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Chapter 15: Problem 73

A reversible heat engine has an efficiency of \(33.3 \%\) removing heat from a hot reservoir and rejecting heat to a cold reservoir at $0^{\circ} \mathrm{C} .$ If the engine now operates in reverse, how long would it take to freeze \(1.0 \mathrm{kg}\) of water at \(0^{\circ} \mathrm{C},\) if it operates on a power of \(186 \mathrm{W} ?\)

Short Answer

Expert verified
Answer: The heat engine operating in reverse would take approximately 897.40 seconds or around 14.96 minutes to freeze 1.0 kg of water at 0°C.

Step by step solution

01

Calculate the heat removed from the cold reservoir

Since the efficiency of the heat engine is 33.3%, the total energy input is \(E_{in} = \frac{186 \mathrm{W}}{0.333} = 558 \mathrm{W}\). Now, since we are working with a reversible heat engine, we know that the fraction of heat removed from the cold reservoir is 1 - efficiency. Therefore, the heat removed from the cold reservoir is \(Q_c = (1 - 0.333)E_{in} = 0.667 \times 558 \mathrm{W} = 372.06 \mathrm{W}\).
02

Determine the heat energy needed to freeze the water

Now, we need to determine the heat energy required to freeze 1.0 kg of water. To freeze the water, the energy required is the product of the mass of water (m), specific heat of water, and the latent heat of fusion for water (L): $$Q = m \times L = 1.0 \mathrm{kg} \times 334 \mathrm{ kJ/kg} = 334000 \mathrm{J}$$.
03

Calculate the time needed to freeze the water

Finally, to find the time it would take to freeze the water at the given power, we can divide the required heat energy by the heat removed from the cold reservoir: $$t = \frac{Q}{Q_c} = \frac{334000 \mathrm{J}}{372.06 \mathrm{W}} = \frac{334000 \mathrm{J}}{372.06 \frac{\mathrm{J}}{\mathrm{s}}} = 897.40 \mathrm{s}$$ The heat engine operating in reverse would take approximately 897.40 seconds or around 14.96 minutes to freeze 1.0 kg of water at 0°C.

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Most popular questions from this chapter

(a) What is the entropy change of 1.00 mol of \(\mathrm{H}_{2} \mathrm{O}\) when it changes from ice to water at \(0.0^{\circ} \mathrm{C} ?\) (b) If the ice is in contact with an environment at a temperature of \(10.0^{\circ} \mathrm{C}\) what is the entropy change of the universe when the ice melts?
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 \(0^{\circ} \mathrm{C} ?\)
The internal energy of a system increases by 400 J while \(500 \mathrm{J}\) of work are performed on it. What was the heat flow into or out of the system?
An air conditioner whose coefficient of performance is 2.00 removes $1.73 \times 10^{8} \mathrm{J}$ of heat from a room per day. How much does it cost to run the air conditioning unit per day if electricity costs \(\$ 0.10\) per kilowatt-hour? (Note that 1 kilowatt-hour \(=3.6 \times 10^{6} \mathrm{J} .\) )
List these in order of increasing entropy: (a) \(0.5 \mathrm{kg}\) of ice and \(0.5 \mathrm{kg}\) of (liquid) water at \(0^{\circ} \mathrm{C} ;\) (b) $1 \mathrm{kg}\( of ice at \)0^{\circ} \mathrm{C} ;\( (c) \)1 \mathrm{kg}$ of (liquid) water at \(0^{\circ} \mathrm{C} ;\) (d) \(1 \mathrm{kg}\) of water at \(20^{\circ} \mathrm{C}\).
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