Question

A 2.599 g sample of a new organic material is combusted in a bomb calorimeter. The temperature of the calorimeter and its contents increase from 25.25 ∘ C to 29.80 ∘ C. The heat capacity (calorimeter constant) of the calorimeter is 31.71 kJ / ∘ C, what is the heat of combustion per gram of the material?

Answer 1

To solve this problem it is necessary to apply the concepts related to thermal transfer given by the thermodynamic definition of heat as a function of mass, specific heat and temperature change.

Mathematically this is equivalent to

`Q = mC_p \Delta T`

Where

m = mass

`C_p =` Specific Heat

`\Delta T =` Change at temperature

In mass terms (KJ / g) this can be expressed as

`(Q)/(g) = (C_p \Delta T)/(m_(grams))`

Our values are given as

`\Delta T = 29.8-25.25 = 4.55\°C`

`C_p = 31.71kJ/\°C`

Replacing,

`(Q)/(m) = ((31.71kJ/\°C) (4.55\°C))/(2.599g)`

`(Q)/(m) =55.51kJ/g`

Therefore the heat of combustion per gram on the material is 55.51KJ/g