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Autor: ciro guilherme | Publicado: 2011-11-28T20:20:00-08:00

   Chapter 20                     1. An isothermal process is one in which Ti = Tf which implies ln(Tf/Ti) = 0. Therefore, with Vf/Vi = 2.00, Eq. 20-4 leads to     <!--[if gte vml 1]>                                                  <![endif]--><!--[if !vml]--><!--[endif]--><!--[if gte mso 9]>     <![endif]-->     2. From Eq. 20-2, we obtain     <!--[if gte vml 1]>   <![endif]--><!--[if !vml]--><!--[endif]--><!--[if gte mso 9]>     <![endif]-->     3. We use the following relation derived in Sample Problem 20-2:     <!--[if gte vml 1]>   <![endif]--><!--[if !vml]--><!--[endif]--><!--[if gte mso 9]>     <![endif]-->     (a) The energy absorbed as heat is given by Eq. 19-14. Using Table 19-3, we find     <!--[if gte vml 1]>   <![endif]--><!--[if !vml]--><!--[endif]--><!--[if gte mso 9]>     <![endif]-->     where we have used the fact that a change in Kelvin temperature is equivalent to a change in Celsius degrees.     (b) With Tf = 373.15 K and Ti = 298.15 K, we obtain     <!--[if gte vml 1]>   <![endif]--><!--[if !vml]--><!--[endif]--><!--[if gte mso 9]>     <![endif]-->     4. (a) This may be considered a reversible process (as well as isothermal), so we use DS = Q/T where Q = Lm with L = 333 J/g from Table 19-4. Consequently,     <!--[if gte vml 1]>   <![endif]--><!--[if !vml]--><!--[endif]--><!--[if gte mso 9]>     <![endif]-->     (b) The situation is similar to that described in part (a), except with L = 2256 J/g, m = 5.00 g, and T = 373 K. We therefore find DS = 30.2 J/K.     5. (a) Since the gas is ideal, its pressure p is given in terms of the number of moles n, the volume V, and the temperature T by p = nRT/V. The work done by the gas during the isothermal expansion is     <!--[if gte vml 1]>   <![endif]--><!--[if !vml]--><!--[endif]--><!--[if gte mso 9]>     <![endif]-->     We substitute V2 = 2.00V1 to obtain     <!--[if gte vml 1]>   <![endif]--><!--[if !vml]--><!--[endif]--><!--[if gte mso 9]>     <![endif]-->     (b) Since the expansion is isothermal, the change in entropy is given by     <!--[if gte vml 1]>   <![endif]--><!--[if !vml]--><!--[endif]--><!--[if gte mso 9]>     <![endif]-->,     where Q is the heat absorbed. According to the first law of thermodynamics, DEint = Q - W. Now the internal energy of an ideal gas depends only on the temperature and not on the pressure and volume. Since the expansion is isothermal, DEint = 0 and Q = W. Thus,     <!--[if gte vml 1]>   <![endif]--><!--[if !vml]--><!--[endif]--><!--[if gte mso 9]>     <![endif]-->     (c) DS = 0 for all reversible adiabatic processes.     6. An isothermal process is one in which Ti = Tf which implies ln (Tf /Ti) = 0. Therefore, Eq. 20-4 leads to  <!--[if gte vml 1]>   <![endif]--><!--[if !vml]--><!--[endif]--><!--[if gte mso 9]>     <![endif]-->     7. (a) The energy that leaves the aluminum as heat has magnitude Q = maca(Tai - Tf), where ma is the mass of the aluminum, ca is the specific heat of aluminum, Tai is the initial temperature of the aluminum, and Tf is the final temperature of the aluminum-water system. The energy that enters the water as heat has magnitude Q = mwcw(Tf - Twi), where mw is the mass of the water, cw is the specific heat of water, and Twi is the initial temperature of the water. The two energies are the same in magnitude since no energy is lost. Thus,  <!--[if gte vml 1]>   <![endif]--><!--[if !vml]--><!--[endif]--><!--[if gte mso 9]>     <![endif]-->     The specific heat of aluminum is 900 J/kg×K and the specific heat of water is 4190 J/kg×K. Thus,     <!--[if gte vml 1]>   <![endif]--><!--[if !vml]--><!--[endif]--><!--[if gte mso 9]>     <![endif]-->     (b) Now temperatures must be given in Kelvins: Tai = 393 K, Twi = 293 K, and Tf = 330 K. For the aluminum, dQ = macadT and the change in entropy is     <!--[if gte vml 1]>   <![endif]--><!--[if !vml]--><!--[endif]--><!--[if gte mso 9]>     <![endif]-->     (c) The entropy change for the water is     <!--[if gte vml 1]>   <![endif]--><!--[if !vml]--><!--[endif]--><!--[if gte mso 9]>     <![endif]--><!--[if gte vml 1]>   <![endif]--><!--[if !vml]--><!--[endif]--><!--[if gte mso 9]>     <![endif]-->     (d) The change in the total entropy of the aluminum-water system is      DS = DSa + DSw = -22.1 J/K + 24.9 J/K = +2.8 J/K.     8. We follow the method shown in Sample Problem 20-2.  Since     DS = <!--[if gte vml 1]>   <![endif]--><!--[if !vml]--><!--[endif]--><!--[if gte mso 9]>     <![endif]--> = mc ln(Tf /Ti) ,     then with DS = 50 J/K, Tf = 380 K, Ti = 280 K and m = 0.364 kg,  we obtain c = 4.5×102 J/kg.K.     9. This problem is similar to Sample Problem 20-2. The only difference is that we need to find the mass m of each of the blocks. Since the two blocks are identical the final temperature Tf is the average of the initial temperatures:     <!--[if gte vml 1]>   <![endif]--><!--[if !vml]--><!--[endif]--><!--[if gte mso 9]>     <![endif]-->     Thus from Q = mcDT we find the mass m:     <!--[if gte vml 1]>   <![endif]--><!--[if !vml]--><!--[endif]--><!--[if gte mso 9]>     <![endif]-->     (a) The change in entropy for block L is      <!--[if gte vml 1]>   <![endif]--><!--[if !vml]--><!--[endif]--><!--[if gte mso 9]>     <![endif]-->     (b) Since the temperature of the reservoir is virtually the same as that of the block, which gives up the same amount of heat as the reservoir absorbs, the change in entropy <!--[if gte vml 1]>   <![endif]--><!--[if !vml]--><!--[endif]--><!--[if gte mso 9]>     <![endif]-->of the reservoir connected to the left block is the opposite of that of the left block: <!--[if gte vml 1]>   <![endif]--><!--[if !vml]--><!--[endif]--><!--[if gte mso 9]>     <![endif]--> = -DSL = +0.710 J/K.     (c) The entropy change for block R is     <!--[if gte vml 1]>   <![endif]--><!--[if !vml]--><!--[endif]--><!--[if gte mso 9]>     <![endif]-->     (d) Similar to the case in part (b) above, the change in entropy <!--[if gte vml 1]>   <![endif]--><!--[if !vml]--><!--[endif]--><!--[if gte mso 9]>     <![endif]--> of the reservoir connected to the right block is given by <!--[if gte vml 1]>   <![endif]--><!--[if !vml]--><!--[endif]--><!--[if gte mso 9]>     <![endif]--> = -DSR = -0.723 J/K.     (e) The change in entropy for the two-block system is      DSL + DSR = -0.710 J/K + 0.723 J/K = +0.013 J/K.     (f) The entropy change for the entire system is given by      DS = DSL + <!--[if gte vml 1]>   <![endif]--><!--[if !vml]--><!--[endif]--><!--[if gte mso 9]>     <![endif]--> + DSR + <!--[if gte vml 1]>   <![endif]--><!--[if !vml]--><!--[endif]--><!--[if gte mso 9]>     <![endif]--> = DSL - DSL + DSR - DSR = 0,     which is expected of a reversible process.     10. We concentrate on the first term of Eq. 20-4 (the second term is zero because the final and initial temperatures are the same, and because ln(1) = 0). Thus, the entropy change is     DS =  nR ln(Vf /Vi)  .     Noting that DS = 0 at  Vf  = 0.40 m3, we are able to deduce that Vi = 0.40 m3.  We now examine the point in the graph where DS = 32 J/K and  Vf  = 1.2 m3; the above expression can now be used to solve for the number of moles.  We obtain n = 3.5 mol.     11. (a) We refer to the copper block as block 1 and the lead block as block 2. The equilibrium temperature Tf satisfies      m1c1(Tf - Ti,1) + m2c2(Tf - Ti2) = 0,     which we solve for Tf :     <!--[if gte vml 1]>   <![endif]--><!--[if !vml]--><!--[endif]--><!--[if gte mso 9]>     <![endif]-->     (b) Since the two-block system in thermally insulated from the environment, the change in internal energy of the system is zero.     (c) The change in entropy is     <!--[if gte vml 1]>   <![endif]--><!--[if !vml]--><!--[endif]--><!--[if gte mso 9]>     <![endif]-->     12. We use Eq. 20-1:           <!--[if gte vml 1]>   <![endif]--><!--[if !vml]--><!--[endif]--><!--[if gte mso 9]>     <![endif]-->     13. The connection between molar heat capacity and the degrees of freedom of a diatomic gas is given by setting f = 5 in Eq. 19-51. Thus, <!--[if gte vml 1]>   <![endif]--><!--[if !vml]--><!--[endif]--><!