Question

A calorimeter contained 75g of water at 16.95 degree Celsius. A 93.3g Iron at 65.58 degrees celsius was placed in it, giving a final temperature of 19.68 degrees Celsius for the system. Calculate the heat capacity of the calorimeter. specific heats are 4.184J/gdegrees Celsius and 0.444J/gdegrees Celsius for Fe.

Answer 1

So the heat gained + heat lost is zero.
75*cwater*(19.68-16.95)+HeatCapictyCalorimeter*(19.68-16.95)+93.3*ciron*(19.68-65.58)=0

solve for heatcapacity of calormeter.

Answer 2

heat lost by the iron was gained by the water and the calorimeter

heat capacity is the same as specific heat