# Determining the Heat Capacity during a combustion in a bomb calorimeter

The combustion of $$\pu{2.350 mmol}$$ of $$\ce{X}$$ caused the temperature of a bomb calorimter to rise from $$23.700 °C$$ to $$26.300 °C$$. Assume $$T = 25.000 °C$$, $$H_\text{comb} = -3310 \frac{kJ}{mol}$$, and $$ng = -3 \frac{\text{mol gas}}{\text{mol X}}$$ for the combustion and determine the heat capacity of the calorimeter.

I am having some difficulty determining if the process I did is correct for this problem. So I used the equation $$∆H=c∆T + ∆ngRT$$. I first determined $$∆H$$ to be $$\pu{-7.7785 kJ}$$ by multiplying $$\pu{0.00235 mol}$$ to $$\pu{-3310kJ/mol}$$. The rest I just plugged into the equation and got $$c$$ to be $$2855.75$$. I am not sure if that's the correct answer but I don't feel very confident with my process. I would appreciate any help and feedback.

• Are you sure of your data ? Your change of temperature is extremely low. It means that the heat capacity of the calorimeter is high. 99% of the heat is used for heating the calorimeter. It seems odd, or even dubious. The heat capacity I get is 2966 J/K. It is enormous ! – Maurice Feb 11 at 21:46