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A freezer has a coefficient of performance of 2.40. The freezer is to convert 1.80 kg of water at to 1.80 kg of ice at in one hour. (a) What amount of heat must be removed from the water at to convert it to ice at ? (b) How much electrical energy is consumed by the freezer during this hour? (c) How much wasted heat is delivered to the room in which the freezer sits?
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We are given a coefficient of performance = 2.40 for a freezer. A mass of water =1.80 kg at temperature convert with the same mass to ice and reach the temperature .
Solution
(a) We want to calculate the removed heat from the water. In this problem, the water has three processes happened. First the water-cooled from temperature to so the heat in this process will be given by
Where is the specific heat capacity of water and equals 4190 (See Table 17.3).
In the second process, the water change from the liquid phase to the solid phase, and the heat will be given by equation 17.20
Where is the heat fusion of water and equals 3.34 J/kg. (See Table 17.4)
The third process, the ice temperature changes from to and the heat in this process will be given by
Where is the specific heat capacity of the ice and equals 2100 (See table 17.3).
Now we can obtain the heat removed from the water by combine the three processes from equations (1), (2) and (3)
The heat removed from the water is
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