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Ch.5 - Thermochemistry
Chapter 5, Problem 60c

Under constant-volume conditions, the heat of combustion of naphthalene (C10H8) is 40.18 kJ/g. A 2.50-g sample of naphthalene is burned in a bomb calorimeter. The temperature of the calorimeter increases from 21.50 to 28.83 °C. (c) Suppose that in changing samples, a portion of the water in the calorimeter were lost. In what way, if any, would this change the heat capacity of the calorimeter?

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insert step 1> Understand that the heat capacity of a calorimeter is the amount of heat required to change its temperature by one degree Celsius.
insert step 2> Recognize that the heat capacity is dependent on the mass of the substances within the calorimeter, including water.
insert step 3> Consider that if water is lost from the calorimeter, the total mass of the calorimeter system decreases.
insert step 4> Realize that with less water, the calorimeter would require less heat to achieve the same temperature change, effectively decreasing its heat capacity.
insert step 5> Conclude that losing water from the calorimeter would decrease its overall heat capacity, meaning it would heat up more for the same amount of heat absorbed.

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Key Concepts

Here are the essential concepts you must grasp in order to answer the question correctly.

Heat Capacity

Heat capacity is the amount of heat required to change the temperature of a substance by one degree Celsius. In calorimetry, it is crucial for determining how much heat is absorbed or released during a chemical reaction. The heat capacity of a calorimeter is influenced by its mass and the specific heat of the materials it contains, which in this case includes water.
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Bomb Calorimeter

A bomb calorimeter is a device used to measure the heat of combustion of a substance under constant volume conditions. It consists of a sealed container (the bomb) where the reaction occurs, surrounded by a known quantity of water. The temperature change of the water is used to calculate the heat released during the combustion, which is essential for understanding the energy content of fuels.
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Effect of Water Loss on Heat Capacity

If water is lost from the calorimeter, the total heat capacity of the system decreases because the remaining water can absorb less heat. This change affects the temperature change observed during the combustion reaction, as a smaller heat capacity would lead to a larger temperature increase for the same amount of heat released, potentially skewing the results of the experiment.
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Related Practice
Textbook Question

A 2.20-g sample of phenol (C6H5OH) was burned in a bomb calorimeter whose total heat capacity is 11.90 kJ/°C. The temperature of the calorimeter plus contents increased from 21.50 to 27.50 °C. (b) What is the heat of combustion per mole of phenol?

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Textbook Question

Under constant-volume conditions, the heat of combustion of benzoic acid (C6H5O6) is 15.57 kJ/g. A 3.500-g sample of sucrose is burned in a bomb calorimeter. The temperature of the calorimeter increases from 20.94 to 24.72 °C. (a) What is the total heat capacity of the calorimeter?

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Textbook Question

Under constant-volume conditions, the heat of combustion of benzoic acid (C6H5O6) is 15.57 kJ/g. A 3.500-g sample of sucrose is burned in a bomb calorimeter. The temperature of the calorimeter increases from 20.94 to 24.72 °C. (b) If the size of the sucrose sample had been exactly twice as large, what would the temperature change of the calorimeter have been?

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Open Question
Can you use an approach similar to Hess’s law to calculate the change in internal energy, _x001F_E, for an overall reaction by summing the _x001F_E values of individual reactions that add up to give the desired overall reaction?
Textbook Question

Consider the following hypothetical reactions: A → B ΔHI = +60 kJ B → C ΔHII = -90 kJ (b) Construct an enthalpy diagram for substances A, B, and C, and show how Hess's law applies.

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Textbook Question

Calculate the enthalpy change for the reaction P4O6(s) + 2 O2(g) → P4O10(s) given the following enthalpies of reaction: P4(s) + 3 O2(g) → P4O6(s) ΔH = -1640.1 kJ P4(s) + 5 O2(g) → P4O10(s) ΔH = -2940.1 kJ

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