HELP HEAT COMBUSTION CHEM

Question

A 2.223 g sample of a new organic material is combusted in a bomb calorimeter. The temperature of the calorimeter and its contents increase from 23.37 °C to  29.52°C. 

The heat capacity (calorimeter constant) of the calorimeter is 28.69 kJ/°C, what is the heat of combustion per gram of the material? (kJ/g)

Answer 1

q = C ΔT

ΔT = T_f - T_i = 29.52 - 23.37 = 6.15 °C

q = 28.69 kJ/°C x 6.15 °C = 176.4 kJ

Heat of combustion per gram = 176.4 kJ / 2.223 g = 79.35 kJ/g

Answer 2

The temperature change in the calorimeter was 29.52 - 23.37^o, = 6.15^o. The heat capacity of the calorimeter is given as 28.69 kJ/^oC, so the energy of the combustion was enough to raise the temperature 6.15^oC, which means the amount of energy (heat of combustion) was 28.69 kJ x 6.15 = 176.44 kJoules for 2.223 grams. That was for 2.223 grams of material combusted. So the heat of combustion per gram = 176.44kJ/2.223 grams = 79.37 kJ/g.