20 THE SECOND LAW OF THERMODYNAMICS 20.1 IDENTIFY: For a heat engine, W = |QH | − |QC | e = W QH > 0, QC < QH SET UP: W = 2200 J |QC | = 4300 J EXECUTE: (a) QH = W + |QC | = 6500 J 2200 J = 0.34 = 34% 6500 J EVALUATE: Since the engine operates on a cycle, the net Q equal the net W But to calculate the efficiency we use the heat energy input, QH (b) e = 20.2 IDENTIFY: For a heat engine, W = |QH | − |QC | e = W QH > 0, QC < QH SET UP: |QH | = 9000 J |QC | = 6400 J EXECUTE: (a) W = 9000 J − 6400 J = 2600 J W 2600 J (b) e = = = 0.29 = 29% QH 9000 J EVALUATE: Since the engine operates on a cycle, the net Q equal the net W But to calculate the efficiency we use the heat energy input, QH 20.3 IDENTIFY and SET UP: The problem deals with a heat engine W = +3700 W and QH = +16,100 J Use Eq (20.4) to calculate the efficiency e and Eq (20.2) to calculate |QC | Power = W/t EXECUTE: (a) e = work output W 3700 J = = = 0.23 = 23, heat energy input QH 16,100 J (b) W = Q = |QH | − |QC | Heat discarded is |QC | = |QH | − W = 16,100 J − 3700 J = 12,400 J (c) QH is supplied by burning fuel; QH = mLc where Lc is the heat of combustion QH 16,100 J = = 0.350 g Lc 4.60 × 104 J/g (d) W = 3700 J per cycle In t = 1.00 s the engine goes through 60.0 cycles P = W/t = 60.0(3700 J)/1.00 s = 222 kW m= P = (2.22 × 105 W)(1 hp/746 W) = 298 hp EVALUATE: QC = −12,400 J In one cycle Qtot = QC + QH = 3700 J This equals Wtot for one cycle 20.4 IDENTIFY: W = |QH | − |QC | e = W QH > 0, QC < QH SET UP: For 1.00 s, W = 180 × 103 J © Copyright 2012 Pearson Education, Inc All rights reserved This material is protected under all copyright laws as they currently exist No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher 20-1 20-2 Chapter 20 EXECUTE: (a) QH = W 180 × 103 J = = 6.43 × 105 J e 0.280 (b) |QC | = |QH | − W = 6.43 × 105 J − 1.80 × 105 J = 4.63 × 105 J EVALUATE: Of the 6.43 × 105 J of heat energy supplied to the engine each second, 1.80 × 105 J is converted to mechanical work and the remaining 4.63 × 105 J is discarded into the low temperature 20.5 reservoir IDENTIFY: This cycle involves adiabatic (ab), isobaric (bc), and isochoric (ca) processes SET UP: ca is at constant volume, ab has Q = 0, and bc is at constant pressure For a constant pressure process W = pΔV and Q = nC p ΔT pV = nRT gives nΔT = ⎛ Cp ⎞ pΔV , so Q = ⎜ ⎟ pΔV If γ = 1.40 the R ⎝ R ⎠ gas is diatomic and C p = 72 R For a constant volume process W = and Q = nCV ΔT pV = nRT gives nΔT = V Δp ⎛C , so Q = ⎜ V R ⎝ R ⎞ 5 ⎟V Δp For a diatomic ideal gas CV = R atm = 1.013 × 10 Pa ⎠ EXECUTE: (a) Vb = 9.0 × 10−3 m3 , pb = 1.5 atm and Va = 2.0 × 10−3 m3 For an adiabatic process γ 1.4 ⎛ 9.0 × 10−3 m3 ⎞ ⎛V ⎞ = 12.3 atm paVa = pbVb pa = pb ⎜ b ⎟ = (1.5 atm ) ⎜ −3 ⎟ ⎜ ⎟ ⎝ Va ⎠ ⎝ 2.0 × 10 m ⎠ (b) Heat enters the gas in process ca, since T increases ⎛C ⎞ ⎛5⎞ Q = ⎜ V ⎟V Δp = ⎜ ⎟ (2.0 × 10−3 m3 )(12.3 atm − 1.5 atm)(1.013 × 105 Pa/atm) = 5470 J QH = 5470 J ⎝ R ⎠ ⎝2⎠ (c) Heat leaves the gas in process bc, since T increases ⎛ Cp ⎞ ⎛7⎞ −3 Q=⎜ ⎟ pΔV = ⎜ ⎟ (1.5 atm)(1.013 × 10 Pa/atm)(−7.0 × 10 m ) = −3723 J QC = −3723 J R ⎝ 2⎠ ⎝ ⎠ γ γ (d) W = QH + QC = +5470 J + (−3723 J) = 1747 J W 1747 J = = 0.319 = 31.9% QH 5470 J EVALUATE: We did not use the number of moles of the gas |Q | IDENTIFY: Apply e = − γ −1 e = − C | QH | r (e) e = 20.6 SET UP: In part (b), QH = 10,000 J The heat discarded is |QC | = 0.594 = 59.4% 9.500.40 (b) |QC | = |QH |(1 − e) = (10,000 J)(1 − 0.594) = 4060 J EXECUTE: (a) e = − EVALUATE: The work output of the engine is W = |QH | − |QC | = 10,000 J − 4060 J = 5940 J 20.7 IDENTIFY: e = − r1−γ SET UP: r is the compression ratio EXECUTE: (a) e = − (8.8)20.40 = 0.581, which rounds to 58% (b) e = − (9.6)20.40 = 0.595 an increase of 1.4% 20.8 EVALUATE: An increase in r gives an increase in e IDENTIFY: Convert coefficient of performance (K) to energy efficiency rating (EER) H H SET UP: K = watts and EER = Btu/h Pwatts Pwatts © Copyright 2012 Pearson Education, Inc All rights reserved This material is protected under all copyright laws as they currently exist No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher The Second Law of Thermodynamics EXECUTE: Btu/h = 0.293 W so H watts = H Btu/h (0.293) K = 0.293 20.9 20-3 H Btu/h = (0.293)EER and Pwatts EER = 3.41K For K = 3.0, EER = (3.41)(3.0) = 10.2 EVALUATE: The EER is larger than K, but this does not mean that the air conditioner is suddenly better at cooling! IDENTIFY and SET UP: For the refrigerator K = 2.10 and QC = +3.4 × 104 J Use Eq (20.9) to calculate |W | and then Eq (20.2) to calculate QH (a) EXECUTE: Performance coefficient K = QC /|W | (Eq 20.9) |W | = QC /K = 3.40 × 104 J/2.10 = 1.62 × 104 J (b) SET UP: The operation of the device is illustrated in Figure 20.9 EXECUTE: W = QC + QH QH = W − QC QH = −1.62 × 104 J − 3.40 × 104 J = −5.02 × 104 J (negative because heat goes out of the system) Figure 20.9 EVALUATE: |QH | = |W | + |QC | The heat |QH | delivered to the high temperature reservoir is greater than 20.10 the heat taken in from the low temperature reservoir |Q | IDENTIFY: K = C and |QH | = |QC | + |W | |W | SET UP: The heat removed from the room is |QC | and the heat delivered to the hot outside is |QH | |W | = (850 J/s)(60.0 s) = 5.10 × 104 J EXECUTE: (a) |QC | = K |W | = (2.9)(5.10 × 104 J) = 1.48 × 105 J (b) |QH | = |QC | + |W | = 1.48 × 105 J + 5.10 × 104 J = 1.99 × 105 J EVALUATE: (c) |QH | = |QC | + |W |, so |QH | > |QC | 20.11 IDENTIFY: The heat Q = mcΔT that comes out of the water to cool it to 5.0°C is QC for the refrigerator SET UP: For water 1.0 L has a mass of 1.0 kg and c = 4.19 × 103 J/kg ⋅ C° P = performance is K = |W | The coefficient of t |QC | |W | EXECUTE: Q = mcΔT = (12.0 kg)(4.19 × 103 J/kg ⋅ C°)(5.0°C − 31°C) = −1.31 × 106 J |QC | = 1.31 × 106 J |QC | |QC | |Q | 1.31 × 106 J = so t = C = = 6129 s = 102 = 1.7 h |W | Pt PK (95 W)(2.25) EVALUATE: 1.7 h seems like a reasonable time to cool down the dozen bottles |Q | IDENTIFY: |QH | = |QC | + |W | K = C W K= 20.12 SET UP: For water, cw = 4190 J/kg ⋅ K and Lf = 3.34 × 105 J/kg For ice, cice = 2010 J/kg ⋅ K EXECUTE: (a) Q = mcice ΔTice − mLf + mcw ΔTw Q = (1.80 kg)([2010 J/kg ⋅ K][−5.0 C°] − 3.34 × 105 J/kg + [4190 J/kg ⋅ K][−25.0 C°]) = −8.08 × 105 J Q = −8.08 × 105 J Q is negative for the water since heat is removed from it (b) |QC | = 8.08 × 105 J W = |QC | 8.08 × 105 J = = 3.37 × 105 J K 2.40 (c) |QH | = 8.08 × 105 J + 3.37 × 105 J = 1.14 ì 106 J â Copyright 2012 Pearson Education, Inc All rights reserved This material is protected under all copyright laws as they currently exist No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher 20-4 Chapter 20 EVALUATE: For this device, QC > and QH < More heat is rejected to the room than is removed from 20.13 the water IDENTIFY: Use Eq (20.2) to calculate |W | Since it is a Carnot device we can use Eq (20.13) to relate the heat flows out of the reservoirs The reservoir temperatures can be used in Eq (20.14) to calculate e (a) SET UP: The operation of the device is sketched in Figure 20.13 EXECUTE: W = QC + QH W = −335 J + 550 J = 215 J Figure 20.13 (b) For a Carnot cycle, TC = TH |QC | TC = (Eq 20.13) |QH | TH ⎛ 335 J ⎞ |QC | = 620 K ⎜ ⎟ = 378 K |QH | ⎝ 550 J ⎠ (c) e(Carnot) = − TC /TH = − 378 K/620 K = 0.390 = 39.0, EVALUATE: We could use the underlying definition of e (Eq 20.4): e = W/QH = (215 J)/(550 J) = 39%, which checks 20.14 IDENTIFY: |W | = |QH | − |QC | QC < 0, QH > e = W Q T For a Carnot cycle, C = − C QH QH TH SET UP: TC = 300 K, TH = 520 K |QH | = 6.45 × 103 J ⎛T ⎞ ⎛ 300 K ⎞ EXECUTE: (a) QC = −QH ⎜ C ⎟ = −(6.45 × 103 J) ⎜ ⎟ = −3.72 × 10 J T 520 K ⎝ ⎠ ⎝ H⎠ (b) |W | = |QH | − |QC | = 6.45 × 103 J − 3.72 × 103 J = 2.73 × 103 J (c) e = W 2.73 × 103 J = = 0.423 = 42.3, QH 6.45 × 103 J EVALUATE: We can verify that e = − TC /TH also gives e = 42.3, 20.15 IDENTIFY: e = W Q T for any engine For the Carnot cycle, C = − C QH QH TH SET UP: TC = 20.0°C + 273.15 K = 293.15 K EXECUTE: (a) QH = W 2.5 × 104 J = = 4.24 × 104 J e 0.59 (b) W = QH + QC so QC = W − QH = 2.5 × 104 J − 4.24 × 104 J = −1.74 × 104 J TH = −TC ⎛ 4.24 × 104 J ⎞ QH = − ( 293.15 K ) ⎜ = 714 K = 441°C ⎜ −1.74 × 104 J ⎟⎟ QC ⎝ ⎠ EVALUATE: For a heat engine, W > 0, QH > and QC < © Copyright 2012 Pearson Education, Inc All rights reserved This material is protected under all copyright laws as they currently exist No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher The Second Law of Thermodynamics 20.16 20-5 IDENTIFY and SET UP: The device is a Carnot refrigerator We can use Eqs (20.2) and (20.13) (a) The operation of the device is sketched in Figure 20.16 TH = 24.0°C = 297 K TC = 0.0°C = 273 K Figure 20.16 The amount of heat taken out of the water to make the liquid → solid phase change is Q = −mLf = −(85.0 kg)(334 × 103 J/kg) = −2.84 × 107 J This amount of heat must go into the working substance of the refrigerator, so QC = +2.84 × 107 J For Carnot cycle |QC |/|QH | = TC /TH EXECUTE: |QH | = |QC |(TH /TC ) = 2.84 × 107 J(297 K/273 K) = 3.09 × 107 J (b) W = QC + QH = +2.84 × 107 J − 3.09 × 107 J = −2.5 × 106 J 20.17 EVALUATE: W is negative because this much energy must be supplied to the refrigerator rather than obtained from it Note that in Eq (20.13) we must use Kelvin temperatures |Q | Q T IDENTIFY: |QH | = |W | + |QC | QH < 0, QC > K = C For a Carnot cycle, C = − C |W | QH TH SET UP: TC = 270 K, TH = 320 K |QC | = 415 J ⎛T ⎞ ⎛ 320 K ⎞ EXECUTE: (a) QH = − ⎜ H ⎟ QC = − ⎜ ⎟ (415 J) = −492 J ⎝ 270 K ⎠ ⎝ TC ⎠ (b) For one cycle, |W | = |QH | − |QC | = 492 J − 415 J = 77 J P = (c) K = (165)(77 J) = 212 W 60 s |QC | 415 J = = 5.4 |W | 77 J EVALUATE: The amount of heat energy |QH | delivered to the high-temperature reservoir is greater than the amount of heat energy |QC | removed from the low-temperature reservoir 20.18 IDENTIFY: The theoretical maximum performance coefficient is K Carnot = |Q | TC K = C |QC | is the TH − TC |W | heat removed from the water to convert it to ice For the water, |Q| = mcw ΔT + mLf SET UP: TC = −5.0°C = 268 K TH = 20.0°C = 293 K cw = 4190 J/kg ⋅ K and Lf = 334 × 103 J/kg EXECUTE: (a) In one year the freezer operates (5 h/day)(365 days) = 1825 h 730 kWh = 0.400 kW = 400 W 1825 h 268 K (b) K Carnot = = 10.7 293 K − 268 K P= (c) |W | = Pt = (400 W)(3600 s) = 1.44 × 106 J |QC | = K |W | = 1.54 × 107 J |Q| = mcw ΔT + mLf gives m= |QC | 1.54 × 107 J = = 36.9 kg cw ΔT + Lf (4190 J/kg ⋅ K)(20.0 K) + 334 × 103 J/kg EVALUATE: For any actual device, K < K Carnot , |QC | is less than we calculated and the freezer makes less ice in one hour than the mass we calculated in part (c) © Copyright 2012 Pearson Education, Inc All rights reserved This material is protected under all copyright laws as they currently exist No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher 20-6 Chapter 20 20.19 IDENTIFY: e = W Q T Q T = − C For a Carnot cycle, C = − C and e = − C QH TH QH QH TH SET UP: TH = 800 K QC = −3000 J EXECUTE: For a heat engine, QH = −QC /(1 − e) = −(−3000 J)/(1 − 0.600) = 7500 J, and then W = eQH = (0.600)(7500 J) = 4500 J EVALUATE: This does not make use of the given value of TH If TH is used, then TC = TH (1 − e) = (800 K)(1 − 0.600) = 320 K and QH = −QCTH /TC , which gives the same result 20.20 IDENTIFY: W = QC + QH For a Carnot cycle, QC T = − C For the ice to liquid water phase transition, QH TH Q = mLf SET UP: For water, Lf = 334 × 103 J/kg EXECUTE: QC = − mLf = −(0.0400 kg)(334 × 103 J/kg) = −1.336 × 104 J QC T = − C gives QH TH QH = −(TH /TC )QC = −( −1.336 × 104 J) [(373.15 K )/(273.15 K) ] = +1.825 × 104 J W = QC + QH = 4.89 × 103 J EVALUATE: For a heat engine, QC is negative and QH is positive The heat that comes out of the engine (Q < 0) goes into the ice (Q > 0) 20.21 IDENTIFY: The power output is P = T W W The theoretical maximum efficiency is eCarnot = − C e = t TH QH SET UP: QH = 1.50 × 104 J TC = 350 K TH = 650 K hp = 746 W EXECUTE: eCarnot = − 350 K TC =1− = 0.4615 W = eQH = (0.4615)(1.50 × 104 J) = 6.923 × 103 J; this is 650 K TH W (240)(6.923 × 103 J) = = 2.77 × 104 W = 37.1 hp 60.0 s t Q T EVALUATE: We could also use C = − C to calculate QH TH the work output in one cycle P = 20.22 ⎛T ⎞ ⎛ 350 K ⎞ 3 QC = − ⎜ C ⎟ QH = − ⎜ ⎟ (1.50 × 10 J) = −8.08 × 10 J Then W = QC + QH = 6.92 × 10 J, the same as T ⎝ 650 K ⎠ ⎝ H⎠ previously calculated IDENTIFY: The immense ocean does not change temperature, but it does lose some entropy because it gives up heat to melt the ice The ice does not change temperature as it melts, but it gains entropy by absorbing heat from the ocean Q SET UP: For a reversible isothermal process ΔS = , where T is the Kelvin temperature at which the T heat flow occurs The heat flows in this problem are irreversible, but since ΔS is path-independent, the entropy change is the same as for a reversible heat flow The heat flow when the ice melts is Q = mLf , with Lf = 334 × 103 J/kg Heat flows out of the ocean (Q < 0) and into the ice (Q > 0) The heat flow for the ice occurs at T = 0°C = 273.15 K The heat flow for the ocean occurs at T = 3.50°C = 276.65 K EXECUTE: Q = mLf = (4.50 kg)(334 × 103 J/kg) = 1.50 × 106 J For the ice, ΔS = Q +1.50 × 106 J Q −1.50 × 106 J = = 5.49 × 103 J/K For the ocean, ΔS = = = −5.42 × 103 J/K The net 273.15 K 276.65 K T T entropy change is 5.49 × 103 J/K + ( −5.42 × 103 J/K) = +70 J/K The entropy of the world increases by 70 J/K EVALUATE: Since this process is irreversible, we expect the entropy of the world to increase, as we have found © Copyright 2012 Pearson Education, Inc All rights reserved This material is protected under all copyright laws as they currently exist No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher The Second Law of Thermodynamics 20.23 IDENTIFY: ΔS = 20-7 Q for each object, where T must be in kelvins The temperature of each object remains constant T SET UP: For water, Lf = 3.34 × 105 J/kg EXECUTE: (a) The heat flow into the ice is Q = mLf = (0.350 kg)(3.34 × 105 J/kg) = 1.17 × 105 J The heat flow occurs at T = 273 K, so ΔS = Q 1.17 × 105 J = = 429 J/K Q is positive and ΔS is positive 273 K T (b) Q = −1.17 × 105 J flows out of the heat source, at T = 298 K ΔS = 20.24 Q −1.17 × 105 J = = −393 J/K 298 K T Q is negative and ΔS is negative (c) ΔS tot = 429 J/K + (−393 J/K) = +36 J/K EVALUATE: For the total isolated system, ΔS > and the process is irreversible IDENTIFY: Apply Qsystem = to calculate the final temperature Q = mcΔT Example 20.6 shows that ΔS = mc ln(T2 /T1 ) when an object undergoes a temperature change SET UP: For water c = 4190 J/kg ⋅ K Boiling water has T = 100.0°C = 373 K EXECUTE: (a) The heat transfer between 100°C water and 30°C water occurs over a finite temperature difference and the process is irreversible (b) (270 kg)c (T2 − 30.0°C) + (5.00 kg)c(T2 − 100°C) = T2 = 31.27°C = 304.42 K ⎛ 304.42 K ⎞ ⎛ 304.42 K ⎞ (c) ΔS = (270 kg)(4190 J/kg ⋅ K)ln ⎜ ⎟ + (5.00 kg)(4190 J/kg ⋅ K)ln ⎜ ⎟ ⎝ 303.15 K ⎠ ⎝ 373.15 K ⎠ ΔS = 4730 J/K + (−4265 J/K) = +470 J/K EVALUATE: ΔSsystem > 0, as it should for an irreversible process 20.25 Q For the