A piece of metal of weight 50 grams is heated to a temperature of 100°C and then introduced into a calorimeter containing water at 14°C, the maximum temperature reached being 20°C. If the weight of the calorimeter is 5 grams, the specific heat of the metal of which it is made is 0.095 and the weight of the water it contains is 92 grams, calculate the specific heat of the given metal.
The Attempt at a Solution
First the metal is added to a known mass of water at a known temperature (50g added to water at 100 degrees) in a calorimeter. The metal and the water eventually reach thermal equilibrium (Te), whereby the metal cools from 100, the water warms from 14. Since heat energy is preserved, heat lost by the metal equals the heat gained by the water (assuming the calorimeter absorbs none of the heat).
The specific heat can be given as the calories per gram of material per degree change. The heat loss of the metal is:
Heat Lost = (specific heat)(50g)(100 C – Te)
Specific heat of water= 1.00cal/g deg
Heat Gained = (1.00cal/g deg)(92g)( Te – 14 C)
Heat Lost = Heat Gained, and the final temperature, Te, can be measured
With Te= 20 C.
Specific heat)(50g)(86 C)= (1.00cal/g deg)(92g)(6 C)
Specific heat=0.1283 cal/g deg
Is this right?