High School

We appreciate your visit to A refrigerator accepts heat at a rate of 10 kW and consumes power work at a rate of 5 kW Determine the coefficient of performance. This page offers clear insights and highlights the essential aspects of the topic. Our goal is to provide a helpful and engaging learning experience. Explore the content and find the answers you need!

A refrigerator accepts heat at a rate of 10 kW and consumes power (work) at a rate of 5 kW. Determine the coefficient of performance (COP) of the refrigerator.

1) 2.00
2) 0.50
3) 15 kW

Answer :

The coefficient of performance (COP) of the refrigerator, which accepts heat at a rate of 10 kW and consumes power at a rate of 5 kW, is 2.00. Therefore the correct answer is 1) 2.00.

The COP of a refrigerator is defined as the amount of heat removed from the cold reservoir Qc divided by the work W done to remove that heat, which is COPref = Qc/W. Given that the refrigerator accepts heat at a rate of 10 kW (this is Qc) and consumes power at a rate of 5 kW (this is W), the COP can be calculated as follows:

COPref = Qc / W = 10 kW / 5 kW = 2.0

The coefficient of performance or COP of a heat pump, refrigerator or air conditioning system is a ratio of useful heating or cooling provided to work required. Higher COPs equate to lower operating costs. The Co-efficient of performance (COP) is an expression of the efficiency of a heat pump. When calculating the COP for a heat pump, the heat output from the condenser (Q) is compared to the power supplied to the compressor (W). COP = |Q|. Therefore the correct answer is 1) 2.00.

Thanks for taking the time to read A refrigerator accepts heat at a rate of 10 kW and consumes power work at a rate of 5 kW Determine the coefficient of performance. We hope the insights shared have been valuable and enhanced your understanding of the topic. Don�t hesitate to browse our website for more informative and engaging content!

Rewritten by : Barada