melting phase T transition, Q = mLf Conservation of energy requires that the quantity of heat that goes into the ice is the IDENTIFY: Both the ice and the room are at a constant temperature, so ΔS = amount of heat that comes out of the room SET UP: For ice, Lf = 334 × 103 J/kg When heat flows into an object, Q > 0, and when heat flows out of an object, Q < EXECUTE: (a) Irreversible because heat will not spontaneously flow out of 15 kg of water into a warm room to freeze the water mLf 2mLf (15.0 kg)(334 × 103 J/kg) −(15.0 kg)(334 × 103 J/kg) (b) ΔS = ΔSice + ΔSroom = + = + Tice Troom 273 K 293 K 20.26 ΔS = +1250 J/K EVALUATE: This result is consistent with the answer in (a) because ΔS > for irreversible processes IDENTIFY: Q = mcΔT for the water Example 20.6 shows that ΔS = mc ln (T2 /T1 ) when an object undergoes a temperature change ΔS = Q/T for an isothermal process SET UP: For water, c = 4190 J/kg ⋅ K 85.0°C = 358.2 K 20.0°C = 293.2 K ⎛T ⎞ ⎛ 293.2 K ⎞ EXECUTE: (a) ΔS = mc ln ⎜ ⎟ = (0.250 kg)(4190 J/kg ⋅ K)ln ⎜ ⎟ = −210 J/K Heat comes out of T ⎝ 358.2 K ⎠ ⎝ 1⎠ the water and its entropy decreases (b) Q = mcΔT = (0.250)(4190 J/kg ⋅ K)(−65.0 K) = −6.81 × 104 J The amount of heat that goes into the air Q +6.81 × 104 J = = +232 J/K T 293.1 K = −210 J/K + 232 J/K = +22 J/K is +6.81 × 104 J For the air, ΔS = ΔSsystem EVALUATE: ΔSsystem > and the process is irreversible © Copyright 2012 Pearson Education, Inc All rights reserved This material is protected under all copyright laws as they currently exist No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher 20-8 Chapter 20 20.27 IDENTIFY: The process is at constant temperature, so ΔS = 20.28 −1850 J = −6.31 J/K 293 K EVALUATE: The entropy change of the gas is negative Heat must be removed from the gas during the compression to keep its temperature constant and therefore the gas is not an isolated system IDENTIFY and SET UP: The initial and final states are at the same temperature, at the normal boiling point of 4.216 K Calculate the entropy change for the irreversible process by considering a reversible isothermal process that connects the same two states, since ΔS is path independent and depends only on the initial and final states For the reversible isothermal process we can use Eq (20.18) The heat flow for the helium is Q = 2mLv , negative since in condensation heat flows out of the helium Q ΔU = Q − W T SET UP: For an isothermal process of an ideal gas, ΔU = and Q = W For a compression, ΔV < and W < EXECUTE: Q = W = −1850 J ΔS = The heat of vaporization Lv is given in Table 17.4 and is Lv = 20.9 × 103 J/kg 20.29 EXECUTE: Q = − mLv = −(0.130 kg)(20.9 × 103 J/kg) = −2717 J ΔS = Q/T = −2717 J/4.216 K = −644 J/K EVALUATE: The system we considered is the 0.130 kg of helium; ΔS is the entropy change of the helium This is not an isolated system since heat must flow out of it into some other material Our result that ΔS < doesn’t violate the 2nd law since it is not an isolated system The material that receives the heat that flows out of the helium would have a positive entropy change and the total entropy change would be positive Q IDENTIFY: Each phase transition occurs at constant temperature and ΔS = Q = mLv T SET UP: For vaporization of water, Lv = 2256 × 103 J/kg Q mLv (1.00 kg)(2256 × 103 J/kg) = = = 6.05 × 103 J/K Note that this is the change (373.15 K) T T of entropy of the water as it changes to steam (b) The magnitude of the entropy change is roughly five times the value found in Example 20.5 EVALUATE: Water is less ordered (more random) than ice, but water is far less random than steam; a consideration of the density changes indicates why this should be so Q IDENTIFY: The phase transition occurs at constant temperature and ΔS = Q = mLv The mass of one T mole is the molecular mass M SET UP: For water, Lv = 2256 × 103 J/kg For N , M = 28.0 × 1023 kg/mol, the boiling point is 77.34 K EXECUTE: (a) ΔS = 20.30 and Lv = 201 × 103 J/kg For silver (Ag), M = 107.9 × 1023 kg/mol, the boiling point is 2466 K and Lv = 2336 × 103 J/kg For mercury (Hg), M = 200.6 × 1023 kg/mol, the boiling point is 630 K and Lv = 272 × 103 J/kg EXECUTE: (a) ΔS = (b) N 2: Q mLv (18.0 × 10−3 kg)(2256 × 103 J/kg) = = = 109 J/K T T (373.15 K) (28.0 × 10−3 kg)(201 × 103 J/kg) (107.9 × 10−3 kg)(2336 × 103 J/kg) = 72.8 J/K Ag: = 102.2 J/K (77.34 K) (2466 K) (200.6 × 10−3 kg)(272 × 103 J/kg) = 86.6 J/K (630 K) (c) The results are the same order or magnitude, all around 100 J/K EVALUATE: The entropy change is a measure of the increase in randomness when a certain number (one mole) goes from the liquid to the vapor state The entropy per particle for any substance in a vapor state is expected to be roughly the same, and since the randomness is much higher in the vapor state (see Exercise 20.29), the entropy change per molecule is roughly the same for these substances Hg: © Copyright 2012 Pearson Education, Inc All rights reserved This material is protected under all copyright laws as they currently exist No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher The Second Law of Thermodynamics 20.31 20-9 IDENTIFY: No heat is transferred, but the entropy of the He increases because it occupies a larger volume and hence is more disordered To calculate the entropy change, we need to find a reversible process that connects the same initial and final states SET UP: The reversible process that connects the same initial and final states is an isothermal expansion at T = 293 K, from V1 = 10.0 L to V2 = 35.0 L For an isothermal expansion of an ideal gas ΔU = and Q = W = nRT ln(V2 /V1 ) EXECUTE: (a) Q = (3.20 mol)(8.315 J/mol ⋅ K)(293 K)ln(35.0 L/10.0 L) = 9767 J ΔS = (b) The isolated system has ΔS > so the process is irreversible 20.32 Q 9767 J = = +33.3 J/K T 293 K EVALUATE: The reverse process, where all the gas in 35.0 L goes through the hole and into the tank does not ever occur IDENTIFY: Apply Eq (20.23) and follow the procedure used in Example 20.11 SET UP: After the partition is punctured each molecule has equal probability of being on each side of the box The probability of two independent events occurring simultaneously is the product of the probabilities of each separate event EXECUTE: (a) On the average, each half of the box will contain half of each type of molecule, 250 of nitrogen and 50 of oxygen (b) See Example 20.11 The total change in entropy is ΔS = kN1ln(2) + kN 2ln(2) = ( N1 + N )k ln(2) = (600)(1.381× 10−23 J/K) ln(2) = 5.74 × 10−21 J/K (c) The probability is (1/2)500 × (1/2)100 = (1/2)600 = 2.4 × 10−181, and is not likely to happen The numerical 20.33 result for part (c) above may not be obtained directly on some standard calculators For such calculators, the result may be found by taking the log base ten of 0.5 and multiplying by 600, then adding 181 and then finding 10 to the power of the sum The result is then 10−181 × 100.87 = 2.4 × 10−181 EVALUATE: The contents of the box constitutes an isolated system ΔS > and the process is irreversible (a) IDENTIFY and SET UP: The velocity distribution of Eq (18.32) depends only on T, so in an isothermal process it does not change (b) EXECUTE: Calculate the change in the number of available microscopic states and apply Eq (20.23) Following the reasoning of Example 20.11, the number of possible positions available to each molecule is altered by a factor of (becomes larger) Hence the number of microscopic states the gas occupies at volume 3V is w2 = (3) N w1, where N is the number of molecules and w1 