--[if gte mso 9]>     <![endif]-->, and <!--[if gte vml 1]>   <![endif]--><!--[if !vml]--><!--[endif]--><!--[if gte mso 9]>     <![endif]-->. In addition to various equations from Chapter 19, we also make use of Eq. 20-4 of this chapter. We note that we are asked to use the ideal gas constant as R and not plug in its numerical value. We also recall that isothermal means constant-temperature, so T2 = T1 for the 1 ® 2 process. The statement (at the end of the problem) regarding “per mole” may be taken to mean that n may be set identically equal to 1 wherever it appears.     (a) The gas law in ratio form (see Sample Problem 19-1) is used to obtain     <!--[if gte vml 1]>   <![endif]--><!--[if !vml]--><!--[endif]--><!--[if gte mso 9]>     <![endif]-->.     (b) The adiabatic relations Eq. 19-54 and Eq. 19-56 lead to     <!--[if gte vml 1]>   <![endif]--><!--[if !vml]--><!--[endif]--><!--[if gte mso 9]>     <![endif]-->.  (c) Similarly, we find   <!--[if gte vml 1]>   <![endif]--><!--[if !vml]--><!--[endif]--><!--[if gte mso 9]>     <![endif]-->  · process 1 ® 2     (d) The work is given by Eq. 19-14:      W = nRT1 ln (V2/V1) = RT1 ln3 =1.10RT1.     Thus, W/ nRT1= ln3 = 1.10.     (e) The internal energy change is DEint = 0 since this is an ideal gas process without a temperature change (see Eq. 19-45). Thus, the energy absorbed as heat is given by the first law of thermodynamics: Q = DEint + W » 1.10RT1, or Q/ nRT1= ln3 = 1.10.     (f) DEint = 0 or DEint / nRT1=0     (g) The entropy change is DS = Q/T1 = 1.10R, or DS/R = 1.10.     · process 2 ® 3     (h) The work is zero since there is no volume change. Therefore, W/nRT1 = 0     (i) The internal energy change is     <!--[if gte vml 1]>   <![endif]--><!--[if !vml]--><!--[endif]--><!--[if gte mso 9]>     <![endif]-->     This ratio (-0.889) is also the value for Q/nRT1 (by either the first law of thermodynamics or by the definition of CV).     (j) DEint /nRT1= -0.889.     (k) For the entropy change, we obtain     <!--[if gte vml 1]>   <![endif]--><!--[if !vml]--><!--[endif]--><!--[if gte mso 9]>     <![endif]-->     · process 3 ® 1     (l) By definition, Q = 0 in an adiabatic process, which also implies an absence of entropy change (taking this to be a reversible process). The internal change must be the negative of the value obtained for it in the previous process (since all the internal energy changes must add up to zero, for an entire cycle, and its change is zero for process 1 ® 2), so DEint = +0.889RT1. By the first law of thermodynamics, then,      W = Q - DEint = -0.889RT1,     or W /nRT1= -0.889.     (m) Q = 0 in an adiabatic process.     (n) DEint /nRT1= +0.889.     (o) DS/nR=0.     14. (a) It is possible to motivate, starting from Eq. 20-3, the notion that heat may be found from the integral (or “area under the curve”) of a curve in a TS diagram, such as this one. Either from calculus, or from geometry (area of a trapezoid), it is straightforward to find the result for a “straight-line” path in the TS diagram:     <!--[if gte vml 1]>   <![endif]--><!--[if !vml]--><!--[endif]--><!--[if gte mso 9]>     <![endif]-->     which could, in fact, be directly motivated from Eq. 20-3 (but it is important to bear in mind that this is rigorously true only for a process which forms a straight line in a graph that plots T versus S). This leads to      Q = (300 K) (15 J/K) = 4.5×103 J     for the energy absorbed as heat by the gas.     (b) Using Table 19-3 and Eq. 19-45, we find     <!--[if gte vml 1]>   <![endif]--><!--[if !vml]--><!--[endif]--><!--[if gte mso 9]>     <![endif]-->     (c) By the first law of thermodynamics,     <!--[if gte vml 1]>   <![endif]--><!--[if !vml]--><!--[endif]--><!--[if gte mso 9]>     <![endif]-->     15. The ice warms to 0°C, then melts, and the resulting water warms to the temperature of the lake water, which is 15°C. As the ice warms, the energy it receives as heat when the temperature changes by dT is dQ = mcI dT, where m is the mass of the ice and cI is the specific heat of ice. If Ti (= 263 K) is the initial temperature and Tf (= 273 K) is the final temperature, then the change in its entropy is     <!--[if gte vml 1]>   <![endif]--><!--[if !vml]--><!--[endif]--><!--[if gte mso 9]>     <![endif]-->     Melting is an isothermal process. The energy leaving the ice as heat is mLF, where LF is the heat of fusion for ice. Thus,      DS = Q/T = mLF/T = (0.010 kg)(333 ´ 103 J/kg)/(273 K) = 12.20 J/K.     For the warming of the water from the melted ice, the change in entropy is     <!--[if gte vml 1]>   <![endif]--><!--[if !vml]--><!--[endif]--><!--[if gte mso 9]>     <![endif]-->     where cw is the specific heat of water (4190 J/kg × K). Thus,     <!--[if gte vml 1]>   <![endif]--><!--[if !vml]--><!--[endif]--><!--[if gte mso 9]>     <![endif]-->     The total change in entropy for the ice and the water it becomes is     <!--[if gte vml 1]>   <![endif]--><!--[if !vml]--><!--[endif]--><!--[if gte mso 9]>     <![endif]-->     Since the temperature of the lake does not change significantly when the ice melts, the change in its entropy is DS = Q/T, where Q is the energy it receives as heat (the negative of the energy it supplies the ice) and T is its temperature. When the ice warms to 0°C,     <!--[if gte vml 1]>   <![endif]--><!--[if !vml]--><!--[endif]--><!--[if gte mso 9]>     <![endif]-->     When the ice melts,     <!--[if gte vml 1]>   <![endif]--><!--[if !vml]--><!--[endif]--><!--[if gte mso 9]>     <![endif]-->     When the water from the ice warms,     <!--[if gte vml 1]>   <![endif]--><!--[if !vml]--><!--[endif]--><!--[if gte mso 9]>     <![endif]-->     The total energy leaving the lake water is      Q = -222 J - 3.33 ´ 103 J - 6.29 ´ 102 J = -4.18 ´ 103 J.     The change in entropy is  <!--[if gte vml 1]>   <![endif]--><!--[if !vml]--><!--[endif]--><!--[if gte mso 9]>     <![endif]-->     The change in the entropy of the ice-lake system is DS = (15.27 - 14.51) J/K = 0.76 J/K.     16. (a) Work is done only for the ab portion of the process. This portion is at constant pressure, so the work done by the gas is     <!--[if gte vml 1]>   <![endif]--><!--[if !vml]--><!--[endif]--><!--[if gte mso 9]>     <![endif]--><!--[if gte vml 1]>   <![endif]--><!--[if !vml]--><!--[endif]--><!--[if gte mso 9]>     <![endif]-->     (b) We use the first law: DEint = Q - W. Since the process is at constant volume, the work done by the gas is zero and Eint = Q. The energy Q absorbed by the gas as heat is Q = nCV DT, where CV is the molar specific heat at constant volume and DT is the change in temperature. Since the gas is a monatomic ideal gas, <!--[if gte vml 1]>   <![endif]--><!--[if !vml]--><!--[endif]--><!--[if gte mso 9]>     <![endif]-->. Use the ideal gas law to find that the initial temperature is      <!--[if gte vml 1]>   <![endif]--><!--[if !vml]--><!--[endif]--><!--[if gte mso 9]>     <![endif]-->     and that the final temperature is      <!--[if gte vml 1]>   <![endif]--><!--[if !vml]--><!--[endif]--><!--[if gte mso 9]>     <![endif]-->.  Thus,  <!--[if gte vml 1]>   <![endif]--><!--[if !vml]--><!--[endif]--><!--[if gte mso 9]>     <![endif]-->     The change in the internal energy is DEint = 6p0V0 or DEint/p0V0=6.00. Since n = 1 mol, this can also be written Q = 6.00RT0.     (c) For a complete cycle, DEint = 0     (d) Since the process is at constant volume, use dQ = nCV dT to obtain     <!--[if gte vml 1]>   <![endif]--><!--[if !vml]--><!--[endif]--><!--[if gte mso 9]>     <![endif]-->     Substituting <!--[if gte vml 1]>   <![endif]--><!--[if !vml]--><!--[endif]--><!--[if gte mso 9]>     <![endif]--> and using the ideal gas law, we write     <!--[if gte vml 1]>   <![endif]--><!--[if !vml]--><!--[endif]--><!--[if gte mso 9]>     <![endif]-->     Thus, <!--[if gte vml 1]>   <![endif]--><!--[if !vml]--><!--[endif]--><!--[if gte mso 9]>     <![endif]-->. Since n = 1, this is <!--[if gte vml 1]>   <![endif]--><!--[if !vml]--><!--[endif]--><!--[if gte mso 9]>     <![endif]-->.     (e) For a complete cycle, DEint = 0 and DS = 0.     