Question

When a 3.673-g sample of a new organic material was combusted in a bomb calorimeter, the temperature of the calorimeter (and its contents) increased from 25.37 °C to 31.02 °C. If the heat capacity (calorimeter constant) of the calorimeter is 41.11 kJ/°C, what is the heat of combustion per gram of the material?

Answer 1

Ans - $Q = C.\Delta T$

detla T = 31.2 - 25.37 = 5.83

Q = 41.11 KJ x 5.83 = 239.67 KJ

So, you know that the combustion of 3.673 g sample, will give off 239.67 kJ of heat.

1 mol / 1 gram x 3.673 grams = 3.673 moles

239.67/3.673 = -65.25 KJ/ mol

239.67 / 3.673 =- 65.251 KJ / g