Chemistry Stoichiometry

Question

A 1.111 g sample of a new organic material is combusted in a bomb calorimeter. The temperature of the calorimeter and its contents increase from 24.19 degrees C to 30.92 degrees C.

The heat capacity (calorimeter constant) of the calorimeter is 41.81 kJ/C what is the heat of combustion per gram of the material?

heat of combustion: _____ kJ/g

Answer 1

Change in temperature = ∆T = 30.92º - 24.19º = 6.73º

q = C∆T

q = heat = ?

C = calorimeter constant = 41.81 kJ/ºC

∆T = 6.73ºC

q = (41.81 kJ/º)(6.73º) = 281.4 kJ

heat of combustion per gram = 281.4 kJ / 1.111 g = 253.3 kJ/g