Solve Thermodynamics Help: Refrigerator Cost/Month

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A household refrigerator with a coefficient of performance of 2.1 removes energy at a rate of 600 Btu/h, leading to confusion about efficiency metrics. The equations for coefficient of performance and work cycle are discussed, with emphasis on calculating Q_in and W_cycle. The user struggles with the concept that a coefficient of performance greater than 1 does not equate to efficiency exceeding 100%. Clarification is provided that thermal efficiency differs from coefficient of performance, which can exceed 1 in refrigeration systems. The discussion aims to resolve the confusion surrounding these thermodynamic principles and their application in calculating monthly electricity costs.
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Homework Statement



A household refrigerator with a coefficient of performance of 2.1 removes energy from the refrigerated space at a rate of 600 Btu/h. Evaluating electricity at $ 0.08 per kW · h determine the cost of electricity in a month when the refrigerator operates for 360 hours.

Homework Equations



coefficient of performance = Q_in / W_cycle
W_cycle = Q_out - Q_in

The Attempt at a Solution



I assumed that Q_out was 600 Btu/hr, since it said that was the energy removed. The coefficient of performance is obviously 2.1, and I tried to find Q_in using the above equations but had some trouble. I also know that you need to find the work and then you can just use the given "$0.08 per kW" to find the money needed, which is the answer. But I am really confused with this one.
 
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I didn't think Q/W could be greater than 1. Since 1 represents 100% effeciency. If an engine had a Q/W value of 2.1, wouldn't it be working at 210% efficiency??
 
thermal efficiency is different than coefficient of performance
 
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