Questions: The energy content of food is typically determined using a bomb calorimeter. Consider the combustion of a 0.34 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 28.0°C, calculate the energy of combustion per gram of butter. Energy of combustion = kJ / g

Solution Steps

First, determine the change in temperature (\(\Delta T\)) of the calorimeter:

\[ \Delta T = T_{\text{final}} - T_{\text{initial}} = 28.0^{\circ} \mathrm{C} - 23.5^{\circ} \mathrm{C} = 4.5^{\circ} \mathrm{C} \]

Next, use the heat capacity of the calorimeter to calculate the total energy released during the combustion:

\[ q = C \times \Delta T \]

where \(C = 2.67 \, \mathrm{kJ}/^{\circ} \mathrm{C}\) is the heat capacity of the calorimeter.

\[ q = 2.67 \, \mathrm{kJ}/^{\circ} \mathrm{C} \times 4.5^{\circ} \mathrm{C} = 12.015 \, \mathrm{kJ} \]

Finally, calculate the energy of combustion per gram of butter. The sample mass is \(0.34 \, \mathrm{g}\).

\[ \text{Energy per gram} = \frac{q}{\text{mass of butter}} = \frac{12.015 \, \mathrm{kJ}}{0.34 \, \mathrm{g}} \]

\[ \text{Energy per gram} = 35.3382 \, \mathrm{kJ/g} \]

Final Answer