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Question

A 151.1-g sample of a metal at 74.3°C is added to 151.1 g  H20 at 14.8°C. The temperature of the water rises to 18.6°C. Calculate the specific heat capacity of the metal, assuming that all the heat lost by the metal is gained by the water. The specific heat capacity of water is 4.18 J/°C·g.

Answer 1

The water gained heat Q = 151.1 x 4.18 x (18.6 - 14.8) = 2400 J

The heat lost by the metal is (the same) = Cv x 151.1 x (74.3 - 18.6) equating this to 2400 and working the algebra gives Cv = 0.285 J/degC g