is the number of possible microscopic states at the start of the process, where the volume is V Then, by Eq (20.23), ΔS = k ln( w2 /w1) = k ln(3) N = Nk ln(3) = nN A k ln(3) = nR ln(3) ΔS = (2.00 mol)(8.3145 J/mol ⋅ K)ln(3) = +18.3 J/K (c) IDENTIFY and SET UP: For an isothermal reversible process ΔS = Q/T EXECUTE: Calculate W and then use the first law to calculate Q ΔT = implies ΔU = 0, since system is an ideal gas Then by ΔU = Q − W , Q = W For an isothermal process, W = ∫ V2 V1 p dV = ∫ V2 (nRT/V ) dV V1 = nRT ln(V2 /V1 ) Thus Q = nRT ln(V2 /V1 ) and ΔS = Q/T = nR ln(V2 /V1 ) ΔS = (2.00 mol)(8.3145 J/mol ⋅ K)ln(3V1/V1 ) = +18.3 J/K 20.34 EVALUATE: This is the same result as obtained in part (b) IDENTIFY: Example 20.8 shows that for a free expansion, ΔS = nR ln(V2 /V1 ) SET UP: V1 = 2.40 L = 2.40 × 10−3 m3 ⎛ 425 m3 EXECUTE: ΔS = (0.100 mol)(8.314 J/mol ⋅ K)ln ⎜ ⎜ 2.40 × 1023 m3 ⎝ EVALUATE: ΔSsystem > and the free expansion is irreversible ⎞ ⎟⎟ = 10.0 J/K ⎠ © Copyright 2012 Pearson Education, Inc All rights reserved This material is protected under all copyright laws as they currently exist No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher 20-10 20.35 Chapter 20 IDENTIFY: The total work that must be done is Wtot = mg Δy |W | = |QH | − |QC | QH > 0, W > and QC < For a Carnot cycle, QC T =− C, QH TH SET UP: TC = 373 K, TH = 773 K |QH |= 250 J ⎛T ⎞ ⎛ 373 K ⎞ EXECUTE: (a) QC = −QH ⎜ C ⎟ = −(250 J) ⎜ ⎟ = −121 J T ⎝ 773 K ⎠ ⎝ H⎠ (b) |W | = 250 J − 121 J = 129 J This is the work done in one cycle Wtot = (500 kg)(9.80 m/s )(100 m) = 4.90 × 105 J The number of cycles required is Wtot 4.90 × 105 J = = 3.80 × 103 cycles |W | 129 J/cycle EVALUATE: In 20.36 QC T = − C , the temperatures must be in kelvins QH TH IDENTIFY: W = QC + QH Since it is a Carnot cycle, QC T = − C The heat required to melt the ice is QH TH Q = mLf SET UP: For water, Lf = 334 × 103 J/kg QH > , QC < QC = − mLf TH = 527°C = 800.15 K EXECUTE: (a) QH = +400 J, W = +300 J QC = W − QH = −100 J TC = −TH (QC /QH ) = −(800.15 K)[(−100 J)/(400 J)] = +200 K = −73°C (b) The total QC required is − mLf = −(10.0 kg)(334 × 103 J/kg) = −3.34 × 106 J QC for one cycle is −100 J, −3.34 × 106 J = 3.34 × 104 cycles −100 J/cycle EVALUATE: The results depend only on the maximum temperature of the gas, not on the number of moles or the maximum pressure IDENTIFY: We know the efficiency of this Carnot engine, the heat it absorbs at the hot reservoir and the temperature of the hot reservoir Q T W SET UP: For a heat engine e = and QH + QC = W For a Carnot cycle, C = − C QC < 0, W > 0, QH QH TH so the number of cycles required is 20.37 and QH > TH = 135°C = 408 K In each cycle, QH leaves the hot reservoir and QC enters the cold reservoir The work done on the water equals its increase in gravitational potential energy, mgh W EXECUTE: (a) e = so W = eQH = (0.22)(150 J) = 33 J QH (b) QC = W − QH = 33 J − 150 J = −117 J (c) ⎛Q ⎞ QC T ⎛ −117 J ⎞ = − C so TC = −TH ⎜ C ⎟ = −(408 K) ⎜ ⎟ = 318 K = 45°C Q QH TH ⎝ 150 J ⎠ ⎝ H⎠ (d) ΔS = Q − QH −150 J 117 J + C = + = The Carnot cycle is reversible and ΔS = TH TC 408 K 318 K W 33 J = = 0.0962 kg = 96.2 g gh (9.80 m/s )(35.0 m) EVALUATE: The Carnot cycle is reversible so ΔS = for the world However some parts of the world (e) W = mgh so m = 20.38 gain entropy while other parts lose it, making the sum equal to zero IDENTIFY: The same amount of heat that enters the person’s body also leaves the body, but these transfers of heat occur at different temperatures, so the person’s entropy changes © Copyright 2012 Pearson Education, Inc All rights reserved This material is protected under all copyright laws as they currently exist No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher 20-12 Chapter 20 (b) process → constant volume (ΔV = 0) Q = nCV ΔT = (0.350 mol)(20.79 J/mol ⋅ K)(600 K − 300 K) = 2180 J ΔV = and W = Then ΔU = Q − W = 2180 J process → Adiabatic means Q = ΔU = nCV ΔT (any process), so ΔU = (0.350 mol)(20.79 J/mol ⋅ K)(492 K − 600 K) = −780 J Then ΔU = Q − W gives W = Q − ΔU = 1780 J (It is correct for W to be positive since ΔV is positive.) process → For constant pressure W = pΔV = (1.013 × 105 Pa)(8.62 × 1023 m3 − 14.1 × 1023 m3 ) = −560 J or W = nRΔT = (0.350 mol)(8.3145 J/mol ⋅ K)(300 K − 492 K) = −560 J, which checks (It is correct for W to be negative, since ΔV is negative for this process.) Q = nC p ΔT = (0.350 mol)(29.10 J/mol ⋅ K)(300 K − 492 K) = −1960 J ΔU = Q − W = −1960 J − ( −560 K) = −1400 J or ΔU = nCV ΔT = (0.350 mol)(20.79 J/mol ⋅ K)(300 K − 492 K) = −1400 J, which checks (c) Wnet = W1→ + W2→3 + W3→1 = + 780 J − 560 J = 1220 J (d) Qnet = Q1→ + Q2→3 + Q3→1 = 2180 J + − 1960 J = +220 J (e) e = work output W 220 J = = = 0.101 = 10.1, heat energy input QH 2180 J e(Carnot) = − TC /TH = − 300 K/600 K = 0.500 EVALUATE: For a cycle ΔU = 0, so by ΔU = Q − W it must be that Qnet = Wnet for a cycle We can also 20.41 check that ΔU net = 0: ΔU net = ΔU1→ + ΔU 2→3 + ΔU 3→1 = 2180 J − 1050 J − 1130 J = e < e(Carnot), as it must IDENTIFY: pV = nRT , so pV is constant when T is constant Use the appropriate expression to calculate Q and W for each process in the cycle e = W QH SET UP: For an ideal diatomic gas, CV = 52 R and C p = 72 R EXECUTE: (a) paVa = 2.0 × 103 J pbVb = 2.0 × 103 J pV = nRT so paVa = pbVb says Ta = Tb (b) For an isothermal process, Q = W = nRT ln(V2 /V1 ) ab is a compression, with Vb < Va , so Q < and heat is rejected bc is at constant pressure, so Q = nC p ΔT = absorbed ca is at constant volume, so Q = nCV ΔT = Cp R pΔV ΔV is positive, so Q > and heat is CV V Δp Δp is negative, so Q < and heat is R rejected (c) Ta = Tc = paVa 2.0 × 103 J pV = = 241 K Tb = b b = Ta = 241 K nR (1.00)(8.314 J/mol ⋅ K) nR pcVc 4.0 × 103 J = = 481 K nR (1.00)(8.314 J/mol ⋅ K) ⎛ 0.0050 m3 ⎞ ⎛V ⎞ = −1.39 × 103 J (d) Qab = nRT ln ⎜ b ⎟ = (1.00 mol)(8.314 J/mol ⋅ K)(241 K)ln ⎜ ⎜ 0.010 m3 ⎟⎟ V ⎝ a⎠ ⎝ ⎠ © Copyright 2012 Pearson Education, Inc All rights reserved This material is protected under all copyright laws as they currently exist No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher The Second Law of Thermodynamics 20-13 ⎛7⎞ Qbc = nC p ΔT = (1.00) ⎜ ⎟ (8.314 J/mol ⋅ K)(241 K) = 7.01 × 103 J ⎝2⎠ ⎛5⎞ Qca = nCV ΔT = (1.00) ⎜ ⎟ (8.314 J/mol ⋅ K)(−241 K) = −5.01 × 103 J Qnet = Qab + Qbc + Qca = 610 J ⎝2⎠ Wnet = Qnet = 610 J (e) e = W 610 J = = 0.087 = 8.7% QH 7.01 × 103 J EVALUATE: We can calculate W for each process in the cycle Wab = Qab = 21.39 × 103 J Wbc = pΔV = (4.0 × 105 Pa)(0.0050 m3 ) = 2.00 × 103 J Wca = Wnet = Wab + Wbc + Wca = 610 J, which does equal Qnet 20.42 (a) IDENTIFY and SET UP: Combine Eqs (20.13) and (20.2) to eliminate QC and obtain an expression for QH in terms of W, TC and TH W = 1.00 J, TC = 268.15 K, TH = 290.15 K For the heat pump QC > and QH < EXECUTE: W = QC + QH ; combining this with QH = QC T = C gives QH TH 1.00 J W = = 13.2 J − TC /TH − (268.15/290.15) (b) Electrical energy is converted directly into heat, so an electrical energy input of 13.2 J would be required W (c) EVALUATE: From part (a), QH = QH decreases as TC decreases The heat pump is less − TC /TH efficient as the temperature difference through which the heat has to be “pumped” increases In an engine, heat flows from TH to TC and work is extracted The engine is more efficient the larger the temperature 20.43 difference through which the heat flows IDENTIFY: Tb = Tc and is equal to the maximum temperature Use the ideal gas law to calculate Ta Apply the appropriate expression to