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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 Δ T shoe = ( m pot c pot + m water c water ) Δ T pot ; (0.40 kg)(450 J/kg · C°)( T – 25°C) = [(0.30 kg)(450 J/kg · C°) + (1.35 L)(1.00 kg/L)(4186 J/kg · C°)](25°C – 20°C), which gives T = 186°C . 19.23 The temperature of the ice will rise to 0°C, 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 ) Δ T Al = m ice ( c ice Δ T ice + L ice + c water Δ T water ); [(0.075 kg)(900 J/kg · C°) + (0.300 kg)(4186 J/kg · C°)](20°C – 17°C) = m ice {(2100 J/kg · C°)[0°C – (– 8.5°C)] + (3.33 × 10 5 J/kg) + (4186 J/kg · C°)(17°C – 0°C)}, which gives m ice = 9.4 × 10 –3 kg = 9.4 g . 19.26 The steam will condense at 100°C, and then the resulting water will cool to the final temperature. The ice will melt at 0°C, 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 Δ T 1 ) = m ice ( L ice + c water Δ T 2 ); m steam [(22.6 × 10 5 J/kg) + (4186 J/kg · C°) (100°C – 20°C)] = (1.00 kg)[(3.33 × 10 5 J/kg) + (4186 J/kg · C°)(20°C – 0°C)], 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: Δ 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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This note was uploaded on 09/09/2008 for the course PHYSICS 7B taught by Professor Packard during the Fall '08 term at Berkeley.
 Fall '08
 Packard
 Work, Heat

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