5. We use the following relation derived in Sample Problem Entropy change of two blocks coming to equilibrium:

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1 Chapter We use the following relation derived in Sample Problem Entropy change of two blocks coming to equilibrium: Tf S = mc ln. Ti (a) The energy absorbed as heat is given by Eq Using Table 19-, we find J 4 Q = cmt = kg75 K = J kg K where we have used the fact that a change in Kelvin temperature is equivalent to a change in Celsius degrees. (b) With T f = 7.15 K and T i = K, we obtain J 7.15 S = 2.00 kg8 ln = 17 J/K. kg K In coming to equilibrium, the heat lost by the 100 cm of liquid water (of mass m w = 100 g and specific heat capacity c w = 4190 J/kgK) is absorbed by the ice (of mass m i, which melts and reaches T f 0C). We begin by finding the equilibrium temperature: warm water cools ice warms to 0 ice melts melted ice warms Q 0 Q + Q + Q + Q = 0 c m T 20 + c m L m + c m T 0 = 0 w w f i i F i w i f which yields, after using L F = 000 J/kg and values cited in the problem, T f = which is equivalent to T f = K. Sample Problem Entropy change of two blocks coming to equilibrium shows that 2 Stemp change = mc ln T T1 for processes where T = T 2 T 1, and Eq gives 87

2 88 CAPTER 20 S melt L m = F T o for the phase change experienced by the ice (with T o = K). The total entropy change is (with T in Kelvins) LF mi Ssystem mwcw ln mici ln micw ln ( )J/K 0.4 J/K. 20. (a) The final pressure is V i V f a 1.00 m 2.00 m 1.00 m p = 5.00 kpa e = 5.00 kpa e 1.84 kpa. f (b) We use the ratio form of the gas law to find the final temperature of the gas: T f pv f f (1.84 kpa)(2.00 m ) Ti 00 K 441 K. pv i i (5.00 kpa)(1.00 m ) For later purposes, we note that this result can be written exactly as T f = T i (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 f V V V a V/ / V f i a Va f i i V i kpae 1.00 m e e W pdv (5.00 kpa) e dv 5.00 kpa e ae.1 kj. (d) Consideration of a two-stage process, as suggested in the hint, brings us simply to Eq Consequently, with C R (see Eq. 19-4), we find (5000 Pa)(1.00 m ) 5 V 2 Vf Tf 1 pv i i 1 S nr ln + n R ln = nr ln2 + ln 2e ln2 + ln2 + ln e Vi 2 Ti 2 Ti 2 2 ln 2 00 K J K. 2. With T L = 290 k, we find Vi

3 89 T L L =1 T = = T T 290 K which yields the (initial) temperature of the high-temperature reservoir: T = 48 K. If we replace = 0.40 in the above calculation with = 0.50, we obtain a (final) high temperature equal to T 580 K. The difference is T T = 580 K 48 K = 97 K. 1. (a) We use WQ. The heat absorbed is Q W 8.2kJ kj (b) The heat exhausted is then QL Q W kj 8.2kJ 25kJ. (c) Now we have Q W 8.2kJ 2kJ. 0.1 (d) Similarly, QC Q W 2 kj 8.2 kj = 18kJ. 4. (a) Using Eq for process D A gives p p V p V V p V 2 0 D D = A A 8 0 = 0 0 which leads to 8 = 2 5/. The result (see Sections 19-9 and 19-11) implies the gas is monatomic. (b) The input heat is that absorbed during process A B: 5 T 5 = = 1 = 2 1 = 5 B Q ncpt n R TA nrta p0v 0 2 TA 2 2 and the exhaust heat is that liberated during process C D: 5 T L QL = ncpt = n R TD 1 = nrtd 1 2 = p0v 0 2 TD where in the last step we have used the fact thattd 4 TA (from the gas law in ratio form). Therefore, Eq leads to

4 840 CAPTER 20 Q 1 Q 4 L %. 7. The coefficient of performance for a refrigerator is given by K = Q L / W, where Q L 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 Q + Q L W = 0 for an integer number of cycles. ere Q is the energy ejected to the hot reservoir as heat. Thus, Q L = W Q. Q is negative and greater in magnitude than W, so Q L = Q W. Thus, Q W K. W The solution for W is W = Q /(K + 1). In one hour, 7.54 MJ W 1.57 MJ..8 1 The rate at which work is done is ( J)/(00 s) = 440 W. 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 2 IV 5 V 4 4 VI 5 VII 2 VIII 1 7 IX 0 8 The multiplicity W is computing using Eq For example, the multiplicity for label IV is 8! 4020 W = = = 5 5!! 120 and the corresponding entropy is (using Eq ) S k W 2 2 = ln = J/K ln 5 = J/K. In this way, we generate the following table:

5 841 Label W S I 1 0 II J/K III J/K IV J/K V J/K VI J/K VII J/K VIII J/K IX 1 0. (a) Equation gives K = 50/150 =.7. (b) Energy conser7. (a) Equation can be written as Q = Q L (1 + 1/K C ) = (5)(1 + 1 ) = 42. kj. 4. (b) Similarly, Eq leads to W = Q L /K = 5/4. = 7.1 kj. 74. The Carnot efficiency (Eq. 20-1) depends linearly on T L so that we can take a derivative TL d 1 =1 = T dt T L and quickly get to the result. With d and T = 400 K, we find dt L T L = 40 K. 75. The gas molecules inside a box can be distributed in many different ways. The number of microstates associated with each distinct configuration is called the multiplicity. In general, if there are N molecules and if the box is divided into two halves, with n L molecules in the left half and n R in the right half, such that nl nr N, there are N! arrangements of the N molecules, but n L! are simply rearrangements of the n L molecules in the left half, and n R! are rearrangements of the n R molecules in the right half. These rearrangements do not produce a new configuration. Therefore, the multiplicity factor associated with this is N! W. n L! n R! The entropy is given by S k lnw. (a) The least multiplicity configuration is when all the particles are in the same half of the box. In this case, for system A with with N, we have

6 842 CAPTER 20! W = = 1.!0! (b) Similarly for box B, with N = 5, W = 5!/(5!0!) = 1 in the least case. (c) The most likely configuration in the particle case is to have 2 on one side and 1 on the other. Thus,! W = =. 2!1! (d) The most likely configuration in the 5 particle case is to have on one side and 2 on the other. Thus, 5! W = = 10.!2! (e) We use Eq with our result in part (c) to obtain S k W 2 2 = ln = ln = J/K. (f) Similarly for the 5 particle case (using the result from part (d)), we find S = k ln 10 = J/K. In summary, the least multiplicity is W = 1; this happens when nl N or nl 0. On the other hand, the greatest multiplicity occurs when nl ( N 1) / 2 or nl ( N 1) / 2. vation requires the exhaust heat to be = 710 J. 72. A metric ton is 1000 kg, so that the heat generated by burning 80 metric tons during one hour is kg 28 MJ kg = MJ. The work done in one hour is W = 750 MJ s00 s = MJ where we use the fact that a watt is a joule-per-second. By Eq , the efficiency is MJ MJ %.

ln. 1 We substitute V 2 = 2.00V 1 to obtain

ln. 1 We substitute V 2 = 2.00V 1 to obtain 1. (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 V2