calculate Q for each process e = W ΔU = for a complete cycle and for QH an isothermal process of an ideal gas SET UP: For helium, CV = 3R/2 and C p = 5R/2 The maximum efficiency is for a Carnot cycle, and eCarnot = − TC /TH EXECUTE: (a) Qin = Qab + Qbc Qout = Qca Tmax = Tb = Tc = 327°C = 600 K paVa pbVb p = → Ta = a Tb = (600 K) = 200 K Ta Tb pb pbVb = nRTb → Vb = nRTb (2 moles)(8.31 J/mol ⋅ K)(600 K) = = 0.0332 m3 pb 3.0 × 105 Pa pbVb pcVc p ⎛ 3⎞ = → Vc = Vb b = (0.0332 m3 ) ⎜ ⎟ = 0.0997 m3 = Va ⎝ 1⎠ Tb Tc pc ⎛ 3⎞ Qab = nCV ΔTab = (2 mol) ⎜ ⎟ (8.31 J/mol ⋅ K)(400 K) = 9.97 × 103 J ⎝ 2⎠ c c nRTb b b Qbc = Wbc = ∫ pdV = ∫ V dV = nRTb ln Vc = nRTb ln Vb Qbc = (2.00 mol)(8.31 J/mol ⋅ K)(600 K)ln = 1.10 × 104 J Qin = Qab + Qbc = 2.10 × 104 J © Copyright 2012 Pearson Education, Inc All rights reserved This material is protected under all copyright laws as they currently exist No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher 20-14 Chapter 20 ⎛5⎞ Qout = Qca = nC p ΔTca = (2.00 mol) ⎜ ⎟ (8.31 J/mol ⋅ K)(400 K) = 1.66 × 104 J ⎝2⎠ (b) Q = ΔU + W = + W → W = Qin − Qout = 2.10 × 104 J − 1.66 × 104 J = 4.4 × 103 J e = W/Qin = 20.44 4.4 × 103 J = 0.21 = 21% 2.10 × 104 J T 200 k = 0.67 = 67% (c) emax = eCarnot = − C = − TH 600 k EVALUATE: The thermal efficiency of this cycle is about one-third of the efficiency of a Carnot cycle that operates between the same two temperatures Q T T IDENTIFY: For a Carnot engine, C = C eCarnot = − C |W | = |QH | − |QC | QH > 0, QC < pV = nRT QH TH TH SET UP: The work done by the engine each cycle is mg Δy, with m = 15.0 kg and Δy = 2.00 m TH = 773 K QH = 500 J EXECUTE: (a) The pV diagram is sketched in Figure 20.44 (b) W = mg Δy = (15.0 kg)(9.80 m/s )(2.00 m) = 294 J |QC | = |QH | − |W | = 500 J − 294 J = 206 J, and QC = 2206 J ⎛Q ⎞ ⎛ −206 J ⎞ TC = −TH ⎜ C ⎟ = −(773 K) ⎜ ⎟ = 318 K = 45°C Q 500 J ⎠ ⎝ ⎝ H⎠ T 318 K = 0.589 = 58.9, (c) e = − C = − TH 773 K (d) |QC | = 206 J (e) The maximum pressure is for state a This is also where the volume is a minimum, so Va = 5.00 L = 5.00 × 10−3 m3 Ta = TH = 773 K pa = nRTa (2.00 mol)(8.315 J/mol ⋅ K)(773 K) = = 2.57 × 106 Pa Va 5.00 × 10−3 m3 EVALUATE: We can verify that e = W gives the same value for e as calculated in part (c) QH Figure 20.44 20.45 IDENTIFY: emax = eCarnot = − TC / TH e = temperature change Q = mcΔT W W/t W QC QH = W = QH + QC so = + For a t t t QH QH /t SET UP: TH = 300.15 K, TC = 279.15 K For water, ρ = 1000 kg/m3 , so a mass of kg has a volume of L For water, c = 4190 J/kg ⋅ K EXECUTE: (a) e = − 279.15K = 7.0% 300.15K © Copyright 2012 Pearson Education, Inc All rights reserved This material is protected under all copyright laws as they currently exist No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher The Second Law of Thermodynamics (b) QH Pout 210 kW Q Q W = = = 3.0 MW C = H − = 3.0 MW − 210 kW = 2.8 MW t e 0.070 t t t (c) m |QC |/t (2.8 × 106 W)(3600 s/h) = = = × 105 kg/h = × 105 L/h t c ΔT (4190 J/kg ⋅ K)(4 K) 20-15 EVALUATE: The efficiency is small since TC and TH don’t differ greatly 20.46 IDENTIFY: Use Eq (20.4) to calculate e SET UP: The cycle is sketched in Figure 20.46 CV = R/2 for an ideal gas C p = CV + R = R/2 Figure 20.46 SET UP: Calculate Q and W for each process process → ΔV = implies W = ΔV = implies Q = nCV ΔT = nCV (T2 − T1 ) But pV = nRT and V constant says p1V = nRT1 and p2V = nRT2 Thus ( p2 − p1 )V = nR (T2 − T1); V Δp = nRΔT (true when V is constant) Then Q = nCV ΔT = nCV (V Δp/nR ) = (CV /R)V Δp = (CV /R )V0 (2 p0 − p0 ) = (CV /R) p0V0 Q > 0; heat is absorbed by the gas.) process → Δp = so W = pΔV = p (V3 − V2 ) = p0 (2V0 − V0 ) = p0V0 (W is positive since V increases.) Δp = implies Q = nC p ΔT = nC p (T2 − T1 ) But pV = nRT and p constant says pV1 = nRT1 and pV2 = nRT2 Thus p (V2 − V1) = nR(T2 − T1 ); pΔV = nRΔT (true when p is constant) Then Q = nC p ΔT = nC p ( pΔV/nR ) = (C p /R ) pΔV = (C p /R )2 p0 (2V0 − V0 ) = (C p /R )2 p0V0 (Q > 0; heat is absorbed by the gas.) process → ΔV = implies W = ΔV = so Q = nCV ΔT = nCV (V Δp/nR ) = (CV /R)(2V0 )( p0 − p0 ) = −2(CV /R) p0V0 (Q < so heat is rejected by the gas.) process → Δp = so W = pΔV = p (V1 − V4 ) = p0 (V0 − 2V0 ) = − p0V0 (W is negative since V decreases) Δp = so Q = nC p ΔT = nC p ( pΔV/nR) = (C p /R) pΔV = (C p /R) p0 (V0 − 2V0 ) = −(C p /R) p0V0 (Q < so heat is rejected by the gas.) © Copyright 2012 Pearson Education, Inc All rights reserved This material is protected under all copyright laws as they currently exist No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher 20-16 Chapter 20 total work performed by the gas during the cycle: Wtot = W1→ + W2→3 + W3→4 + W4→1 = + p0V0 + − p0V0 = p0V0 (Note that Wtot equals the area enclosed by the cycle in the pV-diagram.) total heat absorbed by the gas during the cycle (QH ): Heat is absorbed in processes → and → Cp ⎛ CV + 2C p ⎞ C QH = Q1→ + Q2→3 = V p0V0 + p0V0 = ⎜ ⎟ p0V0 R R R ⎝ ⎠ C + 2(CV + R ) ⎛ 3C + R ⎞ p0V0 = ⎜ V But C p = CV + R so QH = V ⎟ p0V0 R R ⎝ ⎠ total heat rejected by the gas during the cycle (QC ): Heat is rejected in processes → and → Cp ⎛ 2CV + C p ⎞ C QC = Q3→4 + Q4→1 = −2 V p0V0 − p0V0 = − ⎜ ⎟ p0V0 R R R ⎝ ⎠ 2CV + (CV + R) C +R⎞ ⎛ p0V0 = 2⎜ V But C p = CV + R so QC = ⎟ p0V0 R R ⎝ ⎠ efficiency W p0V0 R R e= = = = = QH ([3CV + R]/R )( p0V0 ) 3CV + R 3(5R/2) + R 19 20.47 e = 0.105 = 10.5, EVALUATE: As a check on the calculations note that ⎛ 3C + R ⎞ ⎛ 3CV + R ⎞ QC + QH = 2⎜ V ⎟ p0V0 + ⎜ ⎟ p0V0 = p0V0 = W , as it should R R ⎝ ⎠ ⎝ ⎠ IDENTIFY: Use pV = nRT Apply the expressions for Q and W that apply to each type of process e= W QH SET UP: For O , CV = 20.85 J/mol ⋅ K and C p = 29.17 J/mol ⋅ K EXECUTE: (a) p1 = 2.00 atm, V1 = 4.00 L, T1 = 300 K ⎛T ⎞ V1 V2 ⎛ 450 K ⎞ = V2 = ⎜ ⎟V1 = ⎜ ⎟ (4.00 L) = 6.00 L T1 T2 ⎝ 300 K ⎠ ⎝ T1 ⎠ ⎛T ⎞ p p ⎛ 250 K ⎞ V3 = 6.00 L = p3 = ⎜ ⎟ p2 = ⎜ ⎟ (2.00 atm) = 1.11 atm T2 T3 ⎝ 450 K ⎠ ⎝ T2 ⎠ p2 = 2.00 atm ⎛V ⎞ ⎛ 6.00 L ⎞ V4 = 4.00 L p3V3 = p4V4 p4 = p3 ⎜ ⎟ = (1.11 atm ) ⎜ ⎟ = 1.67 atm V ⎝ 4.00 L ⎠ ⎝ 4⎠ These processes are shown in Figure 20.47 pV (2.00 atm)(4.00 L) = 0.325 mol (b) n = 1 = RT1 (0.08206 L ⋅ atm/mol ⋅ K)(300 K) process → 2: W = pΔV = nRΔT = (0.325 mol)(8.315 J/mol ⋅ K)(150 K) = 405 J Q = nC p ΔT = (0.325 mol)(29.17 J/mol ⋅ K)(150 K) = 1422 J process → 3: W = Q = nCV ΔT = (0.325 mol)(20.85 J/mol ⋅ K)(−200 K) = −1355 J process → 4: ΔU = and ⎛V ⎞ ⎛ 4.00 L ⎞ Q = W = nRT3 ln ⎜ ⎟ = (0.325 mol)(8.315 J/mol ⋅ K)(250 K)ln ⎜ ⎟ = 2274 J V ⎝ 6.00 L ⎠ ⎝ 3⎠ process → 1: W = Q = nCV ΔT = (0.325 mol)(20.85 J/mol ⋅ K)(50 K) = 339 J (c) W = 405 J − 274 J = 131 J © Copyright 2012 Pearson Education, Inc All rights reserved This material is protected under all copyright laws as they currently exist No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher The Second Law of Thermodynamics (d) e = 20-17 W 131 J = = 0.0744 = 7.44, QH 1422 J + 339 J eCarnot = − TC 250 K =1− = 0.444 = 44.4,; eCarnot is much larger TH 450 K EVALUATE: Qtot = 11422 J + ( −1355 J) + (−274 J) + 339 J = 132 J This is equal to Wtot , apart from a slight difference due to rounding For a cycle, Wtot = Qtot , since ΔU = Figure 20.47 20.48 IDENTIFY: The air in the room receives heat radiated from the person at 30.0°C but radiates part of it back to the person at 20.0°C, so it undergoes an entropy change SET UP: A person with surface area A and surface temperature T = 303 K radiates at a rate H = Aeσ T The person absorbs heat from the room at a rate H s = Aeσ Ts , where Ts = 293 K is the temperature of the room In t = 1.0 s, heat Aeσ tT flows