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Question
Hello,

I'm really struggling with this question and I allexperts was my last resort. So I know I have use k = heat delivered / net work input and conservation of energy, but I just have no idea how to proceed. Any help is GREATLY appreciated.

A heat pump delivers heat at a rate of 7.10 kW for 12.0 h. If its coefficient of performance is 6.80, how much heat is taken from the cold reservoir during that time?

Answer
If the pump operates for 12 hours the amount of heat delivered Qout will be equal to:
Qout = power x time = 7.10kW x 12hr = 7100 J/s x 12 * 3600s = 3.07 x 10^8J
The coefficient of performance of 6.8 means that for every Joule of work done by the heat pump 6.8 Joules of heat are delivered to the desired location. Since the pump supplies 1 Joule of work of the 6.8 Joules total an additional 5.8 Joules must be supplied by the cold reservoir!
This means that 1/6.8=14.7% of the energy delivered comes from the heat pump while 5.8/6.8=85.3% of the heat comes from the cold reservoir!
Therefore, 85.3% of the delivered energy comes from the cold reservior:
Qin = 85.3% of 3.07 x 10^8J = 2.62 x 10^8J

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James J. Kovalcin

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I am teaching or have taught AP physics B and C [calculus based mechanics & electricity and magnetism] as well as Lab Physics for college bound students. I have a BS in Physics from the University of Pittsburgh and a Master of Arts in Teaching from same. I have been teaching physics for 34 years. I am constantly updating my skills and have a particular interest in modern physics topics.

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