The heat capacity (calorimeter constant) of the calorimeter is 34.85 kJ/ ∘C, what is the heat of combustion per gram of the material?

Question

A 2.333 g sample of a new organic material is combusted in a bomb calorimeter. The temperature of the calorimeter and its contents increase from 25.17 ∘C to 30.62 ∘C.

The heat capacity (calorimeter constant) of the calorimeter is 34.85 kJ/ ∘C, what is the heat of combustion per gram of the material?

heat of combustion: ? kJ/g

Answer 1

The heat generated by the combustion is

(34.85 kJ/ °C)(30.62 °C – 25.17 °C) = (34.85 kJ/ °C)(5.45 °C) = 189.9 kJ

The heat of generated per gram is (189.9 kJ)/(2.333 g) = 81. 41 kJ/g

Since combustion is an exothermic reaction, I think heat of combustion is supposed to be given as a negative number

q_rxn = –C_cal × ΔT

Answer

heat of combustion: –81. 41 kJ/g