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Unformatted text preview: If m I is the mass of the ice and c I is its specific heat then the ice absorbs heat ( ) . I I f Ii Q c m T T = Since no energy is lost to the environment, these two heats (in absolute value) must be the same. Consequently, ( ) ( ) . W W Wi f I I f Ii c m T T c m T T = The solution for the equilibrium temperature is (4190J / kg K)(0.200kg)(25 C) (2220J/kg K)(0.100kg)( 15 C) (4190J/kg K)(0.200kg) (2220J/kg K)(0.100kg) 16.6 C. W W Wi I I Ii f W W I I c m T c m T T c m c m + = + + = + = This is above the melting point of ice, which invalidates our assumption that no ice has melted. That is, the calculation just completed does not take into account the melting of the ice and is in error. Consequently, we start with a new assumption: that the water and ice reach thermal equilibrium at T f = 0C, with mass m ( < m I ) of the ice melted. The magnitude of the heat rejected by the water is   = , W W Wi Q c m T 41. (a) We work in Celsius temperature, which poses no difficulty for the J/kgK values 41....
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 Spring '08
 Any
 Physics, Work, Heat

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