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Chapter 7: Problem 41

A 1.620 g sample of naphthalene, \(C_{10} \mathrm{H}_{8}(\mathrm{s}),\) is completely burned in a bomb calorimeter assembly and a temperature increase of \(8.44^{\circ} \mathrm{C}\) is noted. If the heat of combustion of naphthalene is \(-5156 \mathrm{kJ} / \mathrm{mol} \mathrm{C}_{10} \mathrm{H}_{8}\) what is the heat capacity of the bomb calorimeter?

Short Answer

Expert verified
The heat capacity of the bomb calorimeter is \(7.71 kJ/^{\circ} C\).

Step by step solution

01

Determine the moles of the sample

The amount of heat released by the naphthalene sample can be calculated by first finding the how many moles are in the sample. Since the molar mass of naphthalene \(C_{10}H_{8}\) is 128.17 g/mol, the number of moles \(n\) can be found by dividing the mass of the sample by the molar mass, i.e., \(n = \frac{1.620 g}{128.17 g/mol} = 0.01263 mol\).
02

Calculate the heat released by the sample

The amount of heat \(\Delta{H}\) released by the naphthalene sample can be calculated by multiplying the number of moles \(n\) by the heat of combustion, i.e., \(\Delta{H} = n \times -5156 kJ/mol = -65.05 kJ\). Note that the heat of combustion is negative because it is an exothermic process (heat is released).
03

Calculate the heat capacity of the bomb calorimeter

As per our initial analysis, the heat absorbed by the calorimeter equals the heat released by the burning sample. Hence, the heat capacity \(C\) of the calorimeter can be found by dividing the absolute value of the heat absorbed (equal to the heat released by the sample) by the change in temperature, i.e., \(C = \frac{\Delta{H}}{\Delta{T}} = \frac{65.05 kJ}{8.44^{\circ} C} = 7.71 kJ/^{\circ} C\).

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Most popular questions from this chapter

Use Hess's law to determine \(\Delta H^{\circ}\) for the reaction \(\mathrm{C}_{3} \mathrm{H}_{4}(\mathrm{g})+2 \mathrm{H}_{2}(\mathrm{g}) \longrightarrow \mathrm{C}_{3} \mathrm{H}_{8}(\mathrm{g}),\) given that $$\mathrm{H}_{2}(\mathrm{g})+\frac{1}{2} \mathrm{O}_{2}(\mathrm{g}) \longrightarrow \mathrm{H}_{2} \mathrm{O}(\mathrm{l}) \quad \Delta H^{\circ}=-285.8 \mathrm{kJ}$$ $$\begin{aligned} \mathrm{C}_{3} \mathrm{H}_{4}(\mathrm{g})+4 \mathrm{O}_{2}(\mathrm{g}) \longrightarrow & 3 \mathrm{CO}_{2}(\mathrm{g})+2 \mathrm{H}_{2} \mathrm{O}(\mathrm{l}) \\ && \Delta H^{\circ}=-1937 \mathrm{kJ} \end{aligned}$$ $$\begin{array}{r} \mathrm{C}_{3} \mathrm{H}_{8}(\mathrm{g})+5 \mathrm{O}_{2}(\mathrm{g}) \longrightarrow 3 \mathrm{CO}_{2}(\mathrm{g})+4 \mathrm{H}_{2} \mathrm{O}(\mathrm{l}) \\ \Delta H^{\circ}=-2219.1 \mathrm{kJ} \end{array}$$

A clay pot containing water at \(25^{\circ} \mathrm{C}\) is placed in the shade on a day in which the temperature is \(30^{\circ} \mathrm{C} .\) The outside of the clay pot is kept moist. Will the temperature of the water inside the clay pot (a) increase; (b) decrease; (c) remain the same?

A 0.205 g pellet of potassium hydroxide, \(\mathrm{KOH}\), is added to \(55.9 \mathrm{g}\) water in a Styrofoam coffee cup. The water temperature rises from 23.5 to \(24.4^{\circ} \mathrm{C}\). [Assume that the specific heat of dilute \(\mathrm{KOH}(aq)\) is the same as that of water.] (a) What is the approximate heat of solution of \(\mathrm{KOH}\) expressed as kilojoules per mole of \(\mathrm{KOH}?\) (b) How could the precision of this measurement be improved without modifying the apparatus?

The method of Exercise 97 is used in some bomb calorimetry experiments. A 1.148 g sample of benzoic acid is burned in excess \(\mathrm{O}_{2}(\mathrm{g})\) in a bomb immersed in 1181 g of water. The temperature of the water rises from 24.96 to \(30.25^{\circ} \mathrm{C}\). The heat of combustion of benzoic acid is \(-26.42 \mathrm{kJ} / \mathrm{g} .\) In a second experiment, a \(0.895 \mathrm{g}\) powdered coal sample is burned in the same calorimeter assembly. The temperature of \(1162 \mathrm{g}\) of water rises from 24.98 to \(29.81^{\circ} \mathrm{C}\). How many metric tons (1 metric ton \(=1000 \mathrm{kg}\) ) of this coal would have to be burned to release \(2.15 \times 10^{9} \mathrm{kJ}\) of heat?

A 465 g chunk of iron is removed from an oven and plunged into \(375 \mathrm{g}\) water in an insulated container. The temperature of the water increases from 26 to \(87^{\circ} \mathrm{C}\). If the specific heat of iron is \(0.449 \mathrm{Jg}^{-1}\) \(^{\circ} \mathrm{C}^{-1},\) what must have been the original temperature of the iron?
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