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

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The discussion focuses on the Coefficient of Performance (COP) of a refrigeration unit with a 1 kW compressor that maintains a temperature of -10°C while the ambient temperature is 20°C. The unit claims a refrigeration capacity of 2 kW, which is consistent with thermodynamic principles. The refrigeration capacity indicates the heat transferred from the cold reservoir to the hot reservoir, and the power required for this process can be significantly lower than the heat transfer, as demonstrated by the principles of thermodynamics and the Carnot cycle.

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natski
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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
[tex]Q_{in}=L+Q_{out}[/tex]

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

If the same thermodynamic cycle is CCW (counter-CW), the system takes [tex]L[/tex] from the electrical engine and [tex]|Q_{out}|[/tex] FROM THE COLD RESERVOIR and gives [tex]|L|+|Q_{out}|[/tex] 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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