A refrigerator operates much like a heat pump, but the desired goal of a refrige

Question

A refrigerator operates much like a heat pump, but the desired goal of a refrigerator is to transfer heat from the cold reservoir as efficiently as possible. For this reason, the coefficient of performance of a refrigerator is measured as Qc/W. A typical commercial refrigerator has a coefficient of performance of about 5.5. Assume such a refrigerator is used by a market to make ice cubes from 2.3 kg of liquid water, starting at 0°C. (a) How much work is done on the refrigerator to freeze this water into ice? (The latent heat of fusion for the water is 3.34×105 J/kg.) (b) Suppose water costs 0.56 cents per kg, and energy costs 10 cents per kWh. If the market charges 99 cents for 2.3 kg of ice, what is its maximum profit on the ice? Express your answer to the nearest cent.

Explanation / Answer

Q = heat removed from ice = 2.3 * 3.34 * 10E5 = 7.68 * 10E5 J W = work done to remove heat = 7.68 * 10E5 / 5.5 = 1.4 * 10E5 J 1 kW-hr = 1000 * 3600 = 3.6 * 10E6 J Cost to freeze = (1.4 / 36) * 10 = .4 cents Cost of water = 2.3 * .56 = 1.29 cents Total cost = 1.69 Sales price = 2.3 * .99 = 228 cents Profit = 226 cents = $2.26

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