part 1152 Heat Engines, Entropy, and the Second Law of Thermodynamics continuously creates entropy as the motion of the falling water turns into molecular motion at the bottom of the falls We humans put turbines into the waterfall, diverting some of the energy stream to our use Water flows spontaneously from high to low elevation and energy spontaneously flows by heat from high to low temperature Into the great flow of solar radiation from Sun to Earth, living things put themselves They live on energy flow, more than just on energy A basking snake diverts energy from a high-temperature source (the Sun) through itself temporarily, before the energy inevitably is radiated from the body of the snake to a low-temperature sink (outer space) A tree builds 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 We 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 CQ22.11 No An engine with no thermal pollution would absorb energy from a reservoir and convert it completely into work; this is a clear violation of the second law of thermodynamics CQ22.12 (a) Shaking opens up spaces between jellybeans The smaller ones more often can fall down into spaces below them (b) The accumulation of larger candies on top and smaller ones on the bottom implies 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 small decrease in gravitational potential energy as the beans settle compactly together CQ22.13 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 © 2014 Cengage Learning All Rights Reserved May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part Chapter 22 1153 SOLUTIONS TO END-OF-CHAPTER PROBLEMS Section 22.1 P22.1 (a) Heat Engines and the Second Law of Thermodynamics   We have e = Weng Qh = 1− Qc Qh → With Qc = 8 000 J, we have Qh = (b) Qc Qh Qc 1− e = 1− e = → Qh = Qc 1− e 8 000 J =  10.7 kJ − 0.250 The work per cycle is Weng = Qh – Qc = 667 J From the definition of output power, P= Weng Δt we have the time for one cycle: Δt = P22.2 (a) Weng 667 J = = 0.533 s P 000 J/s The efficiency of a heat engine is e = Wenv Qh , where Wenv is the work done by the engine and Qh is the energy absorbed from the higher temperature reservoir Thus, if Wenv = Qh , the efficiency is e = = 0.25 or 25% (b) From conservation of energy, the energy exhausted to the lower temperature reservoir is Qc = Qh − Wenv Therefore, if Wenv = Qh , we have Qc = Qh or Qc Qh = P22.3 (a) The efficiency of the engine is e= (b) Weng Qh = 25.0 J = 0.069 4 or 6.94% 360 J The energy expelled to the cold reservoir during each cycle is Qc = Qh − Weng = 360 J − 25.0 J = 335 J © 2014 Cengage Learning All Rights Reserved May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part 1154 P22.4 Heat Engines, Entropy, and the Second Law of Thermodynamics The engine’s output work we identify with the kinetic energy of the bullet: Weng = K = e= 2 mv = ( 0.002 4 kg ) ( 320 m s ) = 123 J 2 Weng Qh Weng 