17. (a) The final mass of ice is (1773 g + 227 g)/2 = 1000 g. This means 773 g of water froze. Energy in the form of heat left the system in the amount mLF, where m is the mass of the water that froze and LF is the heat of fusion of water. The process is isothermal, so the change in entropy is      DS = Q/T = –mLF/T = –(0.773 kg)(333 ´ 103 J/kg)/(273 K) = -943 J/K.     (b) Now, 773 g of ice is melted. The change in entropy is     <!--[if gte vml 1]>   <![endif]--><!--[if !vml]--><!--[endif]--><!--[if gte mso 9]>     <![endif]-->     (c) Yes, they are consistent with the second law of thermodynamics. Over the entire cycle, the change in entropy of the water-ice system is zero even though part of the cycle is irreversible. However, the system is not closed. To consider a closed system, we must include whatever exchanges energy with the ice and water. Suppose it is a constant-temperature heat reservoir during the freezing portion of the cycle and a Bunsen burner during the melting portion. During freezing the entropy of the reservoir increases by 943 J/K. As far as the reservoir-water-ice system is concerned, the process is adiabatic and reversible, so its total entropy does not change. The melting process is irreversible, so the total entropy of the burner-water-ice system increases. The entropy of the burner either increases or else decreases by less than 943 J/K.     18. In coming to equilibrium, the heat lost by the 100 cm3 of liquid water (of mass mw = 100 g and specific heat capacity cw = 4190 J/kg×K) is absorbed by the ice (of mass mi which melts and reaches Tf &gt; 0 °C). We begin by finding the equilibrium temperature:     <!--[if gte vml 1]>   <![endif]--><!--[if !vml]--><!--[endif]--><!--[if gte mso 9]>     <![endif]-->     which yields, after using LF = 333000 J/kg and values cited in the problem, Tf = 12.24 ° which is equivalent to Tf = 285.39 K. Sample Problem 19-2 shows that     <!--[if gte vml 1]>   <![endif]--><!--[if !vml]--><!--[endif]--><!--[if gte mso 9]>     <![endif]-->     for processes where DT = T2 – T1, and Eq. 20-2 gives     <!--[if gte vml 1]>   <![endif]--><!--[if !vml]--><!--[endif]--><!--[if gte mso 9]>     <![endif]-->     for the phase change experienced by the ice (with To = 273.15 K). The total entropy change is (with T in Kelvins)     <!--[if gte vml 1]>   <![endif]--><!--[if !vml]--><!--[endif]--><!--[if gte mso 9]>     <![endif]-->     19. We consider a three-step reversible process as follows: the supercooled water drop (of mass m) starts at state 1 (T1 = 268 K), moves on to state 2 (still in liquid form but at T2 = 273 K), freezes to state 3 (T3 = T2), and then cools down to state 4 (in solid form, with T4 = T1). The change in entropy for each of the stages is given as follows:      DS12 = mcw ln (T2/T1),  DS23 = -mLF/T2,                 DS34 = mcI ln (T4/T3) = mcI ln (T1/T2) = -mcI ln (T2/T1).     Thus the net entropy change for the water drop is     <!--[if gte vml 1]>   <![endif]--><!--[if !vml]--><!--[endif]--><!--[if gte mso 9]>     <![endif]-->     20. (a) We denote the mass of the ice (which turns to water and warms to Tf) as m and the mass of original-water (which cools from 80º down to Tf) as m¢.  From SQ = 0 we have     LF m + cm (Tf – 0º) + cm¢ (Tf  – 80º) = 0 .     Since LF = 333 ´ 103 J/kg, c = 4190 J/(kg·Cº), m¢ = 0.13 kg and m = 0.012 kg, we find Tf = 66.5ºC, which is equivalent to 339.67 K.     (b) Using Eq. 20-2, the process of ice at 0º C turning to water at 0º C involves an entropy change of  <!--[if supportFields]> EQ \F(Q,T) <![endif]--><!--[if supportFields]><![endif]-->  =  <!--[if supportFields]> EQ \F(LF m,273.15 K) <![endif]--><!--[if supportFields]><![endif]-->  =  14.6 J/K .     (c) Using Eq. 20-1, the process of m = 0.012 kg of water warming from 0º C to 66.5º C involves an entropy change of                                                    <!--[if gte vml 1]>   <![endif]--><!--[if !vml]--><!--[endif]--><!--[if gte mso 9]>     <![endif]-->.     (d) Similarly, the cooling of the original-water involves an entropy change of     <!--[if gte vml 1]>   <![endif]--><!--[if !vml]--><!--[endif]--><!--[if gte mso 9]>     <![endif]-->     (e) The net entropy change in this calorimetry experiment is found by summing the previous results; we find (by using more precise values than those shown above) ΔSnet = 4.39 J/K.     21. We note that the connection between molar heat capacity and the degrees of freedom of a monatomic gas is given by setting f = 3 in Eq. 19-51. Thus, <!--[if gte vml 1]>   <![endif]--><!--[if !vml]--><!--[endif]--><!--[if gte mso 9]>     <![endif]-->, and <!--[if gte vml 1]>   <![endif]--><!--[if !vml]--><!--[endif]--><!--[if gte mso 9]>     <![endif]-->.     (a) Since this is an ideal gas, Eq. 19-45 holds, which implies DEint = 0 for this process. Eq. 19-14 also applies, so that by the first law of thermodynamics,      Q = 0 + W = nRT1 ln V2/V1 = p1V1 ln 2 →  Q/p1V1= ln2 = 0.693.     (b) The gas law in ratio form (see Sample Problem 19-1) implies that the pressure decreased by a factor of 2 during the isothermal expansion process to V2=2.00V1, so that it needs to increase by a factor of 4 in this step in order to reach a final pressure of p2=2.00p1. That same ratio form now applied to this constant-volume process, yielding 4.00 = T2T1 which is used in the following:     <!--[if gte vml 1]>   <![endif]--><!--[if !vml]--><!--[endif]--><!--[if gte mso 9]>     <![endif]-->     or <!--[if gte vml 1]>   <![endif]--><!--[if !vml]--><!--[endif]--><!--[if gte mso 9]>     <![endif]-->.     (c) The work done during the isothermal expansion process may be obtained by using Eq. 19-14:   W = nRT1 ln V2/V1= p1V1 ln 2.00  →  W/p1V1= ln2 = 0.693.     (d) In step 2 where the volume is kept constant, W = 0.     (e) The change in internal energy can be calculated by combining the above results and applying the first law of thermodynamics:     <!--[if gte vml 1]>   <![endif]--><!--[if !vml]--><!--[endif]--><!--[if gte mso 9]>     <![endif]-->     or DEint/p1V1 = 9/2 = 4.50.     (f) The change in entropy may be computed by using Eq. 20-4:     <!--[if gte vml 1]>   <![endif]--><!--[if !vml]--><!--[endif]--><!--[if gte mso 9]>     <![endif]-->     The second approach consists of an isothermal (constant T) process in which the volume halves, followed by an isobaric (constant p) process.      (g) Here the gas law applied to the first (isothermal) step leads to a volume half as big as the original. Since <!--[if gte vml 1]>   <![endif]--><!--[if !vml]--><!--[endif]--><!--[if gte mso 9]>     <![endif]-->, the reasoning used above leads to      Q = – p1V1 ln 2.00 <!--[if gte vml 1]>   <![endif]--><!--[if !vml]--><!--[endif]--><!--[if gte mso 9]>     <![endif]-->  <!--[if gte vml 1]>   <![endif]--><!--[if !vml]--><!--[endif]--><!--[if gte mso 9]>     <![endif]-->     (h) To obtain a final volume twice as big as the original, in this step we need to increase the volume by a factor of 4.00. Now, the gas law applied to this isobaric portion leads to a temperature ratio T2/T1 = 4.00. Thus,     <!--[if gte vml 1]>   <![endif]--><!--[if !vml]--><!--[endif]--><!--[if gte mso 9]>     <![endif]-->     or Q/p1V1 = 15/2 = 7.50.     (i) During the isothermal compression process, Eq. 19-14 gives      W = nRT1 ln V2/V1= p1V1 ln (-1/2.00) = -p1V1 ln 2.00  <!--[if gte vml 1]>   <![endif]--><!--[if !vml]--><!--[endif]--><!--[if gte mso 9]>     <![endif]-->  W/p1V1= -ln2 = -0.693.     (j) The initial value of the volume, for this part of the process, is <!--[if gte vml 1]>   <![endif]--><!--[if !vml]--><!--[endif]--><!--[if gte mso 9]>     <![endif]-->, and the final volume is Vf = 2V1. The pressure maintained during this process is p¢ = 2.00p1. The work is given by Eq. 19-16:     <!--[if gte vml 1]>   <![endif]--><!--[if !vml]--><!--[endif]--><!--[if gte mso 9]>     <![endif]-->     (k) Using the first law of thermodynamics, the change in internal energy is     <!--[if gte vml 1]>   <![endif]--><!--[if !vml]--><!--[endif]--><!--[if gte mso 9]>     <![endif]-->     or DEint/p1V1 = 9/2 = 4.50. The result is the same as that obtained in part (e).     (l) Similarly,<!--[if gte vml 1]>   <![endif]--><!--[if !vml]--><!--[endif]--><!