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An air conditioner removes 4 kW of heat from a room, using 1.25 kW of electricity.

a.) What is the coefficient of performance of this air conditioner?

b.) How many kW of heat does this air conditioner transfer to the air outside the room?

Answer :

The coefficient of performance (COP) of an air conditioner can be calculated by dividing the amount of heat removed from the room by the amount of electricity consumed.



a.) To find the coefficient of performance (COP), we divide the heat removed from the room (4 kW) by the electricity consumed (1.25 kW).

COP = heat removed / electricity consumed
COP = 4 kW / 1.25 kW
COP = 3.2

So, the coefficient of performance of this air conditioner is 3.2.

b.) The air conditioner transfers heat from the room to the air outside. In this case, the air conditioner removes 4 kW of heat from the room.

Therefore, the air conditioner transfers 4 kW of heat to the air outside the room.

Learn more about coefficient of performance at https://brainly.com/question/31743908

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