123 J = 1.12 × 10 J e 0.011 Qh = Weng + Qc Qh = = The energy exhaust is Qc = Qh − Weng = 1.12 × 10 J − 123 J = 1.10 × 10 J Q = mcΔT ΔT = P22.5 (a) Q 1.10 × 10 J ⋅ kg°C = = 13.7°C mc (1.80 kg)(448 J) The engine’s efficiency is given by e= Weng Qh = Qh − Qc Q 1.20 kJ = 1− c = 1− Qh Qh 1.70 kJ = 0.294 (or 29.4%) (b) During each cycle, the work done by the engine is Weng = Qh − Qc = 1.70 kJ − 1.20 kJ = 5.00 × 102 J (c) The power transferred out of the engine is P= P22.6 (a) Weng Δt = 5.00 × 102 J = 1.67 × 103 W = 1.67 kW 0.300 s The input energy each hour is (7.89 × 10 ⎛ 60 ⎞ J revolution ) ( 2 500 rev ) ⎜ ⎝ h ⎟⎠ = 1.18 × 109 J h ⎛ 1L ⎞ = 29.4 L h implying fuel input ( 1.18 × 109 J h ) ⎜ ⎟ ⎝ 4.03 × 10 J ⎠ © 2014 Cengage Learning All Rights Reserved May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part Chapter 22 (b) 1155 Qh = Weng + Qc For a continuous-transfer process we may divide by time to have Qh Weng Qc = + Δt Δt Δt Useful power output = Weng Δt = Qh Qc − Δt Δt ⎛ 7.89 × 103 J 4.58 × 103 J ⎞ ⎛ 2 500 rev ⎞ ⎛ ⎞ =⎜ − ⎜ ⎟⎜ ⎟ ⎝ revolution revolution ⎟⎠ ⎝ ⎠ ⎝ 60 s ⎠ = 1.38 × 105 W ⎛ hp ⎞ Peng = ( 1.38 × 105 W ) ⎜ = 185 hp ⎝ 746 W ⎟⎠ (c) (d) P22.7 Peng ⎛ 1.38 × 105 J s ⎞ ⎛ rev ⎞ = τω ⇒ τ = =⎜ ⎜ ⎟ = 527 N ⋅ m ω ⎝ 2 500 rev 60 s ⎟⎠ ⎝ 2π rad ⎠ Peng ⎛ 4.58 × 103 J ⎞ ⎛ 2 500 rev ⎞ = ⎜ ⎟ = 1.91 × 10 W Δt ⎜⎝ revolution ⎟⎠ ⎝ 60 s ⎠ Qc The energy to melt a mass ΔmHg of Hg is Qc = mHg L f The energy absorbed to freeze ΔmAl of aluminum is Qh = mAl L f The efficiency is e = 1− ΔmHg LHg Qc (15.0 g )(1.18 × 104 J kg ) = 1− = 1− Qh ΔmAlLAl (1.00 g )( 3.97 × 105 J kg ) = 0.554 = 55.4%   Section 22.2 P22.8 P22.9 Heat Pumps and Refrigerators COP ( refrigerator ) = Qc W (a) If Qc = 120 J and COP = 5.00, then W = 24.0 J (b) Qh = Qc + W = 120 J + 24 J = 144 J (a) The work done on the refrigerant in each cycle is W = QH − QL = 625 kJ − 550 kJ = 75.0 kJ © 2014 Cengage Learning All Rights Reserved May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part 1156 Heat Engines, Entropy, and the Second Law of Thermodynamics (b) The coefficient of performance of a refrigerator is: COP = QL QL = W QH − QL Solving numerically: COP = P22.10 (a) QL QL 550 kJ = = = 7.33 W QH − QL 625 kJ − 550 kJ The coefficient of performance of a heat pump is COP = Qh W , where Qh is the thermal energy delivered to the warm space and W is the work input required to operate the heat pump Therefore, Qh = W ⋅ COP = ( P ⋅ Δt ) ⋅ COP ⎡⎛ ⎛ 3 600 s ⎞ ⎤ J⎞ = ⎢⎜ 7.03 × 103 ⎟ 8.00 h ⎜ ⎥ 3.80 = 7.69 × 10 J ⎟ s⎠ ⎝ h ⎠⎦ ⎣⎝ ( (b) The energy extracted from the cold space (outside air) is Qc = Qh − W = Qh − or *P22.11 ) Qh ⎞ ⎛ = Qh ⎜ − ⎟ ⎝ COP COP ⎠ ⎞ ⎛ Qc = ( 7.69 × 108 J ) ⎜ − = 5.67 × 108 J ⎟ ⎝ 3.80 ⎠ 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 ) t + ( 0.030 0 kg ) ( 3.33 × 105 J kg ) + ( 0.030 0 kg ) ( 2 090 J kg °C ) ( 20.0°C ) = 1.40 × 10 J = 233 J/s Thus, the work done per second = P = P22.12 (a) 233 J/s = 77.8 W 3.00 The coefficient of performance of a heat pump is COPh.p = QH QH = W QH − QL © 2014 Cengage Learning All Rights Reserved May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part 1184 Heat Engines, Entropy, and the Second Law of Thermodynamics 5.60 ( 10.0 kW ) = 6.60 Qc = 8.48 kW Δt and Qh = Weng + Qc : (b) Weng Δt (c) Qc Δt = Qh Q − c = 10.0 kW − 8.48 kW = 1.52 kW Δt Δt The air conditioner operates in a cycle, so the entropy of the working fluid does not change The hot reservoir increases in entropy by Qh Th (10.0 × 10 = J s ) ( 3 600 s ) 300 K = 1.20 × 105 J K The cold room decreases in entropy by ΔS = − Qc Tc ( 8.48 × 10 =− J s ) ( 3 600 s ) 280 K = −1.09 × 10 J K The net entropy change is positive, as it must be: +1.20 × 105 J K − 1.09 × 105 J K = 1.09 × 10 J K (d) The new ideal COP is COPCarnot = Tc 280 K = = 11.2 Th − Tc 25 K We suppose the actual COP is 0.400(11.2) = 4.48 As a fraction of the original 5.60, this is 4.48 = 0.800 , so the 5.60 fractional change is to drop by 20.0% P22.71 eC = − Tc Weng Weng / Δt = = : Th Qh Qh / Δt Qc Qh = Weng + Qc : Qc Δt = Δt = Qh Δt Qh Δt − = PTh P = (1 − Tc Th ) Th − Tc Weng Δt PTh PTc −P= Th − Tc Th − Tc Qc = mcΔT: PTc ⎛ Δm ⎞ =⎜ cΔT = ⎟ Δt ⎝ Δt ⎠ Th − Tc Qc © 2014 Cengage Learning All Rights Reserved May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part Chapter 22 1185 PTc Δm = Δt (Th − Tc ) cΔT (1.00 × 109 W )( 300 K ) Δm = = 5.97 × 10 kg s Δt ( 200 K ) ( 186 J kg ⋅ °C ) ( 6.00°C ) P22.72 eC = − Tc Weng Weng / Δt = = : Th Qh Qh / Δt Qh Δt = PTh P = − (Tc / Th ) Th − Tc ⎛Q ⎞ PTc =⎜ h ⎟−P= Δt ⎝ Δt ⎠ Th − Tc Qc But Qc = mcΔT, where c is the specific heat of water Therefore, and P22.73 (a) PTc ⎛ Δm ⎞ =⎜ cΔT = ⎟ Δt ⎝ Δt ⎠ Th − Tc Qc Δm = Δt PTc (Th − Tc ) cΔT For the isothermal process AB, the work on the gas is ⎛V ⎞ WAB = −PAVA ln ⎜ B ⎟ ⎝ VA ⎠ ⎛ 50.0 L ⎞ WAB = −5 ( 1.013 × 105 Pa ) ( 10.0 × 10−3 m ) ln ⎜ ⎝ 10.0 L ⎟⎠ WAB = −8.15 × 103 J where we have used 1.00 atm = 1.013 × 105 Pa and 1.00 L = 1.00 × 10–3 m3 WBC = −PB ΔV = − ( 1.013 × 105 Pa ) ⎡⎣( 10.0 − 50.0 ) × 10−3 ⎤⎦ m = +4.05 × 103 J WCA = and Weng = −WAB − WBC = 4.10 × 103 J = 4.10 kJ ANS FIG P22.73 © 2014 Cengage Learning All Rights Reserved May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part 1186 Heat Engines, Entropy, and the Second Law of Thermodynamics (b) Since AB is an isothermal process, ΔEint, AB = and QAB = –WAB = 8.15 × 103 J For an ideal monatomic gas, CV = TB = TA = 5R 3R and CP = 2 −3 PBVB ( 1.013 × 10 Pa ) ( 50.0 × 10 m ) 5.06 × 103 = = nR R R −3 PCVC ( 1.013 × 10 Pa ) ( 10.0 × 10 m ) 1.01 × 103 = = Also, TC = nR R R QCA 3 ⎛ ⎞ ⎛ 5.06 × 10 − 1.01 × 10 ⎞ = nCV ΔT = 1.00 ⎜ R ⎟ ⎜ ⎟⎠ = 6.08 kJ ⎝2 ⎠⎝ R so the total energy absorbed by heat is QAB + QCA = 8.15 kJ + 6.08 kJ = 1.42 × 10 J (c) QBC = nCP ΔT = QBC = 5 ( nRΔT ) = PB ΔVBC 2 1.013 × 105 ) ⎡⎣( 10.0 − 50.0 ) × 10−3 ⎤⎦ = −1.01 × 10 J ( = 1.01 × 10 J P22.74 Weng 4.10 × 103 J = = 0.289 or 28.9% 1.42 × 10 J (d) e= (e) A Carnot engine operating between Thot = TA = 060/R and Tcold = TC = 010/R has eC = − Tc /Th = − 1/5 = 80.0% The efficiency of the cycle is much lower than that of a Carnot engine operating between the same temperature extremes (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 , and Ti represent the state of the gas before the isothermal expansion Let PC , VC , and Ti represent the state after this process, so that PiVi = PCVC Let Pi , 3Vi , and Tf represent the state after the adiabatic compression Qh Then = Weng QAB + QCA PCVCγ = Pi ( 3Vi ) γ © 2014 Cengage Learning All Rights Reserved May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part Chapter 22 PC = Substituting 1187 PiVi VC gives PiViVCγ −1 = Pi ( 3γ Viγ ) Then VCγ −1 = 3γ Viγ −1 and VC = 3γ Vi (γ −1) The work output in the isothermal expansion is C W= C ∫ P dV = nRT ∫ V i i −1 dV i ⎛V ⎞ = nRTi ln ⎜ C ⎟ = nRTi ln 3γ ⎝ Vi ⎠ ( (γ −1) ) = nRT ⎛⎜⎝ γ γ− 1⎞⎟⎠ ln i This is also the input heat, so the entropy change is ΔS = Since CP = γ CV = CV + R, we have and ⎛ γ ⎞ Q = nR ⎜ ln T ⎝ γ − ⎟⎠ (γ − 1) CV = R, CP = R γ −1 γR γ −1 Then the result is (b) CV = ΔS = nCP ln The pair of processes considered here carries the gas from the initial state in P22.77 to the final state here Entropy is a function of state Entropy change does not depend on path Therefore the entropy change in P22.77 equals ΔSisothermal + ΔSadiabatic in this problem Since ΔSadiabatic = 0, the answers to P22.77 and P22.74(a) must be the same P22.75 We recognize that Tc = T1 and Th = T2, and Qc = 350 J and Qh = –1 000 J ΔShot = Q Qh −1000 J =− h = Th T2 600 K ΔScold = Qc Qc +750 J = = Tc T1 350 K © 2014 Cengage Learning All Rights Reserved May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part 1188 Heat Engines, Entropy, and the Second Law of Thermodynamics (a) Δ SU = ΔShot + ΔScold = − (b) eC = − 000 J 750 J − = 0.476 J K 600 K 350 K T1 350 K = 1− = 0.417 T2 600 K Weng = eC Qh = 0.417 ( 1 000 J ) = 417 J (c) ΔW  = WC  − Wreal   = eC Qh  − ( Qh  −  Qc )  ⎛ T⎞ =  ⎜ 1 −  c ⎟ Qh  − ( Qh  −  Qc ) Th ⎠ ⎝ ⎛ ⎞ T =  ⎜ Qh  −  c   Qh ⎟   − ( Qh  −  Qc )  Th ⎝ ⎠ =  Qc  −  ⎛Q Q ⎞ Tc   Qh  = Tc   ⎜ c  −  h ⎟ Th Th ⎠ ⎝ Tc = Tc ΔSU  = T1ΔSU P22.76 At point A, PiVi = nRTi and At point B, 3PiVi = nRTB so n = 1.00 mol TB = 3Ti At point C, ( 3Pi ) ( 2Vi ) = nRTC and TC = 6Ti At point D, Pi ( 2Vi ) = nRTD so TD = 2Ti The heat for each step in the cycle is found 3R 5R using CV = and CP = : 2 QAB = nCV ( 3Ti − Ti ) = 3nRTi ANS FIG P22.76 QBC = nCP ( 6Ti − 3Ti ) = 7.50nRTi QCD = nCV ( 2Ti − 6Ti ) = −6nRTi QDA = nCP (Ti − 2Ti ) = −2.50nRTi (a) Therefore, Qentering = Qh = QAB + QBC = 3nRTi + 7.5nRTi = 10.5nRTi (b) Qleaving = Qc = QCD + QDA = −6nRTi − 2.50nRTi = 8.50nRTi (c) Actual efficiency: e = Qh − Qc 10.5nRTi − 8.5nRTi = = 0.190 Qh 10.5nRTi © 2014 Cengage