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Chapter 15: Problem 39

A reversible refrigerator has a coefficient of performance of \(3.0 .\) How much work must be done to freeze \(1.0 \mathrm{kg}\) of liquid water initially at \(0^{\circ} \mathrm{C} ?\)

Short Answer

Expert verified
Answer: The work required is approximately 111,333.33 Joules.

Step by step solution

01

Calculate energy required to change the water to ice

To determine the required work, we first need to find the amount of energy that the refrigerator has to extract from the water to turn it into ice. This can be calculated using the formula for latent heat of fusion. For water, the latent heat of fusion (Lf) is approximately 334,000 J/kg. Therefore, the energy required to turn 1 kg of water into ice can be given by: Q_cold = mass * Lf Q_cold = (1 kg) * (334,000 J/kg) = 334,000 J
02

Determine the work done by the refrigerator

Now, we have the energy that needs to be extracted from the water. We can use the formula for the coefficient of performance to determine the work done by the refrigerator: Coefficient_of_performance = Q_cold / W Rearranging, we get: W = Q_cold / Coefficient_of_performance W = 334,000 J / 3.0 W = 111,333.33 J
03

Final answer

So, the work that must be done to freeze 1 kg of water initially at 0°C using a reversible refrigerator with a coefficient of performance of 3.0 is approximately 111,333.33 Joules.

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Most popular questions from this chapter

A heat engine uses the warm air at the ground as the hot reservoir and the cooler air at an altitude of several thousand meters as the cold reservoir. If the warm air is at \(37^{\circ} \mathrm{C}\) and the cold air is at $25^{\circ} \mathrm{C},$ what is the maximum possible efficiency for the engine?
(a) How much heat does an engine with an efficiency of \(33.3 \%\) absorb in order to deliver \(1.00 \mathrm{kJ}\) of work? (b) How much heat is exhausted by the engine?
A large block of copper initially at \(20.0^{\circ} \mathrm{C}\) is placed in a vat of hot water \(\left(80.0^{\circ} \mathrm{C}\right) .\) For the first $1.0 \mathrm{J}$ of heat that flows from the water into the block, find (a) the entropy change of the block, (b) the entropy change of the water, and (c) the entropy change of the universe. Note that the temperatures of the block and water are essentially unchanged by the flow of only \(1.0 \mathrm{J}\) of heat.
Consider a heat engine that is not reversible. The engine uses $1.000 \mathrm{mol}\( of a diatomic ideal gas. In the first step \)(\mathrm{A})$ there is a constant temperature expansion while in contact with a warm reservoir at \(373 \mathrm{K}\) from \(P_{1}=1.55 \times 10^{5} \mathrm{Pa}\) and $V_{1}=2.00 \times 10^{-2} \mathrm{m}^{3}$ to \(P_{2}=1.24 \times 10^{5} \mathrm{Pa}\) and $V_{2}=2.50 \times 10^{-2} \mathrm{m}^{3} .$ Then (B) a heat reservoir at the cooler temperature of \(273 \mathrm{K}\) is used to cool the gas at constant volume to \(273 \mathrm{K}\) from \(P_{2}\) to $P_{3}=0.91 \times 10^{5} \mathrm{Pa} .$ This is followed by (C) a constant temperature compression while still in contact with the cold reservoir at \(273 \mathrm{K}\) from \(P_{3}, V_{2}\) to \(P_{4}=1.01 \times 10^{5} \mathrm{Pa}, V_{1} .\) The final step (D) is heating the gas at constant volume from \(273 \mathrm{K}\) to \(373 \mathrm{K}\) by being in contact with the warm reservoir again, to return from \(P_{4}, V_{1}\) to \(P_{1}, V_{1} .\) Find the change in entropy of the cold reservoir in step \(\mathrm{B}\). Remember that the gas is always in contact with the cold reservoir. (b) What is the change in entropy of the hot reservoir in step D? (c) Using this information, find the change in entropy of the total system of gas plus reservoirs during the whole cycle.
The motor that drives a reversible refrigerator produces \(148 \mathrm{W}\) of useful power. The hot and cold temperatures of the heat reservoirs are \(20.0^{\circ} \mathrm{C}\) and \(-5.0^{\circ} \mathrm{C} .\) What is the maximum amount of ice it can produce in \(2.0 \mathrm{h}\) from water that is initially at \(8.0^{\circ} \mathrm{C} ?\)
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