When a 0.740-g sample of trinitrotoluene (TNT), \(\mathrm{C}_{7} \mathrm{H}_{5} \mathrm{N}_{2} \mathrm{O}_{6}\), is bumed in a bomb calorimeter, the temperature increases from \(23.4^{\circ} \mathrm{C}\) to \(26.9^{\circ} \mathrm{C}\). The heat capacity of the calorimeter is \(534 \mathrm{J} /^{\circ} \mathrm{C}\), and it contains \(675 \mathrm{mL}\) of water. How much heat was produced by the combustion of the TNT sample?

Short Answer

Expert verified
The combustion of the TNT sample produced 1,869.0 J of heat.

Step by step solution

01

Understand the Heat Capacity and Temperature Change

Identify the calorimeter's heat capacity and the temperature change. The heat capacity is given as 534 J/°C. The initial temperature is 23.4°C, and the final temperature is 26.9°C. The temperature change (ΔT) is the final temperature minus the initial temperature.
02

Calculate the Temperature Change

Use the initial and final temperatures to calculate the temperature change (ΔT). ΔT = final temperature - initial temperature = 26.9°C - 23.4°C.
03

Calculate the Heat Produced

Use the formula Q = C * ΔT, where Q is the heat absorbed or released, C is the heat capacity of the calorimeter, and ΔT is the temperature change. The heat produced by the combustion is the heat absorbed by the calorimeter.
04

Perform the Calculation

Substitute the values of C and ΔT into the formula to calculate the heat produced. Q = 534 J/°C * (26.9°C - 23.4°C).

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Heat Capacity
Heat capacity is an essential concept in understanding how substances absorb or release heat. It is defined as the amount of heat energy required to raise the temperature of a substance by one degree Celsius (oC). In a bomb calorimetry experiment, the calorimeter itself has a specific heat capacity, which is critical for calculating the heat produced during a combustion reaction. For example, a calorimeter with a heat capacity of 534 joules per degree Celsius (J/oC) will require 534 Joules to increase its temperature by one degree Celsius.

In practical terms, think of heat capacity as a measure of a substance's 'thermal inertia.' A high heat capacity means that a substance won't change temperature very easily when it absorbs or loses heat. By knowing the heat capacity of the calorimeter, we can determine the amount of heat a substance releases when it is burned inside the calorimeter. Remember, the larger the heat capacity, the smaller the temperature change for a given amount of heat absorbed.
Temperature Change
The concept of temperature change is straightforward but critical in calorimetry; it's the difference between the final and initial temperatures of the system. When a substance burns in a bomb calorimeter, it transfers heat to the calorimeter assembly and the surrounding water. In our specific exercise, the calorimeter's temperature increases from 23.4°C to 26.9°C upon the combustion of TNT. This temperature difference, calculated as 26.9°C - 23.4°C, which equals 3.5°C, is an indicator of the heat energy transferred.

To connect this to our everyday experience, consider a hot day which has a high temperature 'peak,' versus a cooler day with a lower 'peak' temperature. The difference in feeling between these days is akin to the temperature change in a calorimeter—except instead of feeling the change, the calorimeter measures it precisely. Understanding the temperature change during an experiment provides the information needed to calculate the heat given off or absorbed during a reaction.
Combustion Analysis
Combustion analysis involves determining the amount of heat released when a substance burns. In a controlled setting such as a bomb calorimeter, this analysis helps us to quantify the energy content of a substance, which is fundamental for fields like material science and engineering. Given that the combustion of TNT in our exercise occurs in a bomb calorimeter, we can accurately measure the heat produced during this chemical reaction.

Role in Calorimetry

Combustion reactions are exothermic, meaning they release heat. By carefully measuring the temperature change within the calorimeter—a device known for its constant volume and constant heat capacity—we can precisely calculate the heat of combustion for the sample. In the exercise, once we have the temperature change, we use the formula \(Q = C \times \Delta T\) to calculate the released heat (\(Q\)). This calculation hinges on the accurate measurement of the temperature increase and the known heat capacity of the calorimeter system.

