# Thermodynamics help

1. Feb 8, 2009

### msteves

1. The problem statement, all variables and given/known data

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. 2. Relevant equations coefficient of performance = Q_in / W_cycle W_cycle = Q_out - Q_in 3. 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.

2. Feb 9, 2009

### greg136

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??

3. Feb 9, 2009

### msteves

thermal efficiency is different than coefficient of performance