Question

A 2.455 g sample of a new organic material is combusted in a bomb calorimeter. The temperature of the calorimeter and its contents increase from 23.61 ∘C to 27.04 ∘C. The heat capacity (calorimeter constant) of the calorimeter is 38.29 kJ/ ∘C, what is the heat of combustion per gram of the material?

Answer 1

Answer: The heat of combustion per gram of the material is 53.5 kJ

Explanation:

Let the heat released during reaction be q.

$q=m\times c\times \Delta T$

$q_{cal}$ = Heat gained by calorimeter

Heat capacity of bomb calorimeter ,C = 38.29 kJ/°C  

Change in temperature = ΔT = (27.04-23.61) °C = 3.43 °C

$q_{cal}=C_{bomb}\times \Delta T=38.29\times 3.43=131.3kJ$

Total heat released during reaction is equal to total heat gained by bomb calorimeter.

$q_{combustion}=-(q_{cal})$

$q_{combustion}=-(131.3)J$

Thus 2.455 g of material releases 131.3 kJ of heat

1 g of material releases =$\frac{131.3}{2.455}\times 1=53.5kJ$ of heat

Thus the heat of combustion per gram of the material is 53.5 kJ