into the room and heat Aeσ tTs flows out of the room The heat flows into and out of the room occur at a temperature of Ts EXECUTE: For the room, ΔS = Aeσ tT Aeσ tTs Aeσ t (T − Ts ) − = Putting in the numbers gives Ts Ts Ts (1.85 m )(1.00)(5.67 × 10−8 W/m ⋅ K )(1.0 s)([303 K]4 − [293 K]4 ) = 0.379 J/K 293 K EVALUATE: The room gains entropy because its disorder increases IDENTIFY: Since there is temperature difference between the inside and outside of your body, you can use it as a heat engine W T For a Carnot engine e = − C Gravitational potential energy is SET UP: For a heat engine e = QH TH ΔS = 20.49 U grav = mgh food-calorie = 1000 cal = 4186 J EXECUTE: (a) e = − TC 303 K =1− = 0.0226 = 2.26% This engine has a very low thermal efficiency TH 310 K (b) U grav = mgh = (2.50 kg)(9.80 m/s )(1.20 m) = 29.4 J This equals the work output of the engine W W 29.4 J so QH = = = 1.30 × 103 J e 0.0226 QH (C) Since 80% of food energy goes into heat, you must eat food with a food energy of 1.30 × 103 J = 1.63 × 103 J Each candy bar gives (350 food-calorie)(4186 J/food-calorie) = 1.47 × 106 J 0.80 e= The number of candy bars required is 1.63 × 103 J 1.47 × 106 J/candy bar = 1.11 × 10−3 candy bars © Copyright 2012 Pearson Education, Inc All rights reserved This material is protected under all copyright laws as they currently exist No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher 20-18 20.50 Chapter 20 EVALUATE: A large amount of mechanical work must be done to use up the energy from one candy bar IDENTIFY: The sun radiates energy into the universe and therefore increases its entropy SET UP: The sun radiates heat energy at a rate H = Aeσ T The rate at which the sun absorbs heat from the surrounding space is negligible, since space is so much colder This heat flows out of the sun at 5800 K and into the surrounding space at K From Appendix F, the radius of the sun is 6.96 × 108 m The surface area of a sphere with radius R is A = 4π R EXECUTE: (a) In s the quantity of heat radiated by the sun is Q = Aeσ tT = 4π R 2eσ tT Putting in the numbers gives Q = 4π (6.96 × 108 m)2 (1.0)(5.67 × 10−8 W/m ⋅ K )(1.0 s)(5800 K) = 3.91 × 1026 J −3.91 × 1026 J +3.91 × 1026 J + = +1.30 × 1026 J/K 5800 K 3K (b) The process of radiation is irreversible; this heat flows from the hot object (sun) to the cold object (space) and not in the reverse direction This is consistent with the answer to part (a) We found ΔSuniverse > and this is the case for an irreversible process ΔS = 20.51 EVALUATE: The entropy of the sun decreases because there is a net heat flow out of it The entropy of space increases because there is a net heat flow into it But the heat flow into space occurs at a lower temperature than the heat flow out of the sun and the net entropy change of the universe is positive IDENTIFY: Use ΔU = Q − W and the appropriate expressions for Q, W and ΔU for each type of process pV = nRT relates ΔT to p and V values e = W , where QH is the heat that enters the gas during the QH cycle SET UP: For a monatomic ideal gas, C p = 52 R and CV = 32 R (a) ab: The temperature changes by the same factor as the volume, and so Cp Q = nC p ΔT = pa (Va − Vb ) = (2.5)(3.00 × 105 Pa)(0.300 m3 ) = 2.25 × 105 J R The work pΔV is the same except for the factor of , so W = 0.90 × 105 J ΔU = Q − W = 1.35 × 105 J bc: The temperature now changes in proportion to the pressure change, and Q = 32 ( pc − pb )Vb = (1.5)(−2.00 × 105 Pa)(0.800 m3 ) = −2.40 × 105 J, and the work is zero (ΔV = 0) ΔU = Q − W = 22.40 × 105 J ca: The easiest way to this is to find the work done first; W will be the negative of area in the p-V plane bounded by the line representing the process ca and the verticals from points a and c The area of this trapezoid is 12 (3.00 × 105 Pa + 1.00 × 105 Pa)(0.800 m3 − 0.500 m3 ) = 6.00 × 104 J and so the work is −0.60 × 105 J ΔU must be 1.05 × 105 J (since ΔU = for the cycle, anticipating part (b)), and so Q must be ΔU + W = 0.45 × 105 J (b) See above; Q = W = 0.30 × 105 J, ΔU = (c) The heat added, during process ab and ca, is 2.25 × 105 J + 0.45 × 105 J = 2.70 × 105 J and the efficiency W 0.30 × 105 = = 0.111 = 11.1, is e = QH 2.70 × 105 EVALUATE: For any cycle, ΔU = and Q = W 20.52 IDENTIFY: Use the appropriate expressions for Q, W and ΔU for each process e = W/QH and eCarnot = − TC /TH SET UP: For this cycle, TH = T2 and TC = T1 © Copyright 2012 Pearson Education, Inc All rights reserved This material is protected under all copyright laws as they currently exist No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher The Second Law of Thermodynamics 20-19 EXECUTE: (a) ab: For the isothermal process, ΔT = and ΔU = W = nRT1 ln(Vb /Va ) = nRT1ln(1/r ) = − nRT1ln(r ) and Q = W = 2nRT1 ln(r ) bc: For the isochoric process, ΔV = and W = Q = ΔU = nCV ΔT = nCV (T2 − T1 ) cd: As in the process ab, ΔU = and W = Q = nRT2ln(r ) da: As in process bc, ΔV = and W = 0; ΔU = Q = nCV (T1 − T2 ) (b) The values of Q for the processes are the negatives of each other (c) The net work for one cycle is Wnet = nR (T2 − T1 )ln(r ), and the heat added is Qcd = nRT2 ln(r ), and the efficiency is e = 20.53 Wnet = − (T1/T2 ) This is the same as the efficiency of a Carnot-cycle engine operating Qcd between the two temperatures EVALUATE: For a Carnot cycle two steps in the cycle are isothermal and two are adiabatic and all the heat flow occurs in the isothermal processes For the Stirling cycle all the heat flow is also in the isothermal steps, since the net heat flow in the two constant volume steps is zero W + W2 IDENTIFY: The efficiency of the composite engine is e12 = , where QH1 is the heat input to the QH1 first engine and W1 and W2 are the work outputs of the two engines For any heat engine, W = QC + QH , and for a Carnot engine, Qlow T = low , where Qlow and Qhigh are the heat flows at the two reservoirs Qhigh Thigh that have temperatures Tlow and Thigh SET UP: Qhigh,2 = 2Qlow,1 Tlow,1 = T ′, Thigh,1 = TH , Tlow,2 = TC and Thigh,2 = T ′ EXECUTE: e12 = e12 = + e12 = − 20.54 Qlow,2 Qhigh,1 W1 + W2 Qhigh,1 + Qlow,1 + Qhigh,2 + Qlow,2 = Since Qhigh,2 = 2Qlow,1, this reduces to QH1 Qhigh,1 Qlow,2 = 2Qhigh,2 Tlow,2 Thigh,2 = Qlow,1 ⎛T ⎞T ⎛ T′ ⎞T TC = 2Qhigh,1 ⎜ low,1 ⎟ C = 2Qhigh,1 ⎜ ⎟ C This gives ⎜ ⎟ T′ ⎝ TH ⎠ T ′ ⎝ Thigh,1 ⎠ T ′ TC The efficiency of the composite system is the same as that of the original engine TH EVALUATE: The overall efficiency is independent of the value of the intermediate temperature T ′ W day = 8.64 × 104 s For the river water, Q = mcΔT , where the heat that goes into IDENTIFY: e = QH the water is the heat QC rejected by the engine The density of water is 1000 kg/m3 When an object undergoes a temperature change, ΔS = mc ln(T2 /T1) SET UP: 18.0°C = 291.1 K 18.5°C = 291.6 K W P 1000 MW EXECUTE: (a) QH = so PH = W = = 2.50 × 103 MW 0.40 e e (b) The heat input in one day is (2.50 × 109 W)(8.64 × 104 s) = 2.16 × 1014 J The mass of coal used per day is 2.16 × 1014 J 2.65 × 107 J/kg = 8.15 × 106 kg (c) |QH | = |W | + |QC | |QC | = |QH | − |W | PC = PH − PW = 2.50 × 103 MW − 1000 MW = 1.50 × 103 MW (d) The heat input to the river is 1.50 × 109 J/s Q = mcΔT and ΔT = 0.5 C° gives m= Q 1.50 × 109 J m = = 7.16 × 105 kg V = = 716 m3 The river flow rate must be 716 m3 /s cΔT (4190 J/kg ⋅ K)(0.5 K) ρ © Copyright 2012 Pearson Education, Inc All rights reserved This material is protected under all copyright laws as they currently exist No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher 20-20 Chapter 20 (e) In