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Chapter 6: Problem 107

The bomb calorimeter in Exercise 106 is filled with \(987 \mathrm{~g}\) water. The initial temperature of the calorimeter contents is \(23.32^{\circ} \mathrm{C}\). A 1.056-g sample of benzoic acid \(\left(\Delta E_{\text {comb }}=-26.42 \mathrm{~kJ} / \mathrm{g}\right.\) ) 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 after the combustion of the 1.056-g benzoic acid sample is approximately 30.06 °C.

Step by step solution

01

Calculate the heat released by the combustion of benzoic acid

First, we need to find the heat released by the combustion of the 1.056 g benzoic acid sample. We are given the energy change during combustion (∆E_comb) of benzoic acid as -26.42 kJ/g. Using the mass and ∆E_comb, we can calculate the heat released by the combustion of benzoic acid: Q_benzoic = mass × ∆E_comb = (1.056 g) × (-26.42 kJ/g) = -27.88 kJ Since the heat released is negative, this means that 27.88 kJ of energy is released during the combustion of benzoic acid.
02

Calculate the heat absorbed by the water

The heat released by the combustion of benzoic acid will be absorbed by the water in the calorimeter. Since heat is transferred, the heat absorbed by the water (Q_water) is equal to the heat released by the benzoic acid (Q_benzoic). Q_water = - Q_benzoic = 27.88 kJ We also need to convert this heat to Joules, as we will use it in the next step along with the specific heat capacity of water: Q_water = 27.88 kJ × 1000 J/kJ = 27880 J
03

Calculate the final temperature of the calorimeter contents

Now we will use the equation Q = mc ∆T to find the final temperature of the calorimeter contents, knowing that the mass (m) of water is 987 g, the initial temperature (T_initial) is 23.32 °C, and the specific heat capacity (c) of water is 4.184 J/(g·°C). We know the value of Q, m, and c, and we need to find the change in temperature ( ∆T = T_final - T_initial ). 27880 J = (987 g) × (4.184 J/(g·°C)) × (T_final - 23.32 °C) Now, we can solve for T_final: T_final - 23.32 °C = 27880 J / [(987 g) × (4.184 J/(g·°C))] T_final - 23.32 °C = 6.737 °C T_final = 30.057 °C
04

Answer

The final temperature of the calorimeter contents after the combustion of the 1.056-g benzoic acid sample is approximately 30.06 °C.

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Most popular questions from this chapter

Consider the following cyclic process carried out in two steps on a gas: Step 1: \(45 \mathrm{~J}\) of heat is added to the gas, and \(10 . \mathrm{J}\) of expansion work is performed. Step \(2: 60 . \mathrm{J}\) of heat is removed from the gas as the gas is compressed back to the initial state. Calculate the work for the gas compression in Step \(2 .\)

Calculate the internal energy change for each of the following. a. One hundred (100.) joules of work is required to compress a gas. At the same time, the gas releases \(23 \mathrm{~J}\) of heat. b. A piston is compressed from a volume of \(8.30 \mathrm{~L}\) to \(2.80 \mathrm{~L}\) against a constant pressure of \(1.90 \mathrm{~atm}\). In the process, there is a heat gain by the system of \(350 . \mathrm{J} .\) c. A piston expands against \(1.00\) atm of pressure from \(11.2 \mathrm{~L}\) to \(29.1 \mathrm{~L}\). In the process, \(1037 \mathrm{~J}\) of heat is absorbed.

The overall reaction in a commercial heat pack can be represented as $$ 4 \mathrm{Fe}(s)+3 \mathrm{O}_{2}(g) \longrightarrow 2 \mathrm{Fe}_{2} \mathrm{O}_{3}(s) \quad \Delta H=-1652 \mathrm{~kJ} $$ a. How much heat is released when \(4.00\) moles of iron are reacted with excess \(\mathrm{O}_{2}\) ? b. How much heat is released when \(1.00\) mole of \(\mathrm{Fe}_{2} \mathrm{O}_{3}\) is produced? c. How much heat is released when \(1.00 \mathrm{~g}\) iron is reacted with excess \(\mathrm{O}_{2}\) ? d. How much heat is released when \(10.0 \mathrm{~g} \mathrm{Fe}\) and \(2.00 \mathrm{~g} \mathrm{O}_{2}\) are reacted?

In a coffee-cup calorimeter, \(50.0 \mathrm{~mL}\) of \(0.100 \mathrm{M} \mathrm{AgNO}_{3}\) and \(50.0 \mathrm{~mL}\) of \(0.100 \mathrm{M} \mathrm{HCl}\) are mixed to yield the following reaction: $$ \mathrm{Ag}^{+}(a q)+\mathrm{Cl}^{-}(a q) \longrightarrow \operatorname{AgCl}(s) $$ The two solutions were initially at \(22.60^{\circ} \mathrm{C}\), and the final temperature is \(23.40^{\circ} \mathrm{C}\). Calculate the heat that accompanies this reaction in \(\mathrm{kJ} / \mathrm{mol}\) of \(\mathrm{AgCl}\) formed. Assume that the combined solution has a mass of \(100.0 \mathrm{~g}\) and a specific heat capacity of \(4.18 \mathrm{~J} /{ }^{\circ} \mathrm{C} \cdot \mathrm{g} .\)

Hess's law is really just another statement of the first law of thermodynamics. Explain.
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