QuestionJune 23, 2025

The energy content of food is typically determined using a bomb calorimeter. Consider the combustion of a 0.46-g sample of butter in a bomb calorimeter having a heat capacity of 2.67kJ/^circ C. If the temperature of the calorimeter increases from 23.5^circ C to 28.5^circ C, calculate the energy of combustion per gram of butter. Energy of combustion=square kJ/g

The energy content of food is typically determined using a bomb calorimeter. Consider the combustion of a 0.46-g sample of butter in a bomb calorimeter having a heat capacity of 2.67kJ/^circ C. If the temperature of the calorimeter increases from 23.5^circ C to 28.5^circ C, calculate the energy of combustion per gram of butter. Energy of combustion=square kJ/g
The energy content of food is typically determined using a bomb calorimeter. Consider the combustion of a 0.46-g sample of butter in a bomb calorimeter having a heat capacity of 2.67kJ/^circ C. If the temperature of the calorimeter increases from 23.5^circ C to 28.5^circ C, calculate the energy of combustion per gram of butter. Energy of combustion=square kJ/g

Solution
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Answer

29.02 \, \text{kJ/g} Explanation 1. Calculate the temperature change \Delta T = 28.5^{\circ}C - 23.5^{\circ}C = 5.0^{\circ}C 2. Calculate total energy absorbed by calorimeter Use formula q = C \cdot \Delta T, where C is heat capacity. q = 2.67 \, \text{kJ/}^{\circ}C \times 5.0^{\circ}C = 13.35 \, \text{kJ} 3. Calculate energy of combustion per gram Divide total energy by mass of butter sample: \frac{13.35 \, \text{kJ}}{0.46 \, \text{g}} Energy per gram = 29.02 \, \text{kJ/g}

Explanation

1. Calculate the temperature change<br /> $\Delta T = 28.5^{\circ}C - 23.5^{\circ}C = 5.0^{\circ}C$<br />2. Calculate total energy absorbed by calorimeter<br /> Use formula $q = C \cdot \Delta T$, where $C$ is heat capacity.<br /> $q = 2.67 \, \text{kJ/}^{\circ}C \times 5.0^{\circ}C = 13.35 \, \text{kJ}$<br />3. Calculate energy of combustion per gram<br /> Divide total energy by mass of butter sample: $\frac{13.35 \, \text{kJ}}{0.46 \, \text{g}}$<br /> Energy per gram = $29.02 \, \text{kJ/g}$
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