Physics Solution Manual for 1100 and 2101

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Unformatted text preview: iven heat to the cold reservoir reversibly. According to the second law of thermodynamics stated in terms of entropy (see Section 15.11), the reversible engine is the one for which ∆S universe = 0 J/K , and the irreversible engine that could exist is the one for which ∆S universe > 0 J/K . The irreversible engine that could not exist is the one for which ∆Suniverse < 0 J/K . SOLUTION Using Equation 15.18, we write the total entropy change of the universe as the sum of the entropy changes of the hot (H) and cold (C) reservoirs. ∆S universe = − QH TH + QC TC 812 THERMODYNAMICS In this expression, we have used − QH for the heat from the hot reservoir because that reservoir loses heat. We have used + QC for the heat rejected to the cold reservoir because that reservoir gains heat. Applying this expression to the three engines gives the following results: − QH QC −1650 J 1120 J Engine I ∆S universe = + = + = +0.4 J/K TH TC 550 K 330 K Engine II ∆Suniverse = Engine III ∆S universe = − QH TH − QH TH + QC + QC TC TC = −1650 J 990 J + = 0 J/K 550 K 330 K = −1650 J 660 J + = −1.0 J/K 550 K 330 K Since ∆S universe = 0 J/K for Engine II, it is reversible. Since ∆S universe > 0 J/K for Engine I, it is irreversible and could exist. Since ∆S universe < 0 J/K for Engine III, it is irreversible and could not exist. 76. REASONING AND SOLUTION Equation 15.19 gives the unavailable work as Wunavailable = T0 ∆S (1) where T0 = 248 K . We also know that Wunavailable = 0.300 Q . Furthermore, we can apply Equation 15.18 to the heat lost from the 394-K reservoir and the heat gained by the reservoir at temperature T, with the result that –Q Q ∆S = + 394 K T With these substitutions for T0, Wunavailable, and ∆S, Equation (1) becomes –Q Q 0.300 Q = ( 248 K ) + 394 K T or T = 267 K 77. SSM REASONING AND SOLUTION The change in entropy ∆S of a system for a process in which heat Q enters or leaves the system reversibly at a constant temperature T is given by Equation 15.18, ∆S = (Q / T ) R . For a phase change, Q = mL , where L is the latent heat (see Section 12.8). Chapter 15 Problems 813 a. If we imagine a reversible process in which 3.00 kg of ice melts into water at 273 K, the change in entropy of the water molecules is ( ) (3.00 kg) 3.35 ×105 J/kg mLf Q ∆S = = = 3.68 × 103 J/K = 273 K T R T R b. Similarly, if we imagine a reversible process in which 3.00 kg of water changes into steam at 373 K, the change in entropy of the water molecules is ( ) (3.00 kg) 2.26 ×106 J/kg mLv Q ∆S = = = 1.82 ×104 J/K = 373 K T R T R c. Since the change in entropy is greater for the vaporization process than for the fusion process, the vaporization process creates more disorder in the collection of water molecules. 78. REASONING According to the discussion on Section 15.11, the change ∆Suniverse in the entropy of the universe is the sum of the change in entropy ∆SC of the cold reservoir and the change in entropy ∆SH of the hot reservoir, or ∆Suniverse = ∆SC + ∆SH. The change in entropy of each reservoir is given by Equation 15.18 as ∆S = (Q/T)R, where Q is the heat removed from or delivered to the reservoir and T is the Kelvin temperature of the reservoir. In applying this equation we imagine a process in which the heat is lost by the house and gained by the outside in a reversible fashion. SOLUTION Since heat is lost from the hot reservoir (inside the house), the change in entropy is negative: ∆SH = −QH/TH. Since heat is gained by the cold reservoir (the outdoors), the change in entropy is positive: ∆SC = +QC/TC. Here we are using the symbols QH and QC to denote the magnitudes of the heats. The change in the entropy of the universe is ∆Suniverse = ∆SC + ∆S H = QC TC − QH 24 500 J 24 500 J = − = 11.6 J/K TH 258 K 294 K In this calculation we have used the fact that TC = 273 − 15 °C = 258 K and TH = 273 + 21 °C = 294 K. 814 THERMODYNAMICS 79. REASONING The change ∆Suniverse in entropy of the universe for this process is the sum of the entropy changes for (1) the warm water (∆Swater) as it cools down from its initial temperature of 85.0 °C to its final temperature Tf , (2) the ice (∆Sice) as it melts at 0 °C, and (3) the ice water (∆Sice water ) as it warms up from 0 °C to the final temperature Tf: ∆Suniverse = ∆Swater + ∆Sice + ∆Sice water. To find the final temperature Tf , we will follow the procedure outlined in Sections 12.7 and 12.8, where we set the heat lost by the warm water as it cools down equal to the heat gained by the melting ice and the resulting ice water as it warms up. The heat Q that must be supplied or removed to change the temperature of a substance of mass m by an amount ∆T is Q = cm∆T (Equation 12.4), where c is the specific heat capacity. The heat that must be supplied to melt a mass m of a substance is Q = mLf (Equation 12.5), where Lf is the latent heat of fusion. SOLUTION a. We begin by finding the final temperature Tf of the water. Setting the heat lost equal to the heat gained gives cmwater ( 85.0 °C − Tf ) = 144424444 4 3 Heat lost by water mice Lf + mice c (Tf − 0.0 °C ) 1 24 43 144 2444 4 3 Heat gained by melting ice Heat gained by ice water Solving this relation for the final temperature Tf yields Tf =...
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