--[if gte mso 9]>     <![endif]--> the same as that obtained in part (f).     22. (a) The final pressure is     <!--[if gte vml 1]>   <![endif]--><!--[if !vml]--><!--[endif]--><!--[if gte mso 9]>     <![endif]-->     (b) We use the ratio form of the gas law (see Sample Problem 19-1) to find the final temperature of the gas:     <!--[if gte vml 1]>   <![endif]--><!--[if !vml]--><!--[endif]--><!--[if gte mso 9]>     <![endif]-->     For later purposes, we note that this result can be written “exactly” as Tf = Ti (2e–1). In our solution, we are avoiding using the “one mole” datum since it is not clear how precise it is.     (c) The work done by the gas is     <!--[if gte vml 1]>   <![endif]--><!--[if !vml]--><!--[endif]--><!--[if gte mso 9]>     <![endif]-->     (d) Consideration of a two-stage process, as suggested in the hint, brings us simply to Eq. 20-4. Consequently, with <!--[if gte vml 1]>   <![endif]--><!--[if !vml]--><!--[endif]--><!--[if gte mso 9]>     <![endif]--> (see Eq. 19-43), we find     <!--[if gte vml 1]>   <![endif]--><!--[if !vml]--><!--[endif]--><!--[if gte mso 9]>     <![endif]-->     23. We solve (b) first.     (b) For a Carnot engine, the efficiency is related to the reservoir temperatures by Eq. 20-13. Therefore,  <!--[if gte vml 1]>   <![endif]--><!--[if !vml]--><!--[endif]--><!--[if gte mso 9]>     <![endif]-->     which is equivalent to 68°C.      (a) The temperature of the cold reservoir is TL = TH – 75 = 341 K – 75 K = 266 K.     24. Eq. 20-13 leads to  <!--[if gte vml 1]>   <![endif]--><!--[if !vml]--><!--[endif]--><!--[if gte mso 9]>     <![endif]-->     quoting more figures than are significant. As a percentage, this is e = 99.99995%.     25. (a) The efficiency is     <!--[if gte vml 1]>   <![endif]--><!--[if !vml]--><!--[endif]--><!--[if gte mso 9]>     <![endif]-->     We note that a temperature difference has the same value on the Kelvin and Celsius scales. Since the temperatures in the equation must be in Kelvins, the temperature in the denominator is converted to the Kelvin scale.     (b) Since the efficiency is given by e = |W|/|QH|, the work done is given by     <!--[if gte vml 1]>   <![endif]--><!--[if !vml]--><!--[endif]--><!--[if gte mso 9]>     <![endif]-->     26. The answers to this exercise do not depend on the engine being of the Carnot design. Any heat engine that intakes energy as heat (from, say, consuming fuel) equal to |QH| = 52 kJ and exhausts (or discards) energy as heat equal to |QL| = 36 kJ will have these values of efficiency e and net work W.     (a) Eq. 20-12 gives  <!--[if gte vml 1]>   <![endif]--><!--[if !vml]--><!--[endif]--><!--[if gte mso 9]>     <![endif]-->     (b) Eq. 20-8 gives  <!--[if gte vml 1]>   <![endif]--><!--[if !vml]--><!--[endif]--><!--[if gte mso 9]>     <![endif]-->     27. With TL = 290 k, we find     <!--[if gte vml 1]>   <![endif]--><!--[if !vml]--><!--[endif]--><!--[if gte mso 9]>     <![endif]-->     which yields the (initial) temperature of the high-temperature reservoir: TH = 483 K. If we replace e = 0.40 in the above calculation with e = 0.50, we obtain a (final) high temperature equal to <!--[if gte vml 1]>   <![endif]--><!--[if !vml]--><!--[endif]--><!--[if gte mso 9]>     <![endif]-->. The difference is     <!--[if gte vml 1]>   <![endif]--><!--[if !vml]--><!--[endif]--><!--[if gte mso 9]>     <![endif]-->     28. (a) Eq. 20-13 leads to  <!--[if gte vml 1]>   <![endif]--><!--[if !vml]--><!--[endif]--><!--[if gte mso 9]>     <![endif]-->     We recall that a Watt is Joule-per-second. Thus, the (net) work done by the cycle per unit time is the given value 500 J/s. Therefore, by Eq. 20-11, we obtain the heat input per unit time:  <!--[if gte vml 1]>   <![endif]--><!--[if !vml]--><!--[endif]--><!--[if gte mso 9]>     <![endif]-->     (b) Considering Eq. 20-8 on a per unit time basis, we find (4.67 – 0.500) kJ/s = 4.17 kJ/s for the rate of heat exhaust.     29. (a) Energy is added as heat during the portion of the process from a to b. This portion occurs at constant volume (Vb), so Qin = nCV DT. The gas is a monatomic ideal gas, so <!--[if gte vml 1]>   <![endif]--><!--[if !vml]--><!--[endif]--><!--[if gte mso 9]>     <![endif]--> and the ideal gas law gives      DT = (1/nR)(pb Vb – pa Va) = (1/nR)(pb – pa) Vb.     Thus, <!--[if gte vml 1]>   <![endif]--><!--[if !vml]--><!--[endif]--><!--[if gte mso 9]>     <![endif]-->. Vb and pb are given. We need to find pa. Now pa is the same as pc and points c and b are connected by an adiabatic process. Thus, <!--[if gte vml 1]>   <![endif]--><!--[if !vml]--><!--[endif]--><!--[if gte mso 9]>     <![endif]--> and     <!--[if gte vml 1]>   <![endif]--><!--[if !vml]--><!--[endif]--><!--[if gte mso 9]>     <![endif]-->     The energy added as heat is     <!--[if gte vml 1]>   <![endif]--><!--[if !vml]--><!--[endif]--><!--[if gte mso 9]>     <![endif]-->     (b) Energy leaves the gas as heat during the portion of the process from c to a. This is a constant pressure process, so     <!--[if gte vml 1]>   <![endif]--><!--[if !vml]--><!--[endif]--><!--[if gte mso 9]>     <![endif]-->     or <!--[if gte vml 1]>   <![endif]--><!--[if !vml]--><!--[endif]--><!--[if gte mso 9]>     <![endif]-->. The substitutions Va – Vc = Va – 8.00 Va = – 7.00 Va and <!--[if gte vml 1]>   <![endif]--><!--[if !vml]--><!--[endif]--><!--[if gte mso 9]>     <![endif]--> were made.     (c) For a complete cycle, the change in the internal energy is zero and      W = Q = 1.47 ´ 103 J – 5.54 ´ 102 J = 9.18 ´ 102 J.     (d) The efficiency is      e = W/Qin = (9.18 ´ 102 J)/(1.47 ´ 103 J) = 0.624 = 62.4%.     30. From Fig. 20-28, we see QH = 4000 J at TH = 325 K.  Combining Eq. 20-11 with Eq. 20-13, we have   <!--[if supportFields]> EQ \f(W, QH ) <![endif]--><!--[if supportFields]><![endif]-->= 1 – <!--[if supportFields]> EQ \f(TC, TH ) <![endif]--><!--[if supportFields]><![endif]-->   Þ   W  = 923 J .     Now, for <!--[if gte vml 1]>   <![endif]--><!--[if !vml]--><!--[endif]--><!--[if gte mso 9]>     <![endif]--> = 550 K, we have                                                     <!--[if gte vml 1]>   <![endif]--><!--[if !vml]--><!--[endif]--><!--[if gte mso 9]>     <![endif]-->     31. (a) The net work done is the rectangular “area” enclosed in the pV diagram:     <!--[if gte vml 1]>   <![endif]--><!--[if !vml]--><!--[endif]--><!--[if gte mso 9]>     <![endif]-->     Inserting the values stated in the problem, we obtain W = 2.27 kJ.     (b) We compute the energy added as heat during the “heat-intake” portions of the cycle using Eq. 19-39, Eq. 19-43, and Eq. 19-46:     <!--[if gte vml 1]>   <![endif]--><!--[if !vml]--><!--[endif]--><!--[if gte mso 9]>     <![endif]-->     where, to obtain the last line, the gas law in ratio form has been used (see Sample Problem 19-1). Therefore, since W = p0V0, we have Qabc = 13W/2 = 14.8 kJ.     (c) The efficiency is given by Eq. 20-11:     <!--[if gte vml 1]>   <![endif]--><!--[if !vml]--><!--[endif]--><!--[if gte mso 9]>     <![endif]-->     (d) A Carnot engine operating between Tc and Ta has efficiency equal to     <!--[if gte vml 1]>   <![endif]--><!--[if !vml]--><!--[endif]--><!--[if gte mso 9]>     <![endif]-->     where the gas law in ratio form has been used.      (e) This is greater than our result in part (c), as expected from the second law of thermodynamics.     32. (a) Using Eq. 19-54 for process D ® A gives     <!--[if gte vml 1]>   <![endif]--><!--[if !vml]--><!--[endif]--><!--[if gte mso 9]>     <![endif]-->     which leads to <!--[if gte vml 1]>   <![endif]--><!--[if !vml]--><!--[endif]--><!--[if gte mso 9]>     <![endif]-->. The result (see §19-9 and §19-11) implies the gas is monatomic.     (b) The input heat is that absorbed during process A ® B:     <!--[if gte vml 1]>   <![endif]--><!--[if !vml]--><!--[endif]--><!--[if gte mso 9]>     <![endif]-->     and the exhaust heat is that liberated during process C ® D:     <!--[if gte vml 1]>   <![endif]--><!--[if !vml]--><!--[endif]--><!--[if gte mso 9]>     <![endif]-->     where in the last step we have used the fact that<!--[if gte vml 1]>   <![endif]--><!--[if !vml]--><!--[endif]--><!