A heat pump is used to heat a house in Minnesota in the winter and has half the ideal coefficient of performance for a reversible heat pump. If the outside air temperature is −15°C and the indoor temperature desired is 20°C, and if it inputs work at a rate of 1,250 W, what is the rate of heat flow into the house?

Question

Quizplus · Accepted Answer

The coefficient of performance (COP) for a heat pump is given by $COP = \frac{Q_H}{W}$, where $Q_H$ is the heat delivered to the house and $W$ is the work input. For a reversible heat pump operating between two temperatures, the ideal COP is given by $COP_{ideal} = \frac{T_H}{T_H - T_L}$, where $T_H$ and $T_L$ are the indoor and outdoor temperatures in Kelvin, respectively. Given that the heat pump has half the ideal COP, we first calculate the ideal COP with $T_H = 293K$ (20°C) and $T_L = 258K$ (-15°C), then divide by 2 for the actual COP, and finally solve for $Q_H$.$$COP_{ideal} = \frac{293}{293 - 258} = \frac{293}{35}$$$$COP_{actual} = \frac{1}{2} \times COP_{ideal} = \frac{1}{2} \times \frac{293}{35}$$Given $W = 1250W$,$$Q_H = COP_{actual} \times W = \frac{1}{2} \times \frac{293}{35} \times 1250$$$$Q_H \approx 5230W$$Therefore, the rate of heat flow into the house is approximately 5,230 W.

A Heat Pump Is Used to Heat a House in Minnesota