Question

A 2.569 g sample of a new organic material is combusted in a bomb calorimeter. The temperature of the calorimeter and its contents increase from 24.77 °C to 28.40 °C. The heat capacity (calorimeter constant) of the calorimeter is 33.79 kJ/ °C, what is the heat of combustion per gram of the material? heat of combustion: kJ/g

Answer 1

Heat absorb by the calorimeter,

= calorimeter constant*temperature change

= 33.79*(28.40-24.77)

= 122.6577 kJ

2.659 gram sample release energy = 122.6577 kJ

So, heat release per gram = 122.6577/2.659

Heat of combustion = 46.13 kJ/g

But it should be negative in number because heat is released by the reaction

So, -46.13 KJ /g

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