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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 reversible engine with an efficiency of \(30.0 \%\) has $T_{\mathrm{C}}=310.0 \mathrm{K} .\( (a) What is \)T_{\mathrm{H}} ?$ (b) How much heat is exhausted for every \(0.100 \mathrm{kJ}\) of work done?
What is the efficiency of an electric generator that produces $1.17 \mathrm{kW}\(.h per \)\mathrm{kg}$ of coal burned? The heat of combustion of coal is \(6.71 \times 10^{6} \mathrm{J} / \mathrm{kg}\).
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