--[if gte mso 9]>     <![endif]--> (from the gas law in ratio form — see Sample Problem 19-1). Therefore, Eq. 20-12 leads to     <!--[if gte vml 1]>   <![endif]--><!--[if !vml]--><!--[endif]--><!--[if gte mso 9]>     <![endif]-->     33. (a) We use <!--[if gte vml 1]>   <![endif]--><!--[if !vml]--><!--[endif]--><!--[if gte mso 9]>     <![endif]-->. The heat absorbed is <!--[if gte vml 1]>   <![endif]--><!--[if !vml]--><!--[endif]--><!--[if gte mso 9]>     <![endif]-->     (b) The heat exhausted is then <!--[if gte vml 1]>   <![endif]--><!--[if !vml]--><!--[endif]--><!--[if gte mso 9]>     <![endif]-->     (c) Now we have <!--[if gte vml 1]>   <![endif]--><!--[if !vml]--><!--[endif]--><!--[if gte mso 9]>     <![endif]-->     (d) Similarly, <!--[if gte vml 1]>   <![endif]--><!--[if !vml]--><!--[endif]--><!--[if gte mso 9]>     <![endif]-->.     34. All terms are assumed to be positive. The total work done by the two-stage system is W1 + W2. The heat-intake (from, say, consuming fuel) of the system is Q1 so we have (by Eq. 20-11 and Eq. 20-8)     <!--[if gte vml 1]>   <![endif]--><!--[if !vml]--><!--[endif]--><!--[if gte mso 9]>     <![endif]-->     Now, Eq. 20-10 leads to  <!--[if gte vml 1]>   <![endif]--><!--[if !vml]--><!--[endif]--><!--[if gte mso 9]>     <![endif]-->     where we assume Q2 is absorbed by the second stage at temperature T2. This implies the efficiency can be written  <!--[if gte vml 1]>   <![endif]--><!--[if !vml]--><!--[endif]--><!--[if gte mso 9]>     <![endif]-->     35. (a) The pressure at 2 is p2 = 3.00p1, as given in the problem statement. The volume is V2 = V1 = nRT1/p1. The temperature is     <!--[if gte vml 1]>   <![endif]--><!--[if !vml]--><!--[endif]--><!--[if gte mso 9]>     <![endif]-->     (b) The process 2 ® 3 is adiabatic, so <!--[if gte vml 1]>   <![endif]--><!--[if !vml]--><!--[endif]--><!--[if gte mso 9]>     <![endif]-->. Using the result from part (a), V3 = 4.00V1, V2 = V1 and g =1.30, we obtain     <!--[if gte vml 1]>   <![endif]--><!--[if !vml]--><!--[endif]--><!--[if gte mso 9]>     <![endif]-->.     (c) The process 4 ® 1 is adiabatic, so <!--[if gte vml 1]>   <![endif]--><!--[if !vml]--><!--[endif]--><!--[if gte mso 9]>     <![endif]-->. Since V4 = 4.00V1, we have      <!--[if gte vml 1]>   <![endif]--><!--[if !vml]--><!--[endif]--><!--[if gte mso 9]>     <![endif]-->     (d) The process 2 ® 3 is adiabatic, so <!--[if gte vml 1]>   <![endif]--><!--[if !vml]--><!--[endif]--><!--[if gte mso 9]>     <![endif]--> or <!--[if gte vml 1]>   <![endif]--><!--[if !vml]--><!--[endif]--><!--[if gte mso 9]>     <![endif]-->. Substituting V3 = 4.00V1, V2 = V1, p2 = 3.00p1 and g =1.30, we obtain     <!--[if gte vml 1]>   <![endif]--><!--[if !vml]--><!--[endif]--><!--[if gte mso 9]>     <![endif]-->     (e) The process 4 ® 1 is adiabatic, so <!--[if gte vml 1]>   <![endif]--><!--[if !vml]--><!--[endif]--><!--[if gte mso 9]>     <![endif]--> and     <!--[if gte vml 1]>   <![endif]--><!--[if !vml]--><!--[endif]--><!--[if gte mso 9]>     <![endif]-->     where we have used V4 = 4.00V1.      (f) The efficiency of the cycle is e = W/Q12, where W is the total work done by the gas during the cycle and Q12 is the energy added as heat during the 1 ® 2 portion of the cycle, the only portion in which energy is added as heat. The work done during the portion of the cycle from 2 to 3 is W23 =  ò p dV. Substitute<!--[if gte vml 1]>   <![endif]--><!--[if !vml]--><!--[endif]--><!--[if gte mso 9]>     <![endif]--> to obtain     <!--[if gte vml 1]>   <![endif]--><!--[if !vml]--><!--[endif]--><!--[if gte mso 9]>     <![endif]-->     Substitute V2 = V1, V3 = 4.00V1, and p3 = 3.00p1 to obtain     <!--[if gte vml 1]>   <![endif]--><!--[if !vml]--><!--[endif]--><!--[if gte mso 9]>     <![endif]-->     Similarly, the work done during the portion of the cycle from 4 to 1 is     <!--[if gte vml 1]>   <![endif]--><!--[if !vml]--><!--[endif]--><!--[if gte mso 9]>     <![endif]-->     No work is done during the 1 ® 2 and 3 ® 4 portions, so the total work done by the gas during the cycle is  <!--[if gte vml 1]>   <![endif]--><!--[if !vml]--><!--[endif]--><!--[if gte mso 9]>     <![endif]-->  The energy added as heat is      Q12 = nCV (T2 – T1) = nCV (3T1 – T1) = 2nCVT1,     where CV is the molar specific heat at constant volume. Now      g = Cp/CV = (CV + R)/CV = 1 + (R/CV),     so CV = R/(g – 1). Here Cp is the molar specific heat at constant pressure, which for an ideal gas is Cp = CV + R. Thus, Q12 = 2nRT1/(g – 1). The efficiency is     <!--[if gte vml 1]>   <![endif]--><!--[if !vml]--><!--[endif]--><!--[if gte mso 9]>     <![endif]-->     With g = 1.30, the efficiency is e = 0.340 or 34.0%.     36. Eq. 20-10 still holds (particularly due to its use of absolute values), and energy conservation implies |W| + QL = QH. Therefore, with TL = 268.15 K and TH = 290.15 K, we find  <!--[if gte vml 1]>   <![endif]--><!--[if !vml]--><!--[endif]--><!--[if gte mso 9]>     <![endif]-->     which (with |W| = 1.0 J) leads to <!--[if gte vml 1]>   <![endif]--><!--[if !vml]--><!--[endif]--><!--[if gte mso 9]>     <![endif]-->     37. A Carnot refrigerator working between a hot reservoir at temperature TH and a cold reservoir at temperature TL has a coefficient of performance K that is given by      <!--[if gte vml 1]>   <![endif]--><!--[if !vml]--><!--[endif]--><!--[if gte mso 9]>     <![endif]-->.     For the refrigerator of this problem, TH = 96° F = 309 K and TL = 70° F = 294 K, so      K = (294 K)/(309 K – 294 K) = 19.6.     The coefficient of performance is the energy QL drawn from the cold reservoir as heat divided by the work done: K = |QL|/|W|. Thus,      |QL| = K|W| = (19.6)(1.0 J) = 20 J.     38. (a) Eq. 20-15 provides     <!--[if gte vml 1]>   <![endif]--><!--[if !vml]--><!--[endif]--><!--[if gte mso 9]>     <![endif]-->     which yields |QH| = 49 kJ when KC = 5.7 and |QL| = 42 kJ.     (b) From §20-5 we obtain     <!--[if gte vml 1]>   <![endif]--><!--[if !vml]--><!--[endif]--><!--[if gte mso 9]>     <![endif]-->     if we take the initial 42 kJ datum to be accurate to three figures. The given temperatures are not used in the calculation; in fact, it is possible that the given room temperature value is not meant to be the high temperature for the (reversed) Carnot cycle — since it does not lead to the given KC using Eq. 20-16.     39. The coefficient of performance for a refrigerator is given by K = |QL|/|W|, where QL is the energy absorbed from the cold reservoir as heat and W is the work done during the refrigeration cycle, a negative value. The first law of thermodynamics yields QH + QL – W = 0 for an integer number of cycles. Here QH is the energy ejected to the hot reservoir as heat. Thus, QL = W – QH. QH is negative and greater in magnitude than W, so |QL| = |QH| – |W|. Thus,  <!--[if gte vml 1]>   <![endif]--><!--[if !vml]--><!--[endif]--><!--[if gte mso 9]>     <![endif]-->     The solution for |W| is |W| = |QH|/(K + 1). In one hour,  <!--[if gte vml 1]>   <![endif]--><!--[if !vml]--><!--[endif]--><!--[if gte mso 9]>     <![endif]-->  <!--[if gte vml 1]>   <![endif]--><!--[if !vml]--><!--[endif]--><!--[if gte mso 9]>     <![endif]-->     The rate at which work is done is (1.57 ´ 106 J)/(3600 s) = 440 W.     40. (a) Using Eq. 20-14 and Eq. 20-16, we obtain     <!--[if gte vml 1]>   <![endif]--><!--[if !vml]--><!--[endif]--><!--[if gte mso 9]>     <![endif]-->     (b) A similar calculation (being sure to use absolute temperature) leads to 0.50 J in this case.     (c) With TL = 100 K, we obtain |W| = 2.0 J.     (d) Finally, with the low temperature reservoir at 50 K, an amount of work equal to |W| = 5.0 J is required.     41. The efficiency of the engine is defined by e = W/Q1 and is shown in the text to be      <!--[if gte vml 1]>   <![endif]--><!--[if !vml]--><!--[endif]--><!--[if gte mso 9]>     <![endif]-->.     The coefficient of performance of the refrigerator is defined by K = Q4/W and is shown in the text to be   <!--[if gte vml 1]>   <![endif]--><!--[if !vml]--><!--[endif]--><!--[if gte mso 9]>     <![endif]-->.     Now Q4 = Q3 – W, so   (Q3 – W)/W = T4/(T3 – T4).     