Learning All Rights Reserved May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part Chapter 22 1189 Tc T = − i = 0.833 Th 6Ti (d) Carnot efficiency: eC = − The Carnot efficiency is much higher f P22.77 ΔS = ∫ i dQ = T f ∫ i f ( nCP dT T = nCP T −1dT = nCP ln T Tf = nCP ln T f − ln Ti i T ∫ ) i ⎛ Tf ⎞ = nCP ln ⎜ ⎟ ⎝ Ti ⎠ ⎡ P ( 3Vi ) nR ⎤ ⎛ PVf nR ⎞ ΔS = nCP ln ⎜ = nCP ln ⎢ ⎥ = nCP ln ⎟ ⎝ nR PVi ⎠ ⎣ nR PVi ⎦ P22.78 (a) water: Twater = 35.0°F → → body: Tbody = 98.6°F ( 98.6 − 32.0) °C ( 37.0 + 273.15) K = 310.15 K → → ΔScold water = ΔSbody = − ∫ dQ = mw c × T ( ( 35.0 − 32.0) °C (1.67 + 273.15) K = 274.82 K Tbody ∫ Twater ⎛ Tbody ⎞ dT = mw c × ln ⎜ T ⎝ Twater ⎟⎠ mw c Tbody − Twater Q =− Tbody Tbody ) ΔSsystem = ΔScold water + ΔSbody ⎛ 310.15 ⎞ = ( 0.454 kg ) ( 4 186 J kg ⋅ K ) × ln ⎜ ⎝ 274.82 ⎟⎠ − ( 0.454 kg ) ( 4 186 J kg ⋅ K ) (b) ( 310.15 − 274.82 ) = 310.15 13.4 J K Conservation of energy, Qhot = −Qcold , gives ( mw c (TF − Twater ) = −mAth c TF − Tbody ( mw (TF − Twater ) = −mAth TF − Tbody ) ) mwTF − mwTwater = −mAthTF + mAthTbody ( mw + mAth )TF = mwTwater + mAthTbody © 2014 Cengage Learning All Rights Reserved May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part 1190 Heat Engines, Entropy, and the Second Law of Thermodynamics Solving for TF, TF = = mwTwater + mAthTbody mw + mAth ( 0.454 kg )( 274.82 K ) + (70.0 kg )( 310.15 K ) 0.454 kg + 70.0 kg = 309.92 K = 310 K (c) ΔS = ΔSice ′ water + ΔSbody ′ ⎛ T ⎞ ⎛ T ⎞ = mw c × ln ⎜ F ⎟ + mAth c × ln ⎜ F ⎟ ⎝ Twater ⎠ ⎝ Tbody ⎠ ⎛ 309.92 ⎞ = ( 0.454 kg ) ( 4 186 J kg ⋅ K ) ln ⎜ ⎝ 274.82 ⎟⎠ ⎛ 309.92 ⎞ + ( 70.0 kg ) ( 4 186 J kg ⋅ K ) ln ⎜ ⎝ 310.15 ⎟⎠ = 11.1 J K (d) Smaller by less than 1% Vf P22.79 (a) W= ∫ Vi P22.80 2Vi PdV = nRT ∫ Vi ⎛ 2V ⎞ dV = ( 1.00 ) RT ln ⎜ i ⎟ = RT ln V ⎝ Vi ⎠ (b) Yes (c) No The second law refers to an engine operating in a cycle, whereas this problem involves only a single process When energy enters a substance by heat, we describe the process with Equation 20.4, Q = mcΔT This is a reversible process; if energy leaves the substance, the temperature drops down again Therefore, the entropy change for one of the samples of water is T f ⎛ Tf ⎞ dQ mcdT ΔS =  ∫  =  ∫  = mc ln ⎜ ⎟ T T ⎝ Ti ⎠ Ti Consequently, the entropy change for both samples of water is ΔStotal  = ΔShot  + ΔScold   ⎡⎛ Tf ⎞ ⎛ Tf ⎞ ⎤ ⎛ Tf ⎞ ⎛ Tf ⎞ = mc ln ⎜ ⎟  + mc ln ⎜ ⎟  = mc ln ⎢⎜ ⎟ ⎜ ⎟ ⎥   ⎝ Thi ⎠ ⎝ Tci ⎠ ⎣⎝ Thi ⎠ ⎝ Tci ⎠ ⎦ (1.00 kg )( 4 186 J/kg ⋅ °C) ln ⎡⎢⎛⎜⎝ 303 K ⎞⎟⎠ ⎛⎜⎝ 283 K ⎞⎟⎠ ⎤⎥  = 4.88 J/K 293 K ⎣ 293 K ⎦ © 2014 Cengage Learning All Rights Reserved May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part Chapter 22 1191 This is not zero While the statements about energy transfer by heat are true, the mixing process is irreversible After the water has come to equilibrium, it will not spontaneously separate again into warm and cool water Therefore, there is an entropy increase of the