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

$\frac{Q}{g} = \frac{C_p \Delta T}{m_{grams}}$

Our values are given as

$\Delta T = 29.8-25.25 = 4.55\°C$

$C_p = 31.71kJ/\°C$

Replacing,

$\frac{Q}{m} = \frac{(31.71kJ/\°C) (4.55\°C)}{2.599g}$

$\frac{Q}{m} =55.51kJ/g$

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