one second, 7.16 × 105 kg of water goes from 291.1 K to 291.6 K 20.55 ⎛T ⎞ ⎛ 291.6 K ⎞ ΔS = mc ln ⎜ ⎟ = (7.16 × 105 kg)(4190 J/kg ⋅ K)ln ⎜ ⎟ = 5.1 × 10 J/K ⎝ 291.1 K ⎠ ⎝ T1 ⎠ EVALUATE: The entropy of the river increases because heat flows into it The mass of coal used per second is huge (a) IDENTIFY and SET UP: Calculate e from Eq (20.6), QC from Eq (20.4) and then W from Eq (20.2) EXECUTE: e = − 1/(r γ −1 ) = − 1/(10.60.4 ) = 0.6111 e = (QH + QC )/QH and we are given QH = 200 J; calculate QC QC = (e − 1)QH = (0.6111 − 1)(200 J) = −78 J (Negative, since corresponds to heat leaving.) Then W = QC + QH = −78 J + 200 J = 122 J (Positive, in agreement with Figure 20.6.) EVALUATE: QH , W > , and QC < for an engine cycle (b) IDENTIFY and SET UP: The stoke times the bore equals the change in volume The initial volume is the final volume V times the compression ratio r Combining these two expressions gives an equation for V For each cylinder of area A = π (d/2) the piston moves 0.864 m and the volume changes from rV to V, as shown in Figure 20.55a l1 A = rV l2 A = V and l1 − l2 = 86.4 × 10−3 m Figure 20.55a EXECUTE: l1 A − l2 A = rV − V and (l1 − l2 ) A = ( r − 1)V V= (l1 − l2 ) A (86.4 × 1023 m)π (41.25 × 10−3 m) = = 4.811 × 1025 m3 10.6 − r −1 At point a the volume is rV = 10.6(4.811 × 1025 m3 ) = 5.10 × 1024 m3 (c) IDENTIFY and SET UP: The processes in the Otto cycle are either constant volume or adiabatic Use the QH that is given to calculate ΔT for process bc Use Eq (19.22) and pV = nRT to relate p, V and T for the adiabatic processes ab and cd EXECUTE: point a: Ta = 300 K, pa = 8.50 × 104 Pa and Va = 5.10 × 1024 m3 point b: Vb = Va /r = 4.81 × 1025 m3 Process a → b is adiabatic, so TaVaγ −1 = TbVbγ −1 Ta (rV )γ −1 = TbV γ −1 Tb = Ta r γ −1 = 300 K(10.6)0.4 = 771 K pV = nRT so pV/T = nR = constant, so paVa /Ta = pbVb /Tb pb = pa (Va /Vb )(Tb /Ta ) = (8.50 × 104 Pa)(rV/V )(771 K/300 K) = 2.32 × 106 Pa point c: Process b → c is at constant volume, so Vc = Vb = 4.81 × 1025 m3 QH = nCV ΔT = nCV (Tc − Tb ) The problem specifies QH = 200 J; use to calculate Tc First use the p, V, T values at point a to calculate the number of moles n pV (8.50 × 104 Pa)(5.10 × 1024 m3 ) n= = = 0.01738 mol (8.3145 J/mol ⋅ K)(300 K) RT © Copyright 2012 Pearson Education, Inc All rights reserved This material is protected under all copyright laws as they currently exist No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher The Second Law of Thermodynamics Then Tc − Tb = 20-21 QH 200 J = = 561.3 K, and nCV (0.01738 mol)(20.5 J/mol ⋅ K) Tc = Tb + 561.3 K = 771 K + 561 K = 1332 K p/T = nR/V = constant so pb /Tb = pc /Tc pc = pb (Tc /Tb ) = (2.32 × 106 Pa)(1332 K/771 K) = 4.01 × 106 Pa point d: Vd = Va = 5.10 × 1024 m3 process c → d is adiabatic, so TdVdγ −1 = TcVcγ −1 Td (rV )γ −1 = TcV γ −1 Td = Tc /r γ −1 = 1332 K/10.60.4 = 518 K pcVc /Tc = pdVd /Td pd = pc (Vc /Vd )(Td /Tc ) = (4.01 × 106 Pa)(V/rV )(518 K/1332 K) = 1.47 × 105 Pa EVALUATE: Can look at process d → a as a check QC = nCV (Ta − Td ) = (0.01738 mol)(20.5 J/mol ⋅ K)(300 K − 518 K) = 278 J, which agrees with part (a) The cycle is sketched in Figure 20.55b Figure 20.55b (d) IDENTIFY and SET UP: The Carnot efficiency is given by Eq (20.14) TH is the highest temperature reached in the cycle and TC is the lowest EXECUTE: From part (a) the efficiency of this Otto cycle is e = 0.611 = 61.1, The efficiency of a Carnot cycle operating between 1332 K and 300 K is e(Carnot) = − TC /TH = − 300 K/1332 K = 0.775 = 77.5%, which is larger EVALUATE: The 2nd law requires that e ≤ e (Carnot), and our result obeys this law 20.56 IDENTIFY: K = |QC | |QH | = |QC | + |W | The heat flows for the inside and outside air occur at constant T, |W | so ΔS = Q/T SET UP: 21.0°C = 294.1 K 35.0°C = 308.1 K EXECUTE: (a) |QC | = K |W | PC = KPW = (2.80)(800 W) = 2.24 × 103 W (b) PH = PC + PW = 2.24 × 103 W + 800 W = 3.04 × 103 W © Copyright 2012 Pearson Education, Inc All rights reserved This material is protected under all copyright laws as they currently exist No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher 20-22 Chapter 20 (c) In h = 3600 s, QH = PHt = 1.094 × 107 J ΔSout = QH 1.094 × 107 J = = 3.55 × 104 J/K 308.1 K TH (d) QC = PCt = 8.064 × 106 J Heat QC is removed from the inside air ΔSin = 20.57 2QC 28.064 × 106 J = = 22.74 × 104 J/K ΔSnet = ΔSout + ΔSin = 8.1 × 103 J/K 294.1 K TC EVALUATE: The increase in the entropy of the outside air is greater than the entropy decrease of the air in the room IDENTIFY and SET UP: Use Eq (20.13) for an infinitesimal heat flow dQH from the hot reservoir and use that expression with Eq (20.19) to relate ΔSH , the entropy change of the hot reservoir, to |QC | (a) EXECUTE: Consider an infinitesimal heat flow dQH that occurs when the temperature of the hot reservoir is T ′: dQC = −(TC /T ′)dQH dQH T′ dQ |QC | = TC ∫ H = TC |ΔS H | T′ (b) The 1.00 kg of water (the high-temperature reservoir) goes from 373 K to 273 K QH = mcΔT = (1.00 kg)(4190 J/kg ⋅ K)(100 K) = 4.19 × 105 J ∫ dQC = 2TC ∫ ΔS H = mc ln(T2 /T1 ) = (1.00 kg)(4190 J/kg ⋅ K)ln(273/373) = −1308 J/K The result of part (a) gives |QC | = (273 K)(1308 J/K) = 3.57 × 105 J QC comes out of the engine, so QC = 23.57 × 105 J Then W = QC + QH = −3.57 × 105 J + 4.19 × 105 J = 6.2 × 104 J (c) 2.00 kg of water goes from 323 K to 273 K QH = 2mcΔT = (2.00 kg)(4190 J/kg ⋅ K)(50 K) = 4.19 × 105 J ΔSH = mc ln(T2 /T1 ) = (2.00 kg)(4190 J/kg ⋅ K)ln(272 / 323) = 21.41 × 103 J/K QC = −TC |ΔS H | = −3.85 × 105 J W = QC + QH = 3.4 × 104 J 20.58 (d) EVALUATE: More work can be extracted from 1.00 kg of water at 373 K than from 2.00 kg of water at 323 K even though the energy that comes out of the water as it cools to 273 K is the same in both cases The energy in the 323 K water is less available for conversion into mechanical work IDENTIFY: The maximum power that can be extracted is the total kinetic energy K of the mass of air that passes over the turbine blades in time t SET UP: The volume of a cylinder of diameter d and length L is (π d /4) L Kinetic energy is 12 mv EXECUTE: (a) The cylinder described contains a mass of air m = ρ (π d /4) L, and so the total kinetic energy is K = ρ (π /8)d Lv This mass of air will pass by the turbine in a time t = L/v, and so the K = ρ (π /8)d 2v3 Numerically, the product ρair (π /8) ≈ 0.5 kg/m3 = 0.5 W ⋅ s3/m5 t This completes the proof maximum power is P = 1/3 1/3 ⎛ (3.2 × 106 W)/(0.25) ⎞ =⎜ = 14 m/s = 50 km/h ⎜ (0.5 W ⋅ s3/m5 )(97 m) ⎟⎟ ⎝ ⎠ (c) Wind speeds tend to be higher in mountain passes EVALUATE: The maximum power is proportional to v3, so increases rapidly with increase in wind speed ⎛ P/e ⎞ (b) v = ⎜ ⎟ ⎝ kd ⎠ © Copyright 2012 Pearson Education, Inc All rights reserved This material is protected under all copyright laws as they currently exist No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher The Second Law of Thermodynamics 20.59 20-23 IDENTIFY: Apply Eq (20.19) From the derivation of Eq (20.6), Tb = r γ −1Ta and Tc = r γ −1Td SET UP: For a constant volume process, dQ = nCV dT EXECUTE: (a) For a constant-volume process for an ideal gas, where the temperature changes from T1 to T2 dT ⎛T ⎞ T2, ΔS = nCV Ñ = nCV ln ⎜ ⎟ The entropy changes are nCV ln(Tc /Tb ) and nCV ln(Ta /Td ) T1 T ⎝ T1 ⎠ (b) The total entropy change for one cycle is the sum of the entropy changes found in part (a); the other processes in the cycle are adiabatic, with Q = and ΔS = The total is then ⎛TT ⎞ TT Tc T r γ −1T T + nCV ln a = nCV ln ⎜ c a ⎟ c a = γ −1 d