mixture during this irreversible process   Challenge Problems P22.81 (a) Given: PA = 25.0 atm and PC = 1.00 atm Use the equation of state for an ideal gas: nRT P 1.00 ( 8.314 J mol ⋅ K ) ( 600 K ) VA = = 1.97 × 10−3 m 25.0 ( 1.013 × 105 Pa ) V= VC = 1.00 ( 8.314 J mol ⋅ K ) ( 400 K ) 1.013 × 105 Pa Since AB is isothermal, = 32.8 × 10−3 m PAVA = PBVB , and since BC is adiabatic, PBVBγ = PCVCγ ANS FIG P22.81 Combining these expressions, ⎡⎛ PC ⎞ VCγ ⎤ VB = ⎢⎜ ⎟ ⎥ ⎣⎝ PA ⎠ VA ⎦ (γ −1) ⎡ 1.00 atm ⎛ ( 32.8 × 10−3 m )1.40 ⎞ ⎤ ⎛ ⎞ = ⎢⎜ ⎟⎥ ⎟⎜ ⎝ ⎢ 25.0 atm ⎠ ⎜⎝ 1.97 × 10−3 m ⎟⎠ ⎥ ⎣ ⎦ (1 0.400) = 11.9 × 10−3 m © 2014 Cengage Learning All Rights Reserved May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part 1192 Heat Engines, Entropy, and the Second Law of Thermodynamics Similarly, ⎡⎛ P ⎞ V γ ⎤ VD = ⎢⎜ A ⎟ A ⎥ ⎣⎝ PC ⎠ VC ⎦ (γ −1) ⎡ 25.0 atm ⎛ ( 1.97 × 10−3 m )1.40 ⎞ ⎤ ⎛ ⎞ = ⎢⎜ ⎟⎥ ⎟⎜ ⎝ ⎢ 1.00 atm ⎠ ⎜⎝ 32.8 × 10−3 m ⎟⎠ ⎥ ⎣ ⎦ (1 0.400) = 5.44 × 10−3 m Since AB is isothermal, PAVA = PBVB ⎛V ⎞ ⎛ 1.97 × 10−3 m ⎞ and PB = PA ⎜ A ⎟ = ( 25.0 atm ) ⎜ = 4.14 atm ⎝ 11.9 × 10−3 m ⎟⎠ ⎝ VB ⎠ Also, CD is an isothermal and ⎛V ⎞ ⎛ 32.8 × 10−3 m ⎞ PD = PC ⎜ C ⎟ = ( 1.00 atm ) ⎜ = 6.03 atm ⎝ 5.44 × 10−3 m ⎟⎠ ⎝ VD ⎠ (b) Energy is added by heat to the gas during the process AB For the isothermal process, ΔEint = 0, and the first law gives ⎛V ⎞ QAB = −WAB = nRTh ln ⎜ B ⎟ ⎝ VA ⎠ or ⎛ 11.9 atm ⎞ Qh = QAB = ( 1.00 mol ) ( 8.314 J mol ⋅ K ) ( 600 K ) ln ⎜ ⎝ 1.97 atm ⎟⎠ = 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 P22.82 The quantity of gas is −6 PAVA ( 100 × 10 Pa ) ( 500 × 10 m ) n= = = 0.020 mol RTA ( 8.314 J mol ⋅ K )( 293 K ) (a) In process A→B, γ ⎛V ⎞ 1.40 PB = PA ⎜ A ⎟ = ( 100 × 103 Pa ) ( 8.00 ) = 1.84 × 106 Pa ⎝V ⎠ B © 2014 Cengage Learning All Rights Reserved May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part Chapter 22 TB = 1193 −6 PBVB ( 1.84 × 10 Pa ) ( 500 × 10 m 8.00 ) = = 673 K nR ( 0.020 mol )( 8.314 J mol ⋅ K ) VA VB = 8.00 → VB = VA 8.00 = 500 = 62.5 cm State C: VC = VB PC = nRTC ( 0.020 mol ) ( 8.314 J mol ⋅ K )( 023 K ) = VC 62.5 × 10−6 m = 2.79 × 106 Pa State D: VD = VA In process C→D: γ 1.40 ⎛ VC ⎞ ⎛ ⎞ PD = PC ⎜ ⎟ = ( 2.79 × 10 Pa ) ⎜ = 1.52 × 105 Pa ⎝ 8.00 ⎟⎠ ⎝ VD ⎠ −6 PDVD ( 1.52 × 10 Pa ) ( 500 × 10 m ) TD = = = 445 K nR ( 0.020 mol )( 8.314 J mol ⋅ K ) TABLE P22.82(a) tabulates these results: T (K) P (kPa) V (cm3) A 293 100 500 B 673 1.84 × 103 62.5 C 1023 2.79 × 103 62.5 D 445 152 500 TABLE P22.82(a) (b) In the adiabatic process A→B, Q = 0, nR (TB − TA ) = ( 0.020 mol ) ( 8.314 J mol ⋅ K )( 673 K – 293 K ) = 162 J ΔEint, A→B = and ΔEint, AB = 162 J = Q − Wout = − Wout → WAB = −162 J © 2014 Cengage Learning All Rights Reserved May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part 1194 Heat Engines, Entropy, and the Second Law of Thermodynamics In the isovolumetric process B→C, W = nR (TC − TB ) = ( 0.020 mol ) ( 8.314 J