Answer
(a) The coefficient of performance (COP) of a refrigerator or freezer is given by the ratio of the heat removed from the cold reservoir (the interior of the freezer) to the work done on the system:
COP = Qc / W
where Qc is the heat removed from the cold reservoir and W is the work done on the system. Since the freezer is operating at steady state, the rate of heat removal from the cold reservoir is equal to the rate of heat discharge to the surrounding environment:
Qc = Qh
where Qh is the rate of heat discharge. Therefore,
COP = Qc / W = Qh / W
Substituting the given values, we get
COP = (Qh / t) / W = (36°C - (-17°C)) / (273°C + 36°C) / (27.3 kW) = 0.127
Therefore, the freezer's COP is 0.127, assuming it is reversible.
(b) The amount of heat required to turn one kilogram of water at 0°C into ice at 0°C is called the latent heat of fusion of water, and is equal to 334 kJ/kg. The rate at which the freezer removes heat from its interior is given by the electrical power consumed, which is 27.3 kW. Therefore, in one hour (3600 seconds), the amount of heat removed from the interior is:
Qc = Wt = (27.3 kW) (3600 s) = 98100 kJ
The amount of water that can be turned to ice in one hour is equal to the amount of heat removed from the interior, divided by the latent heat of fusion:
m = Qc / L = 98100 kJ / 334 kJ/kg = 293 kg
Therefore, the unit can turn 293 kg of water, initially at 0°C, into ice at 0°C in one hour.
Work Step by Step
(a) The coefficient of performance (COP) of a refrigerator or freezer is given by the ratio of the heat removed from the cold reservoir (the interior of the freezer) to the work done on the system:
COP = Qc / W
where Qc is the heat removed from the cold reservoir and W is the work done on the system. Since the freezer is operating at steady state, the rate of heat removal from the cold reservoir is equal to the rate of heat discharge to the surrounding environment:
Qc = Qh
where Qh is the rate of heat discharge. Therefore,
COP = Qc / W = Qh / W
Substituting the given values, we get
COP = (Qh / t) / W = (36°C - (-17°C)) / (273°C + 36°C) / (27.3 kW) = 0.127
Therefore, the freezer's COP is 0.127, assuming it is reversible.
(b) The amount of heat required to turn one kilogram of water at 0°C into ice at 0°C is called the latent heat of fusion of water, and is equal to 334 kJ/kg. The rate at which the freezer removes heat from its interior is given by the electrical power consumed, which is 27.3 kW. Therefore, in one hour (3600 seconds), the amount of heat removed from the interior is:
Qc = Wt = (27.3 kW) (3600 s) = 98100 kJ
The amount of water that can be turned to ice in one hour is equal to the amount of heat removed from the interior, divided by the latent heat of fusion:
m = Qc / L = 98100 kJ / 334 kJ/kg = 293 kg
Therefore, the unit can turn 293 kg of water, initially at 0°C, into ice at 0°C in one hour.
