Solve Thermodynamics Help: Refrigerator Cost/Month

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SUMMARY

The discussion focuses on calculating the monthly electricity cost of a household refrigerator with a coefficient of performance (COP) of 2.1, which removes energy at a rate of 600 Btu/h. The key equations used are the coefficient of performance formula, COP = Q_in / W_cycle, and the work cycle equation, W_cycle = Q_out - Q_in. The user correctly identifies Q_out as 600 Btu/h but struggles to find Q_in and understand the implications of a COP greater than 1, clarifying that COP is distinct from thermal efficiency.

PREREQUISITES
  • Understanding of thermodynamic principles, specifically the coefficient of performance.
  • Familiarity with energy units, particularly Btu and kW.
  • Basic knowledge of work and energy equations in thermodynamics.
  • Ability to perform unit conversions between Btu and kW·h.
NEXT STEPS
  • Study the concept of coefficient of performance in refrigeration systems.
  • Learn how to convert Btu to kW·h for energy cost calculations.
  • Explore the differences between thermal efficiency and coefficient of performance.
  • Investigate real-world applications of thermodynamic principles in household appliances.
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Students studying thermodynamics, engineers working with refrigeration systems, and anyone interested in understanding energy efficiency in household appliances.

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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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