Question

an air conditioner removes heat steadily form a house at a rate of 850 kj/min while drawing electric power at a rate of 6 kW. determine the COP of this air conditioner

Answer 1

The coefficient of performance of the air-conditioner can be calculated as follows:

$$COP_R=\dfrac{Q_L}{W_{net, in}},$$

here, $Q_L$is the total heat removed by the air conditioner, $W_{net, in}$is the power input into the air conditioner.

Then, we get:

$$COP_R=\dfrac{850\ \dfrac{kJ}{min}\cdot\dfrac{1000\ J}{1\ kJ}\cdot\dfrac{1\ min}{60\ s}}{6\ kW\cdot\dfrac{1000\ W}{1\ kW}}=2.36$$

Answer:

$COP_R=2.36$