Question

Ethyl alcohol, \(\mathrm{C}_{2} \mathrm{H}_{5} \mathrm{OH},\) is the intoxicating agent in liquor. Burning \(1.00 \mathrm{~g}\) of ethyl alcohol in an excess of oxygen at \(23.28^{\circ} \mathrm{C}\) and constant volume releases \(29.52 \mathrm{~kJ}\) of heat. When \(7.40 \mathrm{~g}\) of ethyl alcohol is burned in oxygen under the same conditions in a bomb calorimeter, the temperature of the bomb and water rises from \(23.28^{\circ} \mathrm{C}\) to \(48.04^{\circ} \mathrm{C}\). The bomb holds \(0.750 \mathrm{~kg}\) of water. (a) What is \(q\) for the combustion of the ethyl alcohol in the bomb calorimeter? (b) What is \(q_{\mathrm{H}_{2} \mathrm{O}}\) ? (c) What is \(q_{\text {cal }}\) ? (d) What is the heat capacity of the calorimeter?

Short answer

Answer: The heat released during combustion is approximately \(218.048 \, \mathrm{kJ}\), the heat absorbed by water is approximately \(77.81 \, \mathrm{kJ}\), the heat absorbed by the calorimeter is approximately \(140.238 \, \mathrm{kJ}\), and the heat capacity of the calorimeter is approximately \(5.663 \, \mathrm{kJ/K}\).

Step-by-step solution

Calculate Energy Released During the Combustion of \(7.40 \mathrm{~g}\) of Ethyl Alcohol

Using the given data that burning \(1.00 \mathrm{~g}\) of ethyl alcohol releases \(29.52 \mathrm{~kJ}\) of heat, we can find how much energy was released during the combustion of \(7.40 \mathrm{~g}\) of ethyl alcohol. Energy released per gram = 29.52 kJ/g Total mass of ethyl alcohol = 7.40 g \(q = \text{Energy released per gram} \times \text{Total mass of ethyl alcohol} \) \(q = \left(29.52 \, \mathrm{kJ/g} \right) \times \left(7.40 \, \mathrm{g}\right) = 218.048 \, \mathrm{kJ}\)

Calculate the Heat Absorbed by Water

To find the heat absorbed by water \((q_{\mathrm{H}_{2} \mathrm{O}})\), we'll use the specific heat capacity formula: \(q_{\mathrm{H}_{2} \mathrm{O}} = mc\Delta T\) where \(m\) is the mass of water, \(c\) is the specific heat capacity, and \(\Delta{T}\) is the change in temperature. The mass of water is \(0.750 \, \mathrm{kg}\), and the specific heat capacity of water is \(4.184 \, \mathrm{kJ/kg·K}\). The change in temperature is the final temperature minus the initial temperature: \(\Delta{T} = T_{\text{final}} - T_{\text{initial}}\) \(\Delta{T} = 48.04^{\circ} \mathrm{C} - 23.28^{\circ} \mathrm{C} = 24.76^{\circ} \mathrm{C}\) Now, we plug in the values to find the heat absorbed by water: \(q_{\mathrm{H}_{2} \mathrm{O}} = (0.750 \, \mathrm{kg}) \times (4.184 \, \mathrm{kJ/kg·K}) \times (24.76 \, \mathrm{K})\) \(q_{\mathrm{H}_{2} \mathrm{O}} = 77.81 \, \mathrm{kJ}\)

Calculate the Heat Absorbed by the Calorimeter

According to the conservation of energy, the heat released during the combustion \((q)\) is equal to the sum of the heat absorbed by the water \((q_{\mathrm{H}_{2} \mathrm{O}})\) and the heat absorbed by the calorimeter \((q_{\text{cal}})\): \(q = q_{\mathrm{H}_{2} \mathrm{O}} + q_{\text{cal}}\) Now, we solve for the heat absorbed by the calorimeter: \(q_{\text {cal}} = q - q_{\mathrm{H}_{2} \mathrm{O}} = 218.048 \, \mathrm{kJ} - 77.81 \, \mathrm{kJ} = 140.238 \, \mathrm{kJ}\)

Calculate the Heat Capacity of the Calorimeter

The heat capacity of the calorimeter \((C_{\text{cal}})\) can be calculated using the formula: \(q_{\text{cal}} = C_{\text{cal}} \cdot \Delta{T}\) We solve for the heat capacity of the calorimeter: \(C_{\text{cal}} = \frac{q_{\text{cal}}}{\Delta{T}} = \frac{140.238 \, \mathrm{kJ}}{24.76 \, \mathrm{K}} = 5.663 \, \mathrm{kJ/K}\) The heat capacity of the calorimeter is approximately \(5.663 \, \mathrm{kJ/K}\).