Question

A 2.991 g sample of a new organic material is combusted in a bomb calorimeter. The temperature of the calorimeter and its contents increase from 23.95 °C to 27.24 °C. The heat capacity (calorimeter constant) of the calorimeter is 28.85 kJ/°C, what is the heat of combustion per gram of the material? heat of combustion: kJ/g

Answer 1

Given,

mass of substance =m= 2.991 g

Initial Temperature = 23.95 C

Final temperature = 27.24 C

Change in temperature =$\bigtriangleup$T = (27.24 - 23.95) C = 3.29 C

Heat capacity = c = 28.85 KJ/C

heat of combustion per gram = Q_reaction = c$\bigtriangleup$T

Q = 28.85 x 3.29

Q = 94.91 KJ

but Q_reactiion = - Q_calorimeter

Q_calorimeter =-94.91 KJ

heat per gram = -94.91/2.991 = - 37.73 KJ/g