Answer:
a) The Coefficient of Performance of this air conditioner is 2.288.
b) The rate of heat transfer to the outside air is 16.767 kilowatts (1006.02 kilojoules per minute).
Step-by-step explanation:
a) Air conditioners are applications of refrigeration cycles, whose performance is analyzed by means of the Coefficient of Performance (
), dimensionless:
(Eq. 1)
Where:
- Heat removal rate from the house, measured in kilowatts.
- Electric power, measured in kilowatts.
If we know that
and
, the Coefficient of Performance of this air conditioner is:


The Coefficient of Performance of this air conditioner is 2.288.
b) We find the rate of heat transfer to the outside air (
), measured in kilowatts, by applying the First Law of Thermodynamics:
(Eq. 2)
If we get that
and
, then the rate of heat transfer to the outside air is:


The rate of heat transfer to the outside air is 16.767 kilowatts (1006.02 kilojoules per minute).