Physics Solution Manual for 1100 and 2101

2 into equation 4 yields 1 1 suniverse q trest tsun 1

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Unformatted text preview: ations will allow us to determine W . c. The difference in the work produced by the two engines is labeled Wunavailable in Section15.11, where Wunavailable = Wreversible − Wirreversible. The difference in the work is related to the change in the entropy of the universe by Wunavailable = T0 ∆Suniverse (Equation 15.19), where T0 is the Kelvin temperature of the coldest heat reservoir. In this case T0 = TC. SOLUTION a. From part a of the REASONING, the change in entropy of the universe is ∆S universe = − QH TH + QC TC The magnitude QC of the heat rejected to the cold reservoir is related to the magnitude QH of the heat supplied to the engine from the hot reservoir and the magnitude W of the work done by the engine via QC = QH − W (Equation 15.12). Thus, ∆Suniverse becomes ∆S universe = − QH TH + QH − W TC =− 1285 J 1285 J − 264 J + = +1.74 J/K 852 K 314 K 818 THERMODYNAMICS b. The magnitude W of the work done by an engine depends on its efficiency e and input heat QH via W = e QH (Equation 15.11). For a reversible engine, the efficiency is related to the temperatures of its hot and cold reservoirs by e = 1 − (TC/TH), Equation 15.15. The work done by the reversible engine is T Wreversible = e QH = 1 − C T H 314 K QH = 1 − (1285 J ) = 811 J 852 K c. According to the discussion in part c of the REASONING, the difference between the work done by the reversible and irreversible engines is Wreversible − Wirreversible = TC ∆Suniverse = ( 314 K ) (1.74 J/K ) = 546 J 1444 24444 4 3 Wunavailable 82. REASONING An adiabatic process is one for which no heat enters or leaves the system, so Q = 0 J. The work is given as W = +610 J, where the plus sign denotes that the gas does work, according to our convention. Knowing the heat and the work, we can use the first law of thermodynamics to find the change ∆U in internal energy as ∆U = Q – W (Equation 15.1). Knowing the change in the internal energy, we can find the change in the temperature 3 by recalling that the internal energy of a monatomic ideal gas is U = 2 nRT, according to Equation 14.7. As a result, it follows that ∆U = 3 2 nR∆T. SOLUTION Using the first law from Equation 15.5 and the change in internal energy from Equation 14.7, we have ∆U = Q − W Therefore, we find ∆T = bg = 2 Q −W 3nR or 3 2 nR∆T = Q − W b gb g = 3b mol g .31 J / b ⋅ K g 0.50 8 mol 2 0 J − 610 J −98 K The change in temperature is a decrease. 83. SSM REASONING According to the first law of thermodynamics (Equation 15.1), 3 ∆U = Q − W . For a monatomic ideal gas (Equation 14.7), U = 2 nRT . Therefore, for the 3 process in question, the change in the internal energy is ∆U = 2 nR∆T . Combining the last expression for ∆U with Equation 15.1 yields Chapter 15 Problems 3 2 819 nR∆T = Q − W This expression can be solved for ∆T . SOLUTION a. The heat is Q = +1200 J , since it is absorbed by the system. The work is W = +2500 J , since it is done by the system. Solving the above expression for ∆T and substituting the values for the data given in the problem statement, we have ∆T = Q −W 1200 J − 2500 J =3 = –2.1 × 10 2 K 3 2 nR 2 (0.50 mol)[8.31 J / (mol ⋅ K)] b. Since ∆T = Tfinal − Tinitial is negative, Tinitial must be greater than Tfinal ; this change represents a decrease in temperature. Alternatively, one could deduce that the temperature decreases from the following physical argument. Since the system loses more energy in doing work than it gains in the form of heat, the internal energy of the system decreases. Since the internal energy of an ideal gas depends only on the temperature, a decrease in the internal energy must correspond to a decrease in the temperature. 84. REASONING According to Equation 15.11, the efficiency e of a heat engine is e = W / QH , where W is the magnitude of the work done by the engine and QH is the magnitude of the input heat that the engine uses. QH is given, but W is unknown. However, energy conservation requires that QH = W + QC (Equation 15.12), where QC is the magnitude of the heat rejected by the engine and is given. From this equation, therefore, a value for W can be obtained. SOLUTION According to Equation 15.11, the efficiency is e= W QH Since QH = W + QC (Equation 15.12), we can solve for W to show that W = QH − QC . Substituting this result into the efficiency expression gives e= W QH = QH − QC QH = 5.6 × 104 J − 1.8 × 104 J 5.6 × 104 J = 0.68 820 THERMODYNAMICS 85. REASONING AND SOLUTION The work done by the expanding gas is W = Q − ∆U = 2050 J − 1730 J = 320 J The work, according to Equation 6.1, is also the magnitude F of the force exerted on the piston times the magnitude s of its displacement. But the force is equal to the weight mg of the block and piston, so that the work is W = Fs = mgs. Thus, we have s= 320 J W = = 0.24 m mg 135 kg 9.80 m / s 2 b g c h 86. REASONING According to the definition of efficiency given in Equation 15.11, an engine with an efficiency e does work of magnitude W = e QH , where QH is the magnitude of the input hea...
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This note was uploaded on 04/30/2013 for the course PHYS 1100 and 2 taught by Professor Chastain during the Spring '13 term at LSU.

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