Physics Solution Manual for 1100 and 2101

According to equation 161 however we have that f v so

Info iconThis preview shows page 1. Sign up to view the full content.

View Full Document Right Arrow Icon
This is the end of the preview. Sign up to access the rest of the document.

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...
View Full Document

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.

Ask a homework question - tutors are online