Under constant-volume conditions, the heat of combustion of naphthalene \(\left(\mathrm{C}_{10} \mathrm{H}_{8}\right)\) is $40.18 \mathrm{~kJ} / \mathrm{g}$. A 2.50 -g sample of naphthalene is burned in a bomb calorimeter. The temperature of the calorimeter increases from 21.50 to $28.83^{\circ} \mathrm{C}$. (a) What is the total heat capacity of the calorimeter? (b) A 1.50-g sample of a new organic substance is combusted in the same calorimeter. The temperature of the calorimeter increases from 21.14 to $25.08^{\circ} \mathrm{C}$. What is the heat of combustion per gram of the new substance? (c) Suppose that in changing samples, a portion of the water in the calorimeter were lost. In what way, if any, would this change the heat capacity of the calorimeter?

Step by step solution

01

Calculate the heat absorbed by the calorimeter for the naphthalene sample

First, we need to determine the amount of heat absorbed by the calorimeter when the naphthalene sample is burned. To do this, we can use the heat of combustion equation: \(q = m \cdot c\), where \(q\) is the heat absorbed, \(m\) is the mass of the sample, and \(c\) is the heat of combustion per gram. For naphthalene, the heat of combustion is given as \(40.18 \mathrm{~kJ} / \mathrm{g}\). Using the mass of the naphthalene sample (2.50 g), we can find the heat absorbed by the calorimeter: \(q = (2.50 \,\text{g}) \cdot (40.18 \,\text{kJ/g})\) \(q = 100.45 \,\text{kJ}\)

02

Calculate the temperature change of the calorimeter

Now, we need to find the temperature change of the calorimeter during the combustion of the naphthalene sample. The initial and final temperatures are given as \(21.50^{\circ} \mathrm{C}\) and \(28.83^{\circ} \mathrm{C}\), respectively. \(\Delta T = T_{\text{final}} - T_{\text{initial}}\) \(\Delta T = 28.83^{\circ} \mathrm{C} - 21.50^{\circ} \mathrm{C}\) \(\Delta T = 7.33^{\circ} \mathrm{C}\)

03

Calculate the heat capacity of the calorimeter

We can now determine the heat capacity (C) of the calorimeter using the following equation: \(C = \dfrac{q}{\Delta T}\) \(C = \dfrac{100.45 \,\text{kJ}}{7.33^{\circ} \,\text{C}}\) \(C = 13.70 \,\text{kJ/}^{\circ} \mathrm{C}\) So, the heat capacity of the calorimeter is \(13.70 \,\text{kJ/}^{\circ} \mathrm{C}\).

04

Calculate heat absorbed and temperature change for the new substance

For the new organic substance, we have the mass of the sample (1.50 g) and the temperature change: \(\Delta T = 25.08^{\circ} \mathrm{C} - 21.14^{\circ} \mathrm{C} = 3.94^{\circ} \mathrm{C}\). We can now calculate the heat absorbed (q) by the calorimeter during the combustion of the new substance: \(q = C \cdot \Delta T\) \(q = (13.70 \,\text{kJ/}^{\circ} \mathrm{C}) \cdot (3.94^{\circ} \mathrm{C})\) \(q = 53.97 \,\text{kJ}\)

05

Calculate the heat of combustion per gram of the new substance

Finally, we can determine the heat of combustion per gram (c) of the new organic substance using the following equation: \(c = \dfrac{q}{m}\) \(c = \dfrac{53.97 \,\text{kJ}}{1.50 \,\text{g}}\) \(c = 35.98 \,\text{kJ/g}\) So, the heat of combustion per gram of the new substance is \(35.98 \,\text{kJ/g}\).

06

Discuss the effect of water loss on the heat capacity of the calorimeter

If a portion of the water were lost during the change of samples, the total mass of the calorimeter would decrease, affecting its heat capacity. The heat capacity of the calorimeter is directly related to its total mass. If the mass decreases, the heat capacity of the calorimeter would also decrease, which means it would require less heat to increase its temperature by the same amount. As a result, any subsequent measurements using the calorimeter with a lower mass would yield different results, affecting the calculated heat of combustion for other samples.

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