Archived Coefficient of performance for a refrigerator

AI Thread Summary
The discussion revolves around calculating the coefficient of performance (COP) for a refrigerator operating at -10°C with an external temperature of 25°C. The formula used is COP = T(low) / [T(high) - T(low)], which yielded a COP of 7.5. However, this value is questioned as being too high, considering that industry standards achieve only about 50% of the Carnot limit. Participants highlight the discrepancy between the calculated Carnot COP and the practical COP limits in real-world applications. The conversation emphasizes the importance of understanding the difference between theoretical calculations and actual performance in refrigeration systems.
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Homework Statement



The COP for heat pumps and refrigerators achieved in the industry is almost 50% of the Carnot limit. Make estimates for the following case:

For refrigerator to achieve -10°C in the freezer area with an external temperature of 25°C

Homework Equations



COP for refrigerator = T(low) / [T(high)-T(low)]

The Attempt at a Solution



Using this equation, I find that the COP is 7.5, but this seems high. Where have I made a mistake?
 
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You calculated the Carnot COP but the problem statement says..

rmjmu507 said:
The COP for heat pumps and refrigerators achieved in the industry is almost 50% of the Carnot limit
 
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