Question

The energy content of food is typically determined using a bomb calorimeter. Consider the combustion of a 0.22-g sample of butter in a bomb calorimeter having a heat capacity of 2.67 kJ/°C. If the temperature of the calorimeter increases from 23.5°C to 26.1°C, calculate the energy of combustion per gram of butter. Energy of combustion = kJ/g

Answer 1

Answer: The energy of combustion of butter is 31.5 kJ/g

Explanation:

The quantity of heat required to raise the temperature of a substance by one degree Celsius is called the specific heat capacity.

$Q=C\times \Delta T$

Q = Heat absorbed by calorimeter =?

C = heat capacity of calorimeter = $2.67kJ/^0C$

Initial temperature of the calorimeter  = $T_i$ = $23.5^0C$

Final temperature of the calorimeter  = $T_f$  = $26.1^0C$

Change in temperature ,$\Delta T=T_f-T_i=(26.1-23.5)^0C=2.6^0C$

Putting in the values, we get:

$Q=2.67kJ/^0C\times 2.6^0C=6.94kJ$

As heat absorbed by calorimeter is equal to heat released by combustion of butter

$Q=q$

Heat released by 0.22 g of butter = 6.94 kJ

Heat released by 1g of butter = $\frac{6.94}{0.22}\times 1=31.5kJ$

The energy of combustion of butter is 31.5 kJ/g