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

Step-by-step explanation:

Let the heat released during reaction be q.

`q=m* c* \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)* \Delta T=38.29* 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 =
`(131.3)/(2.455)* 1=53.5kJ` of heat

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