A 140.0-g sample of water at 25.0°C is mixed with 110.0 g of a certain metal at 100.0°C?
After thermal equilibrium is established, the (final) temperature of the mixture is 29.6°C. What is the heat capacity of the metal, assuming it is constant over the temperature range concerned?
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heat energy gained by the water = heat energy lost by the metal
heat energy gained by the water = 140.0 g x 4.184 J/g C x (29.6-25.0) C
heat energy gained by the water = 2694 J
heat energy lost by the metal = 2694 J = heat capacity 110.0 g x (100.0 - 29.6) C
heat capacity of the metal = 2694 J / (110.0 g x 70.4 C) = 0.348 J/g C