help with chem please

Question

A 1.207 g1.207 g sample of a new organic material is combusted in a bomb calorimeter. The temperature of the calorimeter and its contents increase from 23.57 ∘C23.57⁢ ∘C to 28.72 ∘C.28.72⁢ ∘C.

The heat capacity (calorimeter constant) of the calorimeter is 43.19 kJ/ ∘C,43.19 kJ/ ∘C, what is the heat of combustion per gram of the material?

Answer 1

q = Cal∆T

q = heat = ?

Ccal = calorimeter constant = 43.19 kJ/º

∆T = change in temperature = 28.72º - 23.57º = 5.15º

q = (43.19 kJ/º)(5.15º0 = 222.4 kJ

Heat of combustion per gram = 222.4 kJ / 1.207 g = 184.3 kJ / g