Physics Solution Manual for 1100 and 2101

# According to equation 161 however we have that f v so

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Unformatted text preview: st is the average temperature of the rest of the universe. The explicit algebraic signs (− and +) in Equations (2) indicate, respectively, that the sun loses heat and the rest of the universe gains heat. The heat Q radiated by the sun to the rest of the universe in a time t = 1.0 s is given by Equation 13.2: 4 (13.2) Q = eσ Tsun At In Equation 13.2, e is the emissivity of the sun (assumed to be 1, because the sun is taken to ( be a perfect blackbody), σ = 5.67 × 10−8 J/ s ⋅ m 2 ⋅ K 4 ) is the Stefan-Boltzmann constant, and A is the sun’s surface area. Because the sun is a sphere with a radius r, its surface area is ( A = 4π r 2 = 4π 6.96 × 108 m ) 2 = 6.09 × 1018 m 2 (3) SOLUTION Substituting Equations (2) into Equation (1), we obtain ∆S universe = ∆Ssun + ∆Srest = −Q Tsun + Q Trest 1 1 =Q − Trest Tsun (4) Substituting Equation 13.2 into Equation (4) yields 1 1 ∆Suniverse = Q − Trest Tsun 1 1 4 = eσ Tsun At T − T sun rest Therefore, the net entropy change of the entire universe due to one second of thermal radiation from the sun is 828 THERMODYNAMICS ∆S universe = ( ) ( ) 1 1 4 = (1) 5.67 × 10−8 J/ s ⋅ m 2 ⋅ K 4 ( 5800 K ) 6.09 × 1018 m 2 (1.0 s ) − 2.73 K 5800 K = 1.4 × 10 26 J/K 97. REASONING The power rating P of the heater is equal to the heat Q supplied to the gas divided by the time t the heater is on, P = Q / t (Equation 6.10b). Therefore, t = Q / P . The heat required to change the temperature of a gas under conditions of constant pressure is given by Q = CP n ∆T ( Equation 15.6 ) , where CP is the molar specific heat capacity at constant pressure, n is the number of moles, and ∆T = Tf − Ti is the change in temperature. For a monatomic ideal gas, the specific heat capacity at constant pressure is CP = 5 R, 2 Equation (15.7), where R is the universal gas constant. We do not know n, Tf and Ti, but we can use the ideal gas law, PV = nRT , (Equation 14.1) to replace nRTf by PfVf and to replace nRTi by PiVi, quantities that we do know. SOLUTION Substituting Q = CP n ∆T = CP n (Tf − Ti ) into t = Q / P and using the fact that CP = 5 R give 2 t= Q CP n ( Tf − Ti ) = = P P 5 2 R n (Tf − Ti ) P Replacing RnTf by PfVf and RnTi by PiVi and remembering that Pi = Pf , we find t= 5P 2i (Vf − Vi ) P Since the volume of the gas increases by 25.0%, Vf = 1.250Vi. The time that the heater is on is 5 P V −V 5 P 1.250V − V 5 P 0.250 V ( )i ( ( i) i i) t= 2 i f =2 i =2 i P P P = 5 2 ( 7.60 ×105 Pa ) ( 0.250) (1.40 ×10−3 m3 ) = 44.3 s 15.0 W Chapter 15 Problems 829 98. REASONING AND SOLUTION The total heat generated by the students is 7 Q = (200)(130 W)(3000 s) = 7.8 × 10 J For the isochoric process. Q = Cvn∆T = (5R/2)n ∆T The number of moles of air in the room is found from the ideal gas law to be ( )( ) 1.01 × 105 Pa 1200 m3 PV n= = = 5.0 × 104 mol RT 8.31 J/ ( mol ⋅ K ) ( 294 K ) Now ∆T = 99. SSM ( Q = 5 Rn 2 REASONING ) 7.8 × 107 J 5 2 ( 8.31 J/ ( mol ⋅ K ) 5.0 × 104 mol ) = 75 K The efficiency of either engine is given by Equation 15.13, e = 1 − QC / QH . Since engine A receives three times more input heat, produces five times more work, and rejects two times more heat than engine B, it follows that QHA = 3 QHB , WA = 5 WB , and QCA = 2 QCB . As required by the principle of energy conservation for engine A (Equation 15.12), QHA = QCA + W A 123 123 { 3 QHB 2 Q CB 5W B Thus, 3 QHB = 2 QCB + 5 WB (1) Since engine B also obeys the principle of energy conservation (Equation 15.12), QHB = QCB + WB Substituting QHB from Equation (2) into Equation (1) yields 3( QCB + WB ) = 2 QCB + 5 WB Solving for WB gives WB = 1 2 QCB (2) 830 THERMODYNAMICS Therefore, Equation (2) predicts for engine B that 3 QHB = QCB + WB = 2 QCB SOLUTION a. Substituting QCA = 2 QCB and QHA = 3 QHB into Equation 15.13 for engine A, we have QCA eA = 1 − QHA b. Substituting QHB = 3 2 = 1− 2 QCB 3 QHB = 1− 2 QCB 3 ( 3 2 QCB ) = 1− 4 5 = 9 9 QCB into Equation 15.13 for engine B, we have eB = 1 − QCB QHB = 1− QCB 3 2 = 1− QCB 2 1 = 3 3 100. REASONING AND SOLUTION a. The final temperature of the adiabatic process is given by solving Equation 15.4 for Tf. Tf = Ti − W 3 nR 2 = 393 K – 3 (1.00 2 825 J = 327 K mol) [8.31 J/(mol ⋅ K)] b. According to Equation 15.5 for the adiabatic expansion of an ideal gas, PViγ = Pf Vfγ . i Therefore, P Vfγ = Viγ i P f From the ideal gas law, PV = nRT ; therefore, the ratio of the pressures is given by T = i Pf Tf Pi Vf Vi Combining the previous two equations gives T Vfγ = Viγ i T f Solving for Vf we obtain Vf Vi Chapter 15 Problems Vfγ Ti Viγ = Vf Tf Vi T Vf = i Tf Vf( or 1/ ( γ −1) (γ −1) Vi γ −1) T = Vi i T f T = i T f (γ −1) Vi 1/ ( γ −1) Therefore, T Vf = Vi i T f 1 /(γ −1) 1/(2 / 3) 393 K = (0.100 m3 ) 327 K 393 K = (0.100 m3 ) 327 K 3/ 2 = 0.132 m3 831 CHAPTER 16 WAVES AND SOUND ANSWERS TO FOCUS ON CONCEPTS QUESTIONS 1. (c) It is indeed the direction in which the disturbance occurs that distinguishe...
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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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