What is the Coefficient of Performance (COP) for this refigeration unit?

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The discussion centers on the Coefficient of Performance (COP) of a refrigeration unit with a 1 kW compressor that maintains a temperature of -10°C in a 20°C environment, claiming a refrigeration capacity of 2 kW. It is confirmed that this claim is consistent with thermodynamic principles, as the refrigeration capacity indicates the heat removed from the cold space, which can exceed the electrical power input. The conversation highlights the efficiency of heat pumps, where the heat extracted from the cold reservoir can supplement the energy used by the compressor. The use of Carnot cycle principles is mentioned to explain how heat transfer can occur without a direct one-to-one relationship with electrical power consumption. Understanding these thermodynamic concepts is crucial for evaluating refrigeration unit performance.
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I don't know how to get started on this question, what equations should I be using?

A freezer unit has a 1kW compressor, and can maintain the refigerated space at -10 C while the ambient temperature is 20 C. It is claimed that the unit has a refigeration capacity of 2 kW. Is this claim consistent with the laws of thermodynamics?

I just don't recall any equations for power in themodynamics.
Thanks for any help.

natksi
 
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natski said:
A freezer unit has a 1kW compressor, and can maintain the refigerated space at -10 C while the ambient temperature is 20 C. It is claimed that the unit has a refigeration capacity of 2 kW. Is this claim consistent with the laws of thermodynamics?
It is. The refrigeration capacity refers to the amount of heat per unit time that is transferred from the cold reservoir (the air in the freezer) to the hot reservoir (the air outside the freezer). Power (work/time) is required to do this, but the amount of power needed can be much less than the heat transfer. That is why geothermal heating with heat pumps is so popular.

AM
 
Hi natski,

If you have a Carnot engine, then you have
Q_{in}=L+Q_{out}

Because the thermodynamic cycle is CW (clockwise) in PV, the system take Q_{in} from the hot source and gives L and Q_{out} back.

If the same thermodynamic cycle is CCW (counter-CW), the system takes L from the electrical engine and |Q_{out}| FROM THE COLD RESERVOIR and gives |L|+|Q_{out}| to the HOT RESERVOIR. So, you get more heat than the spent electrical power, because a part of your heat comes from OUTSIDE (the cold source) ...and you don't pay for that :biggrin:
 
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