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Chapter 6: Problem 113

The bomb calorimeter in Exercise 112 is filled with 987 \(\mathrm{g}\) water. The initial temperature of the calorimeter contents is $23.32^{\circ} \mathrm{C} .\( A \)1.056-\mathrm{g}\( sample of benzoic acid \)\left(\Delta E_{\mathrm{comb}}=\right.\( \)-26.42 \mathrm{kJ} / \mathrm{g}$ ) is combusted in the calorimeter. What is the final temperature of the calorimeter contents?

Short Answer

Expert verified
The final temperature of the calorimeter contents is approximately \(30.059^\circ C\).

Step by step solution

01

Write down the equations

Write down the heat capacity equation for water (q = mcΔT) and the energy conservation equation (q_water = -q_benzoic).
02

Calculate the heat produced by benzoic acid combustion

Using the given combustion energy of benzoic acid, \(\Delta E_{comb} = -26.42\ kJ/g\) and the mass of benzoic acid, \(m_{benzoic}=1.056\ g\), calculate the heat produced by benzoic acid combustion using the formula \(q_{benzoic} = m_{benzoic} \times \Delta E_{comb}\).
03

Calculate the heat absorbed by water

According to the energy conservation equation, the heat absorbed by water is equal to the negative of the heat produced by the benzoic acid combustion: \(q_{water} = -q_{benzoic}\).
04

Calculate the final temperature of the calorimeter contents

First, let's write the heat capacity equation for water: \(q_{water} = m_{water}c_{water}\Delta T_{water}\). We know the initial temperature, \(T_{initial} = 23.32^\circ C\), mass of water, \(m_{water} = 987\ g\), and the specific heat capacity of water, \(c_{water} = 4.184\ \frac{J}{g\cdot K}\). We will now substitute \(q_{water}\) from step 3 into this equation and solve for the change in temperature, \(\Delta T_{water}\). From there, we will find the final temperature of the calorimeter contents by adding the initial temperature to the change in temperature.
05

Write down the equations Heat capacity equation for water: \(q_{water} = m_{water}c_{water}\Delta T_{water}\) Energy conservation equation: \(q_{water} = -q_{benzoic}\)

Step 2: Calculate the heat produced by benzoic acid combustion Using the given combustion energy and mass of benzoic acid: \(q_{benzoic} = m_{benzoic} \times \Delta E_{comb}\) \(q_{benzoic} = 1.056\ g \times -26.42\ \frac{kJ}{g}\) \(q_{benzoic} = -27.918\ kJ\)
06

Calculate the heat absorbed by water Using the energy conservation equation: \(q_{water} = -q_{benzoic}\) \(q_{water} = -(-27.918\ kJ)\) \(q_{water} = 27.918\ kJ\)

Step 4: Calculate the final temperature of the calorimeter contents Using the heat capacity equation and substituting \(q_{water}\): \(m_{water}c_{water}\Delta T_{water} = q_{water}\) \(987\ g \times 4.184\ \frac{J}{g\cdot K} \times \Delta T_{water} = 27,918\ J\) Calculating the change in temperature: \(\Delta T_{water} = \frac{27,918\ J}{987\ g \times 4.184\ \frac{J}{g\cdot K}}\) \(\Delta T_{water} = 6.739\ K\) Finding the final temperature: \(T_{final} = T_{initial} + \Delta T_{water}\) \(T_{final} = 23.32^\circ C + 6.739\ K\) \(T_{final} = 30.059^\circ C\) Thus, the final temperature of the calorimeter contents is approximately \(30.059^\circ C\).

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Most popular questions from this chapter

Hydrogen gives off \(120 . \mathrm{J} / \mathrm{g}\) of energy when burned in oxygen, and methane gives off \(50 . \mathrm{J} / \mathrm{g}\) under the same circumstances. If a mixture of 5.0 \(\mathrm{g}\) hydrogen and \(10 . \mathrm{g}\) methane is burned, and the heat released is transferred to 50.0 \(\mathrm{g}\) water at \(25.0^{\circ} \mathrm{C},\) what final temperature will be reached by the water?

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