Energy in chemical reaction?

Question

A temperature rise of 1.78oC was observed when 1.00*10^-3 mol of propane gas was burnt in a calorimeter. The calibration factor of the calorimeter was previouly determined to be 1250J oC^-1

1. calculate the heat of combustion of propane in kJ mol-1

Answer 1

It takes 1250 J to raise the temperature of the calorimeter by 1ºC. So it took 1.78 x 1250 J = 2225 J to raise it by 1.78ºC. This energy was released by burning 1.00*10^-3 mol of propane.

Heat of combustion of propane = 2225 x 1000 = 2225 kJ/mol