The work done by the engine is used to drive the refrigerator, so W is the same for the two. Solve the engine equation for W and substitute the resulting expression into the refrigerator equation. The engine equation yields W = (T1 – T2)Q1/T1 and the substitution yields     <!--[if gte vml 1]>   <![endif]--><!--[if !vml]--><!--[endif]--><!--[if gte mso 9]>     <![endif]-->     Solving for Q3/Q1, we obtain     <!--[if gte vml 1]>   <![endif]--><!--[if !vml]--><!--[endif]--><!--[if gte mso 9]>     <![endif]-->     With T1 = 400 K, T2 = 150 K, T3 = 325 K, and T4 = 225 K, the ratio becomes Q3/Q1=2.03.     42. (a) Eq. 20-13 gives the Carnot efficiency as 1 – TL /TH .  This gives 0.222 in this case.  Using this value with Eq. 20-11 leads to      W = (0.222)(750 J) = 167 J.     (b) Now, Eq. 20-16 gives KC = 3.5.  Then, Eq. 20-14 yields |W| = 1200/3.5 = 343 J.       43. We are told K = 0.27KC where     <!--[if gte vml 1]>   <![endif]--><!--[if !vml]--><!--[endif]--><!--[if gte mso 9]>     <![endif]-->     where the Fahrenheit temperatures have been converted to Kelvins. Expressed on a per unit time basis, Eq. 20-14 leads to     <!--[if gte vml 1]>   <![endif]--><!--[if !vml]--><!--[endif]--><!--[if gte mso 9]>     <![endif]-->     Appendix D indicates 1 But/h = 0.0003929 hp, so our result may be expressed as |W|/t = 0.25 hp.     44. The work done by the motor in t = 10.0 min is |W| = Pt = (200 W)(10.0 min)(60 s/min) = 1.20 ´ 105 J. The heat extracted is then     <!--[if gte vml 1]>   <![endif]--><!--[if !vml]--><!--[endif]--><!--[if gte mso 9]>     <![endif]-->     45. We need nine labels:               Label         Number of molecules on side 1         Number of molecules on side 2             I         8         0             II         7         1             III         6         2             IV         5         3             V         4         4             VI         3         5             VII         2         6             VIII         1         7             IX         0         8             The multiplicity W is computing using Eq. 20-20. For example, the multiplicity for label IV is  <!--[if gte vml 1]>   <![endif]--><!--[if !vml]--><!--[endif]--><!--[if gte mso 9]>     <![endif]-->     and the corresponding entropy is (using Eq. 20-21)     <!--[if gte vml 1]>   <![endif]--><!--[if !vml]--><!--[endif]--><!--[if gte mso 9]>     <![endif]-->     In this way, we generate the following table:               Label         W         S             I         1         0             II         8         2.9 ´ 10–23 J/K             III         28         4.6 ´ 10–23 J/K             IV         56         5.6 ´ 10–23 J/K             V         70         5.9 ´ 10–23 J/K             VI         56         5.6 ´ 10–23 J/K             VII         28         4.6 ´ 10–23 J/K             VIII         8         2.9 ´ 10–23 J/K             IX         1         0             46. (a) We denote the configuration with n heads out of N trials as (n; N). We use Eq. 20-20:  <!--[if gte vml 1]>   <![endif]--><!--[if !vml]--><!--[endif]--><!--[if gte mso 9]>     <![endif]-->     (b) There are 2 possible choices for each molecule: it can either be in side 1 or in side 2 of the box. If there are a total of N independent molecules, the total number of available states of the N-particle system is     <!--[if gte vml 1]>   <![endif]--><!--[if !vml]--><!--[endif]--><!--[if gte mso 9]>     <![endif]-->     With N  = 50, we obtain Ntotal = 250 =1.13 ´ 1015.     (c) The percentage of time in question is equal to the probability for the system to be in the central configuration:     <!--[if gte vml 1]>   <![endif]--><!--[if !vml]--><!--[endif]--><!--[if gte mso 9]>     <![endif]-->     With N = 100, we obtain      (d) W(N/2, N) = N!/[(N/2)!]2 = 1.01 ´ 1029,     (e) Ntotal = 2N =1.27 ´ 1030,     (f) and p(N/2;N) = W(N/2, N)/ Ntotal = 8.0%.     Similarly, for N = 200, we obtain      (g) W(N/2, N) = 9.25 ´ 1058,      (h) Ntotal =1.61 ´ 10 60,      (i) and p(N/2; N) = 5.7%.     (j) As N increases the number of available microscopic states increase as 2N, so there are more states to be occupied, leaving the probability less for the system to remain in its central configuration. Thus, the time spent in there decreases with an increase in N.     47. (a) Suppose there are nL molecules in the left third of the box, nC molecules in the center third, and nR molecules in the right third. There are N! arrangements of the N molecules, but nL! are simply rearrangements of the nL molecules in the right third, nC! are rearrangements of the nC molecules in the center third, and nR! are rearrangements of the nR molecules in the right third. These rearrangements do not produce a new configuration. Thus, the multiplicity is     <!--[if gte vml 1]>   <![endif]--><!--[if !vml]--><!--[endif]--><!--[if gte mso 9]>     <![endif]-->     (b) If half the molecules are in the right half of the box and the other half are in the left half of the box, then the multiplicity is     <!--[if gte vml 1]>   <![endif]--><!--[if !vml]--><!--[endif]--><!--[if gte mso 9]>     <![endif]-->     If one-third of the molecules are in each third of the box, then the multiplicity is     <!--[if gte vml 1]>   <![endif]--><!--[if !vml]--><!--[endif]--><!--[if gte mso 9]>     <![endif]-->  The ratio is  <!--[if gte vml 1]>   <![endif]--><!--[if !vml]--><!--[endif]--><!--[if gte mso 9]>     <![endif]-->     (c) For N = 100,  <!--[if gte vml 1]>   <![endif]--><!--[if !vml]--><!--[endif]--><!--[if gte mso 9]>     <![endif]-->     48. Using Hooke’s law, we find the spring constant to be     <!--[if gte vml 1]>   <![endif]--><!--[if !vml]--><!--[endif]--><!--[if gte mso 9]>     <![endif]-->.     To find the rate of change of entropy with a small additional stretch, we use Eq. 20-7 (see also Sample Problem 20-3) and obtain     <!--[if gte vml 1]>   <![endif]--><!--[if !vml]--><!--[endif]--><!--[if gte mso 9]>     <![endif]-->.     49. Using Eq. 19-34 and Eq. 19-35, we arrive at     Dv = (<!--[if supportFields]> EQ \r(3) <![endif]--><!--[if supportFields]><![endif]--> - <!--[if supportFields]> EQ \r(2) <![endif]--><!--[if supportFields]><![endif]-->)<!--[if supportFields]> EQ \r(RT/M) <![endif]--><!--[if supportFields]><![endif]-->          (a) We find, with M = 28 g/mol = 0.028 kg/mol (see Table 19-1), Dvi=87 m/s at 250 K,     (b) and Dvf =122 »1.2×102 m/s at 500 K.     (c) The expression above for Dv implies      T  =  <!--[if supportFields]> EQ \f(M,R( EQ \r(3)  -  EQ \r(2) )2) <![endif]--><!--[if supportFields]><![endif]-->(Dv)2     which we can plug into Eq. 20-4 to yield      DS =  nR ln(Vf /Vi) + nCV ln(Tf /Ti) = 0 + nCV ln[(Dvf)2/(Dvi)2] = 2nCV ln(Dvf /Dvi).     Using Table 19-3 to get CV = 5R/2 (see also Table 19-2) we then find, for n = 1.5 mol, DS = 22 J/K.     50. The net work is figured from the (positive) isothermal expansion (Eq. 19-14) and the (negative) constant-pressure compression (Eq. 19-48).  Thus,     Wnet = nRTH ln(Vmax/Vmin) +  nR(TL – TH)     where n = 3.4, TH = 500 K, TL = 200 K and Vmax/Vmin = 5/2  (same as the ratio TH /TL ).  Therefore, Wnet = 4468 J.  Now, we identify the “input heat” as that transferred in steps 1 and 2:   Qin = Q1 + Q2 = nCV (TH – TL)  + nRTH ln(Vmax/Vmin)     where CV  = 5R/2 (see Table 19-3).  Consequently, Qin = 34135 J.  Dividing these results gives the efficiency:  Wnet /Qin = 0.131 (or about 13.1%).     51. (a) If TH is the temperature of the high-temperature reservoir and TL is the temperature of the low-temperature reservoir, then the maximum efficiency of the engine is  <!--[if gte vml 1]>   <![endif]--><!--[if !vml]--><!--[endif]--><!--[if gte mso 9]>     <![endif]-->     (b) The efficiency is defined by e = |W|/|QH|, where W is the work done by the engine and QH is the heat input. W is positive. Over a complete cycle, QH = W + |QL|, where QL is the heat output, so e = W/(W + |QL|) and |QL| = W[(1/e) – 1]. Now e = (TH – TL)/TH, where TH is the temperature of the high-temperature heat reservoir and TL is the temperature of the low-temperature reservoir. Thus,     <!--[if gte vml 1]>   <![endif]--><!--[if !vml]--><!--[endif]--><!