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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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Key Concepts

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

Benzoic Acid
Benzoic acid is an organic compound with the chemical formula \( C_7H_6O_2 \). It is widely used as a standard for calibrating bomb calorimeters due to its reliable and well-documented combustion properties.
During combustion, benzoic acid reacts with oxygen to release a specific amount of heat energy. This characteristic makes measuring other reaction energies simpler when using a bomb calorimeter.
  • It is commonly found in solid form at room temperature and is known for its role in food preservation and the synthesis of various chemicals.
  • The compound's high combustion energy (-26.42 kJ/g) is precisely measured, making it ideal for calorimetric standards.
Understanding benzoic acid's role helps in grasping how energy measurements work in controlled settings like a bomb calorimeter.
Heat Transfer
Heat transfer is the process of energy movement from a hotter object to a cooler one. In a bomb calorimeter, heat released during a chemical reaction is transferred from the combustion chamber to the surrounding water.
This transfer allows the energy change from a reaction to be quantified, as the surrounding water captures the released heat, leading to an increase in its temperature.
  • There are three main types of heat transfer: conduction, convection, and radiation, but in bomb calorimeters, primarily conduction is involved.
  • The calorimeter ensures minimal heat loss to the environment, making it an isolated system ideal for precise measurements.
The efficiency of a bomb calorimeter relies heavily on the effective transfer and containment of heat within its system, which is why it is extensively used in thermodynamic studies.
Specific Heat Capacity
Specific heat capacity (\( c \)) is a property of a substance that indicates the amount of heat required to raise the temperature of 1 gram of the substance by 1°C (or 1 K). Water, with a specific heat capacity of 4.184 J/(g°C), is commonly used in calorimetry because of its high value, allowing it to absorb large amounts of heat without a significant change in temperature.
The specific heat capacity helps in calculating the temperature change of the water in the bomb calorimeter during a reaction.
  • It is a crucial factor when determining the thermal properties of substances.
  • The large specific heat capacity of water acts as a buffer, allowing small energy changes to be observed accurately through temperature differences.
Mastery of specific heat capacity is fundamental for understanding how different substances respond to heat input and energy changes.
Combustion Reaction
A combustion reaction is a high-energy chemical process that occurs when a substance reacts with oxygen, releasing energy in the form of heat and light. In a bomb calorimeter, these reactions are contained, allowing for precise measurements of the energy released.
Benzoic acid undergoes complete combustion inside the calorimeter, providing a controlled environment where energy changes can be directly measured.
  • Combustion reactions are generally characterized by the production of gases such as carbon dioxide and water vapor.
  • These reactions are essential in studying both energy changes in chemistry and practical applications in energy production.
Understanding the dynamics of combustion reactions is key to connecting theoretical concepts to real-world energy transformations.

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Most popular questions from this chapter

The enthalpy change for a reaction is a state function and it is an extensive property. Explain.

Consider a balloon filled with helium at the following conditions. $$ \begin{array}{l} 313 \text { g He } \\ 1.00 \text { atm } \\ \text { 1910. L } \\ \text { Molar Heat Capacity }=20.8 \mathrm{~J} /{ }^{\circ} \mathrm{C} \cdot \mathrm{mol} \end{array} $$ The temperature of this balloon is decreased by \(41.6^{\circ} \mathrm{C}\) as the volume decreases to \(1643 \mathrm{~L}\), with the pressure remaining constant. Determine \(q, w\), and \(\Delta E\) (in \(\mathrm{kJ}\) ) for the compression of the balloon.

Calculate \(\Delta H\) for the reaction $$ \mathrm{N}_{2} \mathrm{H}_{4}(l)+\mathrm{O}_{2}(g) \longrightarrow \mathrm{N}_{2}(g)+2 \mathrm{H}_{2} \mathrm{O}(l) $$ given the following data: $$ \begin{array}{cr} \text { Equation } & \Delta H(\mathrm{k}\rfloor) \\ 2 \mathrm{NH}_{3}(g)+3 \mathrm{~N}_{2} \mathrm{O}(g) \longrightarrow 4 \mathrm{~N}_{2}(g)+3 \mathrm{H}_{2} \mathrm{O}(I) & -1010 \\ \mathrm{~N}_{2} \mathrm{O}(g)+3 \mathrm{H}_{2}(g) \longrightarrow \mathrm{N}_{2} \mathrm{H}_{4}(I)+\mathrm{H}_{2} \mathrm{O}(I) & -317 \\ 2 \mathrm{NH}_{3}(g)+\frac{1}{2} \mathrm{O}_{2}(g) \longrightarrow \mathrm{N}_{2} \mathrm{H}_{4}(I)+\mathrm{H}_{2} \mathrm{O}(I) & -143 \\ \mathrm{H}_{2}(\mathrm{~g})+\frac{1}{2} \mathrm{O}_{2}(g) \longrightarrow \mathrm{H}_{2} \mathrm{O}(I) & -286 \end{array} $$

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?

Consider the reaction \(2 \mathrm{HCl}(a q)+\mathrm{Ba}(\mathrm{OH})_{2}(a q) \longrightarrow \mathrm{BaCl}_{2}(a q)+2 \mathrm{H}_{2} \mathrm{O}(l)\) \(\Delta H=-118 \mathrm{~kJ}\) Calculate the heat when \(100.0 \mathrm{~mL}\) of \(0.500 \mathrm{M} \mathrm{HCl}\) is mixed with \(300.0 \mathrm{~mL}\) of \(0.100 \mathrm{M} \mathrm{Ba}(\mathrm{OH})_{2}\). Assuming that the temperature of both solutions was initially \(25.0^{\circ} \mathrm{C}\) and that the final mixture has a mass of \(400.0 \mathrm{~g}\) and a specific heat capacity of \(4.18 \mathrm{~J} /{ }^{\circ} \mathrm{C} \cdot \mathrm{g}\), calculate the final temperature of the mixture.

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