A 3.283 g sample of a new organic material is combusted in a bomb calorimeter. The temperature of the calorimeter and its contents increase from 25.67 *C to 31.40 *C.

Question

A 3.283 g sample of a new organic material is combusted in a bomb calorimeter. The temperature of the calorimeter and its contents increase from 25.67 *C to 31.40 *C.

The heat capacity (calorimeter constant) of the calorimeter is 33.61 kJ/ *C, what is the heat of combustion per gram of the material?

heat of combustion: __________ kJ/ g

Answer 1

q = Ccal∆T

q = heat = ?

Ccal = calorimeter constant = 33.62 kJ/ºC

∆T = change in temperature = 31.40 - 25.67 = 5.73ºC

q = (33.62 kJ/ºC)(5.73ºC) = 192.6 kJ ... heat released from 3.283 g sample

heat of combustion = 192.6 kJ / 3.283 g = 58.68 kJ/g