Real-World Applications

Information gleaned from combustion analysis is also valuable in various industries, for instance, to assess the energy content of fuels or materials. This kind of data is crucial for designing engines, assessing environmental impact, and even calculating nutritional content in food science. Being adept with combustion analysis means having the ability to make informed decisions with respect to material usage, energy efficiency, and safety protocols, especially involving explosive substances.

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Most popular questions from this chapter

Propane, \(C_{3} \mathrm{H}_{8}\), is a hydrocarbon that is commonly used as a fuel. (a) Write a balanced equation for the complete combustion of propane gas. (b) Calculate the volume of air at \(25^{\circ} \mathrm{C}\) and 1.00 atmosphere that is needed to completely combust 25.0 grams of propane. Assume that air is 21.0 percent \(\mathrm{O}_{2}\) by volume. (Hint: We will see how to do this calculation in a later chapter on gases - for now use the information that \(1.00 \mathrm{L}\) of air at \(25^{\circ} \mathrm{C}\) and 1.00 atm contains \(0.275 \mathrm{g}\) of \(\mathrm{O}_{2}\) per liter.) (c) The heat of combustion of propane is \(-2,219.2 \mathrm{kJ} / \mathrm{mol}\). Calculate the heat of formation, \(\Delta H_{\mathrm{f}}^{\circ}\) of propane given that \(\Delta H_{\mathrm{f}}^{\circ} \quad\) of \(\mathrm{H}_{2} \mathrm{O}(l)=-285.8 \mathrm{kJ} / \mathrm{mol}\) and \(\Delta H_{\mathrm{f}}^{\circ} \quad\) of \(\mathrm{CO}_{2}(g)=-393.5 \mathrm{kJ} / \mathrm{mol}\) (d) Assuming that all of the heat released in burning 25.0 grams of propane is transferred to 4.00 kilograms of water, calculate the increase in temperature of the water.

Which compound in each of the following pairs has the larger lattice energy? Note: \(\mathrm{Mg}^{2+}\) and \(\mathrm{Li}^{+}\) have similar radii; O \(^{2-}\) and \(\mathrm{F}^{-}\) have similar radii. Explain your choices. (a) MgO or MgSe (b) LiF or MgO (c) \(\mathrm{Li}_{2} \mathrm{O}\) or \(\mathrm{LiCl}\) (d) Li_se or MgO

Calculate \(\Delta H^{\circ}\) for the process \(\mathrm{Zn}(s)+\mathrm{S}(s)+2 \mathrm{O}_{2}(g) \longrightarrow \mathrm{ZnSO}_{4}(s)\) from the following information: \(\mathrm{Zn}(s)+\mathrm{S}(s) \longrightarrow \mathrm{ZnS}(s) \quad \Delta H^{\circ}=-206.0 \mathrm{kJ}\) \(\mathrm{ZnS}(s)+2 \mathrm{O}_{2}(g) \longrightarrow \mathrm{ZnSO}_{4}(s) \quad \Delta H^{\circ}=-776.8 \mathrm{kJ}\)

Calculate \(\Delta H^{\circ}\) for the process \(\mathrm{Sb}(s)+\frac{5}{2} \mathrm{Cl}_{2}(g) \rightarrow \mathrm{SbCl}_{5}(s)\) from the following information: \(\mathrm{Sb}(s)+\frac{3}{2} \mathrm{Cl}_{2}(g) \longrightarrow \mathrm{SbCl}_{3}(s) \quad \Delta H^{\circ}=-314 \mathrm{kJ}\) \(\mathrm{SbCl}_{3}(s)+\mathrm{Cl}_{2}(g) \longrightarrow \mathrm{SbCl}_{5}(s) \quad \Delta H^{\circ}=-80 \mathrm{kJ}\)

Explain the difference between heat capacity and specific heat of a substance.

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