A heat pump delivers heat at a rate of \(7.81 \mathrm{kW}\) for $10.0 \mathrm{h} .\( If its coefficient of performance is \)6.85,$ how much heat is taken from the cold reservoir during that time?

Short Answer

Expert verified
Answer: Approximately 1,200.6 kWh.

Step by step solution

01

Determine the work input

The formula for the coefficient of performance (COP) of a heat pump is given by: COP = Q_H / W where, COP = Coefficient of performance (6.85) Q_H = Amount of heat delivered by the heat pump (7.81 kW) W = Work input Let's find the amount of work input, using this formula: W = Q_H / COP

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