An air conditioner removes heat steadily from a house at a rate of 750 kJ/min while drawing electric power at a rate of 5.25 kW. Determine (a) the COP of this air conditioner and (b) the rate of heat transfer to the outside air. Answers:(a) 2.38, (b) 1065 kJ/min

a) the COP of this air conditioner is 2.38

b) the rate of heat transfer to the outside air is 1065 kJ/min

Explanation:

Given the data in the question;

[ Outdoor ] ←  Q [ W ] Q ← [ House ]

Rate of heat removed from the house; Q  = 750 kJ/min = ( 750 kJ/min × ( 1 kW / 60 kJ/min ) ) = 12.5 kW

Net-work input; W = 5.25 kW

a) The coefficient of performance of the air conditioner; COP.

COP = Q / W

we substitute

COP = 12.5 kW / 5.25 kW

COP = 2.38

Therefore, the COP of this air conditioner is 2.38

b) the rate of heat transfer to the outside air.

Q = Q + W

we substitute

Q = 12.5 kW + 5.25 kW

Q = 17.75 kW

Q = ( 17.75 × 60 ) kJ/min

Q = 1065 kJ/min

Therefore, the rate of heat transfer to the outside air is 1065 kJ/min

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