general chem chapter 6

Question

A 3.921 sample of a new organic material is combusted in a bomb calorimeter. The temperature of the calorimeter and its contents increase from 24.89 degrees c  to 28.74 degrees c

The heat capacity (calorimeter constant) of the calorimeter is  28.91 kJ/c*what is the heat of combustion per gram of the material?

Answer 1

They ask for the heat of combustion PER GRAM, and we are combusting 3.921 g, so this must be taken into account at the end of the calculation.

The heat generated is calculated by multiplying the change in temperature by the calorimeter constant. Note how the units cancel to leave us with kJ (units of heat).

∆T = 28.74º - 24.89º = 3.85º

3.85º x 28.91 kJ/º = 111.3 kJ of heat generated by combustion of 3.921 g of material

Heat per gram = 111.3 kJ / 3.921 g = 28.39 kJ/g