--[if gte mso 9]>     <![endif]-->     The heat output is used to melt ice at temperature Ti = – 40°C. The ice must be brought to 0°C, then melted, so   |QL| = mc(Tf – Ti) + mLF,     where m is the mass of ice melted, Tf is the melting temperature (0°C), c is the specific heat of ice, and LF is the heat of fusion of ice. Thus,      WTL/(TH – TL) = mc(Tf – Ti) + mLF.     We differentiate with respect to time and replace dW/dt with P, the power output of the engine, and obtain      PTL/(TH – TL) = (dm/dt)[c(Tf – Ti) + LF].     Therefore,  <!--[if gte vml 1]>   <![endif]--><!--[if !vml]--><!--[endif]--><!--[if gte mso 9]>     <![endif]-->     Now, P = 100 ´ 106 W, TL = 0 + 273 = 273 K, TH = 800 + 273 = 1073 K, Ti = –40 + 273 = 233 K, Tf = 0 + 273 = 273 K, c = 2220 J/kg·K, and LF = 333 ´ 103 J/kg, so     <!--[if gte vml 1]>   <![endif]--><!--[if !vml]--><!--[endif]--><!--[if gte mso 9]>     <![endif]-->     We note that the engine is now operated between 0°C and 800°C.     52. (a) Combining Eq. 20-11 with Eq. 20-13, we obtain     <!--[if gte vml 1]>   <![endif]--><!--[if !vml]--><!--[endif]--><!--[if gte mso 9]>     <![endif]-->     (b) Combining Eq. 20-14 with Eq. 20-16, we find     <!--[if gte vml 1]>   <![endif]--><!--[if !vml]--><!--[endif]--><!--[if gte mso 9]>     <![endif]-->     53. (a) Starting from <!--[if gte vml 1]>   <![endif]--><!--[if !vml]--><!--[endif]--><!--[if gte mso 9]>     <![endif]--> (for calorimetry problems) we can derive (when no phase changes are involved)     <!--[if gte vml 1]>   <![endif]--><!--[if !vml]--><!--[endif]--><!--[if gte mso 9]>     <![endif]-->  which is equivalent to 314 K.     (b) From Eq. 20-1, we have     <!--[if gte vml 1]>   <![endif]--><!--[if !vml]--><!--[endif]--><!--[if gte mso 9]>     <![endif]-->     (c) For water, the change in entropy is     <!--[if gte vml 1]>   <![endif]--><!--[if !vml]--><!--[endif]--><!--[if gte mso 9]>     <![endif]-->     (d) The net result for the system is (30.5 – 27.1) J/K = 3.4 J/K. (Note: these calculations are fairly sensitive to round-off errors. To arrive at this final answer, the value 273.15 was used to convert to Kelvins, and all intermediate steps were retained to full calculator accuracy.)     54. For an isothermal ideal gas process, we have Q = W = nRT ln(Vf /Vi ).  Thus,      DS = Q/T = W/T = nR ln(Vf /Vi )      (a) Vf /Vi = (0.800)/(0.200) = 4.00, DS = (0.55)(8.31)ln(4.00) = 6.34 J/K.     (b) Vf /Vi = (0.800)/(0.200) = 4.00, DS = (0.55)(8.31)ln(4.00) = 6.34 J/K.     (c) Vf /Vi = (1.20)/(0.300) = 4.00, DS = (0.55)(8.31)ln(4.00) = 6.34 J/K.     (d) Vf /Vi = (1.20)/(0.300) = 4.00, DS = (0.55)(8.31)ln(4.00) = 6.34 J/K.     55. Except for the phase change (which just uses Eq. 20-2), this has some similarities with Sample Problem 20-2.  Using constants available in the Chapter 19 tables, we compute  DS = m[cice ln(273/253) + <!--[if supportFields]> EQ \f(Lf,273) <![endif]--><!--[if supportFields]><![endif]--> + cwater ln(313/273)] =  1.18 ´ 103 J/K.     56. Eq. 20-4 yields      DS =  nR ln(Vf /Vi)  +   nCV ln(Tf /Ti)  =  0 + nCV ln(425/380)     where n = 3.20 and CV  =  <!--[if supportFields]> EQ \f(3,2) <![endif]--><!--[if supportFields]><![endif]-->R  (Eq. 19-43). This gives 4.46 J/K.     57. (a) It is a reversible set of processes returning the system to its initial state; clearly, DSnet = 0.     (b) Process 1 is adiabatic and reversible (as opposed to, say, a free expansion) so that Eq. 20-1 applies with dQ = 0 and yields DS1 = 0.     (c) Since the working substance is an ideal gas, then an isothermal process implies Q = W, which further implies (regarding Eq. 20-1) dQ = p dV. Therefore,     <!--[if gte vml 1]>   <![endif]--><!--[if !vml]--><!--[endif]--><!--[if gte mso 9]>     <![endif]-->     which leads to <!--[if gte vml 1]>   <![endif]--><!--[if !vml]--><!--[endif]--><!--[if gte mso 9]>     <![endif]-->     (d) By part (a), DS1 + DS2 + DS3 = 0. Then, part (b) implies DS2 = -DS3. Therefore, DS2 = 23.0 J/K.     58. (a) The most obvious input-heat step is the constant-volume process. Since the gas is monatomic, we know from Chapter 19 that <!--[if gte vml 1]>   <![endif]--><!--[if !vml]--><!--[endif]--><!--[if gte mso 9]>     <![endif]-->. Therefore,     <!--[if gte vml 1]>   <![endif]--><!--[if !vml]--><!--[endif]--><!--[if gte mso 9]>     <![endif]-->     Since the heat transfer during the isothermal step is positive, we may consider it also to be an input-heat step. The isothermal Q is equal to the isothermal work (calculated in the next part) because DEint = 0 for an ideal gas isothermal process (see Eq. 19-45). Borrowing from the part (b) computation, we have     <!--[if gte vml 1]>   <![endif]--><!--[if !vml]--><!--[endif]--><!--[if gte mso 9]>     <![endif]-->     Therefore, QH = QV + Qisotherm = 7.2 ´ 103 J.     (b) We consider the sum of works done during the processes (noting that no work is done during the constant-volume step). Using Eq. 19-14 and Eq. 19-16, we have     <!--[if gte vml 1]>   <![endif]--><!--[if !vml]--><!--[endif]--><!--[if gte mso 9]>     <![endif]-->     where (by the gas law in ratio form, as illustrated in Sample Problem 19-1) the volume ratio is  <!--[if gte vml 1]>   <![endif]--><!--[if !vml]--><!--[endif]--><!--[if gte mso 9]>     <![endif]-->  Thus, the net work is     <!--[if gte vml 1]>   <![endif]--><!--[if !vml]--><!--[endif]--><!--[if gte mso 9]>     <![endif]-->     (c) Eq. 20-11 gives  <!--[if gte vml 1]>   <![endif]--><!--[if !vml]--><!--[endif]--><!--[if gte mso 9]>     <![endif]-->     59. (a) Processes 1 and 2 both require the input of heat, which is denoted QH. Noting that rotational degrees of freedom are not involved, then, from the discussion in Chapter 19, <!--[if gte vml 1]>   <![endif]--><!--[if !vml]--><!--[endif]--><!--[if gte mso 9]>     <![endif]-->, and <!--[if gte vml 1]>   <![endif]--><!--[if !vml]--><!--[endif]--><!--[if gte mso 9]>     <![endif]-->. We further note that since the working substance is an ideal gas, process 2 (being isothermal) implies Q2 = W2. Finally, we note that the volume ratio in process 2 is simply 8/3. Therefore,     <!--[if gte vml 1]>   <![endif]--><!--[if !vml]--><!--[endif]--><!--[if gte mso 9]>     <![endif]-->     which yields (for T = 300 K and T' = 800 K) the result QH = 25.5 ´ 103 J.     (b) The net work is the net heat (Q1 + Q2 + Q3). We find Q3 from nCp (T - T') = -20.8 ´ 103 J. Thus, W = 4.73 ´ 103 J.     (c) Using Eq. 20-11, we find that the efficiency is     <!--[if gte vml 1]>   <![endif]--><!--[if !vml]--><!--[endif]--><!--[if gte mso 9]>     <![endif]-->     60. (a) Starting from <!--[if gte vml 1]>   <![endif]--><!--[if !vml]--><!--[endif]--><!--[if gte mso 9]>     <![endif]-->  (for calorimetry problems) we can derive (when no phase changes are involved)  <!--[if gte vml 1]>   <![endif]--><!--[if !vml]--><!--[endif]--><!--[if gte mso 9]>     <![endif]-->     which is equivalent to 229 K.     (b) From Eq. 20-1, we have     <!--[if gte vml 1]>   <![endif]--><!--[if !vml]--><!--[endif]--><!--[if gte mso 9]>     <![endif]-->  (c) Also,  <!--[if gte vml 1]>   <![endif]--><!--[if !vml]--><!--[endif]--><!--[if gte mso 9]>     <![endif]-->     (d) The net result for the system is (2.38 – 1.69) J/K = 0.69 J/K. (Note: these calculations are fairly sensitive to round-off errors. To arrive at this final answer, the value 273.15 was used to convert to Kelvins, and all intermediate steps were retained to full calculator accuracy.)     61. From the formula for heat conduction, Eq. 19-32, using Table 19-6, we have     H  =  kA <!--[if supportFields]> EQ \F(TH - TC,L) <![endif]--><!--[if supportFields]><![endif]--> = (401) <!--[if supportFields]> EQ \B(p(0.02)2) <![endif]--><!--[if supportFields]><![endif]-->270/1.50      which yields H = 90.7 J/s.  Using Eq. 20-2, this is associated with an entropy rate-of-decrease of the high temperature reservoir (at 573 K) equal to      ΔS/t = –90.7/573 = –0.158 (J/K)/s.     And it is associated with an entropy rate-of-increase of the low temperature reservoir (at 303 K) equal to      ΔS/t = +90.7/303 = 0.299 (J/K)/s.     The net result is (0.299 – 0.158) (J/K)/s = 0.141 (J/K)/s.     62. (a) Eq. 20-14 gives K = 560/150 = 3.73.     (b) Energy conservation requires the exhaust heat to be 560 + 150 = 710 J.     63. (a) Eq. 20-15 can be written as |QH| = |QL|(1 + 1/KC ) = (35)(1 + <!--[if supportFields]> EQ \f(1,4.6) <![endif]--><!--[if supportFields]><![endif]-->) = 42.6 kJ.     (b) Similarly, Eq. 20-14 leads to |W| = |QL|/K = 35/4.6 = 7.61 kJ.     64. (a) A good way to (mathematically) think of this is: consider the terms when you expand   (1 + x)4 = 1 + 4x + 6x2 + 4x3 + x4.     The coefficients correspond to the multiplicities.  Thus, the smallest coefficient is 1.     (b) The largest coefficient is 6.     (c) Since the logarithm of 1 is zero, then Eq. 20-21 gives S = 0 for the least case.     (d) S = k ln(6) = 2.47 ´ 10-23 J/K.     65. (a)  Eq. 20-2 gives the entropy change for each reservoir (each of which, by definition, is able to maintain constant temperature conditions within itself).  The net entropy change is therefore  DS   =   <!--[if supportFields]> EQ \f(+|Q|,273 + 24) <![endif]--><!--[if supportFields]><![endif]--> + <!--[if supportFields]> EQ \f(-|Q|  ,273 + 130) <![endif]--><!--[if supportFields]><![endif]--> = 4.45 J/K     where we set |Q| = 5030 J.       (b) We have assumed that the conductive heat flow in the rod is “steady-state”; that is, the situation described by the problem has existed and will exist for “long times.”  Thus there are no entropy change terms included in the calculation for elements of the rod itself.     66. Eq. 20-10 gives  <!--[if gte vml 1]>   <![endif]--><!--[if !vml]--><!--[endif]--><!--[if gte mso 9]>     <![endif]--><!--[if gte vml 1]>   <![endif]--><!--[if !vml]--><!--[endif]--><!--[if gte mso 9]>     <![endif]-->     67. We adapt the discussion of §20-7 to 3 and 5 particles (as opposed to the 6 particle situation treated in that section).     (a) The least multiplicity configuration is when all the particles are in the same half of the box. In this case, using Eq. 20-20, we have     <!--[if gte vml 1]>   <![endif]--><!--[if !vml]--><!--[endif]--><!--[if gte mso 9]>     <![endif]-->     (b) Similarly for box B, W = 5!/(5!0!) = 1 in the “least” case.     (c) The most likely configuration in the 3 particle case is to have 2 on one side and 1 on the other. Thus,  <!--[if gte vml 1]>   <![endif]--><!--[if !vml]--><!--[endif]--><!--[if gte mso 9]>     <![endif]-->     (d) The most likely configuration in the 5 particle case is to have 3 on one side and 2 on the other. Thus,  <!--[if gte vml 1]>   <![endif]--><!--[if !vml]--><!--[endif]--><!--[if gte mso 9]>     <![endif]-->     (e) We use Eq. 20-21 with our result in part (c) to obtain     <!--[if gte vml 1]>   <![endif]--><!--[if !vml]--><!--[endif]--><!--[if gte mso 9]>     <![endif]-->     (f) Similarly for the 5 particle case (using the result from part (d)), we find      S = k ln 10 = 3.2 ´ 10-23 J/K.     68. A metric ton is 1000 kg, so that the heat generated by burning 380 metric tons during one hour is<!--[if gte vml 1]>   <![endif]--><!--[if !vml]--><!--[endif]--><!--[if gte mso 9]>     <![endif]--> The work done in one hour is     <!--[if gte vml 1]>   <![endif]--><!--[if !vml]--><!--[endif]--><!--[if gte mso 9]>     <![endif]-->     where we use the fact that a Watt is a Joule-per-second. By Eq. 20-11, the efficiency is     <!--[if gte vml 1]>   <![endif]--><!--[if !vml]--><!--[endif]--><!--[if gte mso 9]>     <![endif]-->     69. Since the volume of the monatomic ideal gas is kept constant it does not do any work in the heating process. Therefore the heat Q it absorbs is equal to the change in its inertial energy: <!--[if gte vml 1]>   <![endif]--><!--[if !vml]--><!--[endif]--><!--[if gte mso 9]>     <![endif]-->. Thus,     <!--[if gte vml 1]>   <![endif]--><!--[if !vml]--><!--[endif]--><!--[if gte mso 9]>     <![endif]-->     70. With the pressure kept constant,     <!--[if gte vml 1]>   <![endif]--><!--[if !vml]--><!--[endif]--><!--[if gte mso 9]>     <![endif]-->     so we need to replace the factor 3/2 in the last problem by 5/2. The rest is the same. Thus the answer now is     <!--[if gte vml 1]>   <![endif]--><!--[if !vml]--><!--[endif]--><!--[if gte mso 9]>     <![endif]-->     71. The change in entropy in transferring a certain amount of heat Q from a heat reservoir at T1 to another one at T2 is DS = DS1 + DS2 = Q(1/T2 - 1/T1).     (a) DS = (260 J)(1/100 K – 1/400 K) = 1.95 J/K.     (b) DS = (260 J)(1/200 K – 1/400 K) = 0.650 J/K.     (c) DS = (260 J)(1/300 K – 1/400 K) = 0.217 J/K.     (d) DS = (260 J)(1/360 K – 1/400 K) = 0.072 J/K.     (e) We see that as the temperature difference between the two reservoirs decreases, so does the change in entropy.     72. The Carnot efficiency (Eq. 20-13) depends linearly on TL so that we can take a derivative  <!--[if gte vml 1]>   <![endif]--><!--[if !vml]--><!--[endif]--><!--[if gte mso 9]>     <![endif]-->     and quickly get to the result. With <!--[if gte vml 1]>   <![endif]--><!--[if !vml]--><!--[endif]--><!--[if gte mso 9]>     <![endif]--> and TH = 400 K, we find dTL ® DTL = -40 K.     73. (a) We use Eq. 20-16. For configuration A     <!--[if gte vml 1]>   <![endif]--><!--[if !vml]--><!--[endif]--><!--[if gte mso 9]>     <![endif]-->     (b) For configuration B     <!--[if gte vml 1]>   <![endif]--><!--[if !vml]--><!--[endif]--><!--[if gte mso 9]>     <![endif]-->     (c) Since all microstates are equally probable,     <!--[if gte vml 1]>   <![endif]--><!--[if !vml]--><!--[endif]--><!--[if gte mso 9]>     <![endif]-->     We use these formulas for N = 100. The results are     (d) <!--[if gte vml 1]>   <![endif]--><!--[if !vml]--><!--[endif]--><!--[if gte mso 9]>     <![endif]-->     (e) <!--[if gte vml 1]>   <![endif]--><!--[if !vml]--><!--[endif]--><!--[if gte mso 9]>     <![endif]-->     (f) and f  WB/WA » 0.14.     Similarly, using the same formulas for N = 200, we obtain     (g) WA = 9.05 ´ 1058,      (h) WB = 1.64 ´ 1057,      (i) and f = 0.018.     (j) We see from the calculation above that f decreases as N increases, as expected.     74. (a) From Eq. 20-1, we infer Q = ∫ T dS, which corresponds to the “area under the curve” in a T-S diagram.  Thus, since the area of a rectangle is (height)´(width), we have Q1→2 = (350)(2.00) = 700J.     (b) With no “area under the curve” for process 2 → 3, we conclude Q2→3 = 0.     (c) For the cycle, the (net) heat should be the “area inside the figure,” so using the fact that the area of a triangle is ½ (base) ´ (height), we find     Qnet =  <!--[if supportFields]> EQ \F(1,2)<![endif]--><!--[if supportFields]><![endif]--> (2.00)(50) = 50 J  .     (d) Since we are dealing with an ideal gas (so that ΔEint = 0 in an isothermal process), then     W1→2  = Q1→2  = 700 J   .     (e) Using Eq. 19-14 for the isothermal work, we have     W1→2 = nRT ln <!--[if supportFields]> EQ \F(V2,V1)<![endif]--><!--[if supportFields]><![endif]-->   .     where T = 350 K.  Thus, if V1 = 0.200 m3, then we obtain     V2 = V1 exp (W/nRT)  = (0.200) e0.12  = 0.226 m3   .     (f) Process 2 → 3 is adiabatic; Eq. 19-56 applies with γ = 5/3 (since only translational degrees of freedom are relevant, here).     T2V2γ-1 = T3V3γ-1     This yields V3 = 0.284 m3.     (g) As remarked in part (d), ΔEint = 0 for process 1 → 2.     (h) We find the change in internal energy from Eq. 19-45 (with CV = <!--[if supportFields]> EQ \F(3,2)<![endif]--><!--[if supportFields]><![endif]--> R):     ΔEint  =  nCV (T3 – T2) = –1.25 ´ 103 J   .     (i) Clearly, the net change of internal energy for the entire cycle is zero.  This feature of a closed cycle is as true for a T-S diagram as for a p-V diagram.     (j) For the adiabatic (2 → 3) process, we have W = -ΔEint.  Therefore, W = 1.25 ´ 103 J.  Its positive value indicates an expansion.      75. Since the inventor’s claim implies that less heat (typically from burning fuel) is needed to operate his engine than, say, a Carnot engine (for the same magnitude of net work), then QH¢ &lt; QH (See Fig. 20-35(a)) which implies that the Carnot (ideal refrigerator) unit is delivering more heat to the high temperature reservoir than engine X draws from it.  This (using also energy conservation) immediately implies Fig. 20-35(b) which violates the second law.