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Chapter 6: Problem 112

In a bomb calorimeter, the reaction vessel is surrounded by water that must be added for each experiment. Since the amount of water is not constant from experiment to experiment, the mass of water must be measured in each case. The heat capacity of the calorimeter is broken down into two parts: the water and the calorimeter components. If a calorimeter contains 1.00 \(\mathrm{kg}\) water and has a total heat capacity of \(10.84 \mathrm{kJ} / \mathrm{C},\) what is the heat capacity of the calorimeter components?

Short Answer

Expert verified
Heat capacity (water) = \(4.18 \frac{kJ}{C}\) #tag_title# Step 2: Calculate the heat capacity of calorimeter components #tag_content# To find the heat capacity of the calorimeter components, subtract the heat capacity of water from the total heat capacity given: Heat capacity (calorimeter components) = Total heat capacity - Heat capacity (water) Heat capacity (calorimeter components) = \(10.84 \frac{kJ}{C} - 4.18 \frac{kJ}{C}\) Heat capacity (calorimeter components) = \(6.66 \frac{kJ}{C}\)

Step by step solution

01

Calculate the heat capacity of water

To calculate the heat capacity of water, we will use the mass of water and the specific heat capacity of water. The specific heat capacity of water is constant and equal to \(4.18 \frac{kJ}{kg \cdot C}\). The formula to find the heat capacity of water is: Heat capacity (water) = Mass (water) × Specific heat capacity (water) Heat capacity (water) = \(1.00 kg \times 4.18 \frac{kJ}{kg \cdot C}\)

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Most popular questions from this chapter

In which of the following systems is (are) work done by the surroundings on the system? Assume pressure and temperature are constant. a. $2 \operatorname{SO}_{2}(g)+\mathrm{O}_{2}(g) \longrightarrow 2 \operatorname{SO}_{3}(g)$ b. \(\mathrm{CO}_{2}(s) \longrightarrow \mathrm{CO}_{2}(g)\) c. $4 \mathrm{NH}_{3}(g)+7 \mathrm{O}_{2}(g) \longrightarrow 4 \mathrm{NO}_{2}(g)+6 \mathrm{H}_{2} \mathrm{O}(g)$ d. \(\mathrm{N}_{2} \mathrm{O}_{4}(g) \longrightarrow 2 \mathrm{NO}_{2}(g)\) e. \(\mathrm{CaCO}_{3}(s) \longrightarrow \mathrm{CaCO}(s)+\mathrm{CO}_{2}(g)\)

Combustion of table sugar produces \(\mathrm{CO}_{2}(g)\) and $\mathrm{H}_{2} \mathrm{O}(l) .$ When 1.46 \(\mathrm{g}\) table sugar is combusted in a constant-volume (bomb) calorimeter, 24.00 \(\mathrm{kJ}\) of heat is liberated. a. Assuming that table sugar is pure sucrose, $\mathrm{C}_{12} \mathrm{H}_{22} \mathrm{O}_{11}(s)$ write the balanced equation for the combustion reaction. b. Calculate \(\Delta E\) in $\mathrm{kJ} / \mathrm{mol} \mathrm{C}_{12} \mathrm{H}_{22} \mathrm{O}_{11}$ for the combustion reaction of sucrose.

The best solar panels currently available are about 19\(\%\) efficient in converting sunlight to electricity. A typical home will use about \(40 .\) kWh of electricity per day \((1 \mathrm{kWh}=1 \text { kilowatt }\) hour; \(1 \mathrm{kW}=1000 \mathrm{J} / \mathrm{s}\) ). Assuming 8.0 hours of useful sunlight per day, calculate the minimum solar panel surface area necessary to provide all of a typical home's electricity. (See Exercise 132 for the energy rate supplied by the sun.)

Calculate \(\Delta H\) for the reaction: $$ 2 \mathrm{NH}_{3}(g)+\frac{1}{2} \mathrm{O}_{2}(g) \longrightarrow \mathrm{N}_{2} \mathrm{H}_{4}(l)+\mathrm{H}_{2} \mathrm{O}(l) $$ given the following data: $$ 2 \mathrm{NH}_{3}(g)+3 \mathrm{N}_{2} \mathrm{O}(g) \longrightarrow 4 \mathrm{N}_{2}(g)+3 \mathrm{H}_{2} \mathrm{O}(l) $$ \(\Delta H=-1010 . \mathrm{kJ}\) $$ \mathrm{N}_{2} \mathrm{O}(g)+3 \mathrm{H}_{2}(g) \longrightarrow \mathrm{N}_{2} \mathrm{H}_{4}(l)+\mathrm{H}_{2} \mathrm{O}(l) $$ \(\Delta H=-317 \mathrm{kJ}\) $$ \mathrm{N}_{2} \mathrm{H}_{4}(l)+\mathrm{O}_{2}(g) \longrightarrow \mathrm{N}_{2}(g)+2 \mathrm{H}_{2} \mathrm{O}(l) $$ \(\Delta H=-623 \mathrm{kJ}\) $$ \mathrm{H}_{2}(g)+\frac{1}{2} \mathrm{O}_{2}(g) \longrightarrow \mathrm{H}_{2} \mathrm{O}(l) $$ \(\Delta H=-286 \mathrm{kJ}\)

For the process $\mathrm{H}_{2} \mathrm{O}(l) \longrightarrow \mathrm{H}_{2} \mathrm{O}(g)\( at 298 \)\mathrm{K}\( and \)1.0 \mathrm{atm},\( \)\Delta H$ is more positive than \(\Delta E\) by 2.5 \(\mathrm{kJ} / \mathrm{mol}\) . What does the 2.5 \(\mathrm{kJ} / \mathrm{mol}\) quantity represent?
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