mol ⋅ K )( 023 K − 673 K ) = 149 J = 149 J = Q − Wout = Q − → QBC = 149 J ΔEint, B→C = ΔEint, B→C In the adiabatic process C→D, Q = nR (TD − TC )      = ( 0.020 mol ) ( 8.314 J mol ⋅ K )( 445 K − 023 K ) = −246 J ΔEint, C→D = −246 J = Q − Wout = − Wout → WCD = 246 J ΔEint, C→D = For the entire cycle, ΔEint, net = nRΔT = 0: Weng = −162 J + + 246.3 J + = 84.3 J ΔEint = Qnet + Weng = → Qnet = −Weng = 84.3 J TABLE P22.82(b) tabulates these results: Q Weng ΔEint A→B –162 162 B→C 149 149 C→D 246 –246 D→A –65.0 –65.0 ABCD 84.3 84.3 TABLE P22.82(b) (c) From B→C, the input energy is Qh = 149 J (d) From D→A, the energy exhaust is Qc = 65.0 J (e) From ABCDA, Weng = 84.3 J © 2014 Cengage Learning All Rights Reserved May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part Chapter 22 Weng 84.3 J = 0.565 149 J (f) The efficiency is: e = (g) Let f represent the angular speed of the crankshaft Then Qh = 1195 f is the frequency at which we obtain work in the amount of 84.3 J/cycle: ⎛ f⎞ 000 J s = ⎜ ⎟ ( 84.3 J cycle ) ⎝ 2⎠ f= 000 J s = 23.7 rev s = 1.42 × 103 rev 84.3 J cycle     © 2014 Cengage Learning All Rights Reserved May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part 1196 Heat Engines, Entropy, and the Second Law of Thermodynamics ANSWERS TO EVEN-NUMBERED PROBLEMS P22.2 (a) 0.25 or 25%; (b) QC / Qh = 3/4 P22.4 13.7°C P22.6 (a) 29.4 L/h; (b) 185 hp; (c) 527 N · m; (d) 1.91 × 105 W P22.8 (a) 24.0 J; (b) 144 J P22.10 (a) 7.69 × 108 J; (b) 5.67 × 108 J P22.12 (a) 2.65 × 107 J; (b) 3.20 P22.14 0.540 or 54.0% P22.16 The efficiency of a Carnot engine operating between these temperatures is 6.83% Therefore, there is no way that the inventor’s engine can have an efficiency of 0.110 = 11.0% P22.18 ⎛ Th ⎞ ⎛ Tc ⎞ (a) PΔt ⎜ ; (b) PΔt ⎜ ⎟ ⎝ Th − Tc ⎠ ⎝ Th − Tc ⎟⎠ P22.20 (a) See P22.20(a) for the full solution; (b) See P22.20(b) for the full solution P22.22 72.2 J P22.24 0.330 or 33.0% P22.26 (a) 0.300; (b) 1.40 × 10−3 K−1; (c) −2.00 ×10−3 K−1; (d) No The derivative in part (c) depends only on Th P22.28 (a) 5.12%; (b) 5.27 TJ/h; (c) As fossil-fuel prices rise, this way to use solar energy will become a good buy P22.30 (a) See P22.30(a) for full explanation; (b) − P22.32 (a) See TABLE P22.32(a); (b) See TABLE P22.32(b); (c) 23.7%; (d) 23.7% P22.34 23.1 mW P22.36 (a) 51.2%; (b) 36.2% P22.38 See P22.38 for the full derivation Tc ; (c) The combination of Th reversible engines is itself a reversible engine so it has the Carnot efficiency No improvement in net efficiency has resulted; (d) Ti = (Th + Tc ) ; 1/2 (e) Ti = (ThTc ) © 2014 Cengage Learning All Rights Reserved May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part Chapter 22 1197 P22.40 (a) See TABLE P22.40; (b) heads and tails P22.42 −610 J/K P22.44 717 J/K P22.46 2.70 kJ/K P22.48 (a) 5.76 J/K; (b) no change in temperature P22.50 244 J/K P22.52 W/K P22.54 (a) ΔSh = − P22.56 +1.18 J/K P22.58 (a) 