Fa06Soln2 - Physics 7B Fall, 2006 Homework 2 Solutions...

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Unformatted text preview: Physics 7B Fall, 2006 Homework 2 Solutions 19.13 We find the temperature from heat lost = heat gained; m shoe c shoe T shoe = ( m pot c pot + m water c water ) T pot ; (0.40 kg)(450 J/kg C)( T 25C) = [(0.30 kg)(450 J/kg C) + (1.35 L)(1.00 kg/L)(4186 J/kg C)](25C 20C), which gives T = 186C . 19.23 The temperature of the ice will rise to 0C, at which point melting will occur, and then the resulting water will rise to the final temperature. We find the mass of the ice cube from heat lost = heat gained; ( m Al c Al + m water c water ) T Al = m ice ( c ice T ice + L ice + c water T water ); [(0.075 kg)(900 J/kg C) + (0.300 kg)(4186 J/kg C)](20C 17C) = m ice {(2100 J/kg C)[0C ( 8.5C)] + (3.33 10 5 J/kg) + (4186 J/kg C)(17C 0C)}, which gives m ice = 9.4 10 3 kg = 9.4 g . 19.26 The steam will condense at 100C, and then the resulting water will cool to the final temperature. The ice will melt at 0C, and then the resulting water will rise to the final temperature. We find the mass of the steam required from heat lost = heat gained; m steam ( L steam + c water T 1 ) = m ice ( L ice + c water T 2 ); m steam [(22.6 10 5 J/kg) + (4186 J/kg C) (100C 20C)] = (1.00 kg)[(3.33 10 5 J/kg) + (4186 J/kg C)(20C 0C)], which gives m steam = 0.16 kg . 19.36 ( a ) In an adiabatic process there is no heat flow: Q = . ( b ) We use the first law of thermodynamics to find the change in internal energy: U = Q W = 0 ( 2350 J) 0 = + 2350 J. ( c ) The internal energy of an ideal gas depends only on the temperature, U = 8 nRT , so we see that an increase in the internal energy means that the temperature must rise . 19.39 The work done during an isothermal process is W = nRT ln( V 2 / V 1 ) = (2.00 mol)(8.315 J/mol= (2....
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Fa06Soln2 - Physics 7B Fall, 2006 Homework 2 Solutions...

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