a = ln(1) = 0, so ΔS = Tb Td ⎝ TbTd ⎠ TbTd r Td Ta (c) The system is not isolated, and a zero change of entropy for an irreversible system is certainly possible EVALUATE: In an irreversible process for an isolated system, ΔS > But the entropy change for some of the components of the system can be negative or zero Q IDENTIFY: For a reversible isothermal process, ΔS = For a reversible adiabatic process, Q = and T ΔS = The Carnot cycle consists of two reversible isothermal processes and two reversible adiabatic processes SET UP: Use the results for the Stirling cycle from Problem 20.52 EXECUTE: (a) The graph is given in Figure 20.60 dQ (b) For a reversible process, dS = , and so dQ = T dS , and Q = ∫ dQ = ∫ T dS , which is the area under T the curve in the TS plane (c) QH is the area under the rectangle bounded by the horizontal part of the rectangle at TH and the ΔS = nCV ln 20.60 verticals |QC | is the area bounded by the horizontal part of the rectangle at TC and the verticals The net work is then QH − |QC |, the area bounded by the rectangle that represents the process The ratio of the areas is the ratio of the lengths of the vertical sides of the respective rectangles, and the efficiency is W TH − TC e= = QH TH (d) As explained in Problem 20.52, the substance that mediates the heat exchange during the isochoric expansion and compression does not leave the system, and the diagram is the same as in part (a) As found in that problem, the ideal efficiency is the same as for a Carnot-cycle engine EVALUATE: The derivation of eCarnot using the concept of entropy is much simpler than the derivation in Section 20.6, but yields the same result Figure 20.60 © Copyright 2012 Pearson Education, Inc All rights reserved This material is protected under all copyright laws as they currently exist No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher 20-24 20.61 Chapter 20 IDENTIFY: The temperatures of the ice-water mixture and of the boiling water are constant, so ΔS = Q T The heat flow for the melting phase transition of the ice is Q = + mLf SET UP: For water, Lf = 3.34 × 105 J/kg EXECUTE: (a) The heat that goes into the ice-water mixture is Q = mLf = (0.120 kg)(3.34 × 105 J/kg) = 4.008 × 104 J This same amount of heat leaves the boiling water, so ΔS = Q −4.008 × 104 J = = −107 J/K 373 K T Q 4.008 × 104 J = = +147 J/K 273 K T (c) For any segment of the rod, the net heat flow is zero, so ΔS = (d) ΔS tot = −107 J/K + 147 J/K = +39.4 J/K (Carry extra figures when subtraction is involved.) (b) ΔS = 20.62 EVALUATE: The heat flow is irreversible, since the system is isolated and the total entropy change is positive IDENTIFY: Use the expression derived in Example 20.6 for the entropy change in a temperature change SET UP: For water, c = 4190 J/kg ⋅ K 20°C = 293.15 K, 78°C = 351.15 K and 120°C = 393.15 K EXECUTE: (a) ΔS = mcln(T2 /T1 ) = (250 × 10−3 kg)(4190 J/kg ⋅ K)ln(351.15 K/293.15 K) = 189 J/K (b) ΔS = − mcΔT −(250 × 10−3 kg)(4190 J/kg ⋅ K)(351.15 K − 293.15 K) = = −155 J/K 393.15 K Telement (c) The sum of the result of parts (a) and (b) is ΔSsystem = 34.6 J/K (Carry extra figures when subtraction 20.63 is involved.) EVALUATE: (d) Heating a liquid is not reversible Whatever the energy source for the heating element, heat is being delivered at a higher temperature than that of the water, and the entropy loss of the source will be less in magnitude than the entropy gain of the water The net entropy change is positive IDENTIFY: Use the expression derived in Example 20.6 for the entropy change in a temperature change For the value of T for which ΔS is a maximum, d ( ΔS )/dT = SET UP: The heat flow for a temperature change is Q = mcΔT EXECUTE: (a) As in Example 20.10, the entropy change of the first object is m1c1ln(T/T1) and that of the second is m2c2ln(T ′/T2 ), and so the net entropy change is as given Neglecting heat transfer to the surroundings, Q1 + Q2 = , m1c1 (T − T1) + m2c2 (T ′ − T2 ) = , which is the given expression (b) Solving the energy-conservation relation for T ′ and substituting into the expression for ΔS gives ⎛ ⎛T ⎞ m c ⎛ T T ⎞⎞ ΔS = m1c1ln ⎜ ⎟ + m2c21n ⎜⎜1 − 1 ⎜ − ⎟ ⎟⎟ Differentiating with respect to T and setting the ⎝ T1 ⎠ ⎝ m2c2 ⎝ T2 T2 ⎠ ⎠ mc ( m2c2 )(m1c1/m2c2 )(−1/T2 ) This may be solved for derivative equal to gives = 1 + ⎛ T ⎛ T T1 ⎞ ⎞ ( / ) m c m c − − ⎟ ⎟⎟ 1 2 ⎜ ⎜⎜ ⎝ T2 T2 ⎠ ⎠ ⎝ T= m1c1T1 + m2c2T2 Using this value for T in the conservation of energy expression in part (a) and m1c1 + m2c2 m1c1T1 + m2c2T2 Therefore, T = T ′ when ΔS is a maximum m1c1 + m2c2 EVALUATE: (c) The final state of the system will be that for which no further entropy change is possible If T < T ′, it is possible for the temperatures to approach each other while increasing the total entropy, but when T = T ′, no further spontaneous heat exchange is possible solving for T ′ gives T ′ = © Copyright 2012 Pearson Education, Inc All rights reserved This material is protected under all copyright laws as they currently exist No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher The Second Law of Thermodynamics 20.64 20-25 IDENTIFY: Calculate QC and QH in terms of p and V at each point Use the ideal gas law and the pressure-volume relation for adiabatic processes for an ideal gas e = − |QC | |QH | SET UP: For an ideal gas, C p = CV + R, and taking air to be diatomic, C p = 72 R, CV = 52 R and γ = 75 EXECUTE: Referring to Figure 20.7 in the textbook, QH = n 72 R (Tc − Tb ) = 72 ( pcVc − pbVb ) Similarly, QC = n 52 R( paVa − pdVd ) What needs to be done is to find the relations between the product of the pressure and the volume at the four points For an ideal gas, ⎛T ⎞ pcVc pbVb so pcVc = paVa ⎜ c ⎟ For a compression = Tc Tb ⎝ Ta ⎠ γ −1 ⎛V ⎞ ratio r, and given that for the Diesel cycle the process ab is adiabatic, pbVb = paVa ⎜ a ⎟ ⎝ Vb ⎠ = paVa r γ −1 γ −1 ⎛V ⎞ Similarly, pdVd = pcVc ⎜ c ⎟ ⎝ Va ⎠ Note that the last result uses the fact that process da is isochoric, and ⎛T ⎞ Vd = Va ; also, pc = pb (process bc is isobaric), and so Vc = Vb ⎜ c ⎟ Then, ⎝ Ta ⎠ −γ Vc Tc Vb Tb Ta Va Tc ⎛ TaVaγ −1 ⎞ ⎛ Va ⎞ T = ⋅ = ⋅ ⋅ = ⋅⎜ = c rγ ⎟ ⎜ ⎟ γ − ⎜ ⎟ Va Tb Va Ta Tb Vb Ta ⎝ TbVb ⎠ ⎝ Vb ⎠ Ta γ ⎛T ⎞ Combining the above results, pdVd = paVa ⎜ c ⎟ r γ −γ Substitution of the above results into Eq (20.4) ⎝ Ta ⎠ γ ⎡⎛ T ⎞ ⎤ ⎢ ⎜ c ⎟ r γ −γ − ⎥ ⎥ 5⎢ T gives e = − ⎢ ⎝ a ⎠ ⎥ ⎢ ⎛ Tc ⎞ γ −1 ⎥ r − ⎢ ⎜⎝ Ta ⎟⎠ ⎥ ⎣ ⎦ (b) e = − ⎡ (5.002)r −0.56 − ⎤ T ⎢ ⎥ , where c = 3.167 and γ = 1.40 have been used Substitution of r = 21.0 1.4 ⎣⎢ (3.167) − r 0.40 ⎦⎥ Ta yields e = 0.708 = 70.8, EVALUATE: The efficiency for an Otto cycle with r = 21.0 and γ = 1.40 is e = − r1−γ = − (21.0) −0.40 = 70.4, This is very close to the value for the Diesel cycle © Copyright 2012 Pearson Education, Inc All rights reserved This material is protected under all copyright laws as they currently exist No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher ... of the device is sketched in Figure 20 .13 EXECUTE: W = QC + QH W = −335 J + 550 J = 215 J Figure 20 .13 (b) For a Carnot cycle, TC = TH |QC | TC = (Eq 20 .13) |QH | TH ⎛ 335 J ⎞ |QC | = 620 K ⎜... 20-17 W 131 J = = 0.0744 = 7.44, QH 1422 J + 339 J eCarnot = − TC 250 K =1− = 0.444 = 44.4,; eCarnot is much larger TH 450 K EVALUATE: Qtot = 11422 J + ( 135 5 J) + (−274 J) + 339 J = 132 J This... )(12.3 atm − 1.5 atm)(1. 013 × 105 Pa/atm) = 5470 J QH = 5470 J ⎝ R ⎠ ⎝2⎠ (c) Heat leaves the gas in process bc, since T increases ⎛ Cp ⎞ ⎛7⎞ −3 Q=⎜ ⎟ pΔV = ⎜ ⎟ (1.5 atm)(1. 013 × 10 Pa/atm)(−7.0 ×