0.268; (b) 0.423 P22.60 8.36 × 106 J/K · s P22.62 (a) 2.93; (b) (COP)refrigerator; (c) with EER 5, $510, with EER 10, $255; Thus, the cost for air conditioning is half as much for an air conditioner with EER 10 compared with an air conditioner with EER P22.64 (a) 2.49 kJ; (b) 1.50 kJ; (c) –990 J P22.66 (a) P22.68 (a) 214 J and 64.3 J; (b) 35.7 J and 35.7 J The total energy the firebox puts out equals to the total energy transferred to the environment; (c) The net flow of energy by heat from the cold to the hot reservoir without work input is possible; (d) Weng S = 233 J, Qh, S = 333 J; ⎛ 1⎞ Q Q ; (b) ΔSc = − ; (c) Q ⎜ − ⎟ Th Tc ⎝ Tc Th ⎠ Tc ; (b) smaller; (c) 14.6 Th − Tc (e) 83.3 J; (f) 83.3 J; (g) 0; (h) The output of 83.8 J of energy from the heat engine by work in a cyclic process without any exhaust by heat is impossible; (i) −0.111 J/K; (j) A decrease in total entropy is impossible P22.70 (a) 8.48 kW; (b) 1.52 kW; (c) 1.09 × 104 J/K; (d) drop by 20.0% P22.72 PTc (Th − Tc ) cΔT P22.74 (a) ΔS = nCp ln 3; (b) The pair of processes considered here carries the gas from the initial state in P22.77 to the final state here Entropy is a function of state Entropy change does not depend on path Therefore, the entropy change in P22.77 equals ΔSisothermal + ΔSadiabatic in this problem Since ΔSadiabatic = 0, the answers to P22.77 and P22.74(a) must be the same © 2014 Cengage Learning All Rights Reserved May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part 1198 Heat Engines, Entropy, and the Second Law of Thermodynamics P22.76 (a) 10.5 nRTi; (b) 8.50 nRTi; (c) 0.190; (d) 0.833; The Carnot efficiency is much higher P22.78 (a) 13.4 J/K; (b) 310 K; (c) 11.1 J/K; (d) smaller by less than 1% P22.80 The computed change in entropy is 4.88 J/K, which is not zero While the statements about energy transfer by heat are true, the mixing process is irreversible After the water has come to equilibrium, it will not spontaneously separate again into warm and cool water Therefore, there is an entropy increase of the mixture during the irreversible process P22.82 (a) See TABLE P22.82(a); (b) See TABLE P22.82(b); (c) 149 J; (d) 65.0 J; (e) 84.3 J; (f) 0.565; (g) 1.42 × 103 rev/min © 2014 Cengage Learning All Rights Reserved May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part ... reduces to ⎝ VB ⎠ ⎝ VCVD ⎠ VB = VAVC ( 10.0 L ) ( 24.0 L ) = = 16.0 L VD 15.0 L ⎛V ⎞ ⎛ 10.0 L ⎞ Finally, PB = PA ⎜ A ⎟ = ( 1 400 kPa ) ⎜ = 875 kPa ⎝ 16.0 L ⎟⎠ ⎝ VB ⎠ State P (kPa) V (L) T (K)... D ⎟ ⎝ VC ⎠ ⎛ 15.0 L ⎞ = ( 2.34 mol ) ( 8.314 J mol ⋅ K ) ( 549 K ) ln ⎜ ⎝ 24.0 L ⎟⎠ = −5.02 kJ Finally, for the adiabatic process D → A: Q = ΔEint = nCV (TA − TD ) ⎡3 ⎤ = ( 2.34 mol ) ⎢ ( 8.314... contribution to entropy production is much greater than the above estimate, based only on metabolism P22.53 The change in entropy of a reservoir is ΔS = Qr T , where Qr is the energy absorbed (Qr