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).
Explanation:
a) Air conditioners are applications of refrigeration cycles, whose performance is analyzed by means of the Coefficient of Performance ([tex]COP_{R}[/tex]), dimensionless:
[tex]COP_{R} = \frac{\dot Q_{L}}{\dot W}[/tex] (Eq. 1)
Where:
[tex]\dot Q_{L}[/tex] - Heat removal rate from the house, measured in kilowatts.
[tex]\dot W[/tex]- Electric power, measured in kilowatts.
If we know that [tex]\dot W = 5.10\,kW[/tex] and [tex]\dot Q_{L} = 11.667\,kW[/tex], the Coefficient of Performance of this air conditioner is:
[tex]COP_{R} = \frac{11.667\,kW}{5.10\,kW}[/tex]
[tex]COP_{R} = 2.288[/tex]
The Coefficient of Performance of this air conditioner is 2.288.
b) We find the rate of heat transfer to the outside air ([tex]\dot Q_{H}[/tex]), measured in kilowatts, by applying the First Law of Thermodynamics:
[tex]\dot Q_{H} = \dot Q_{L}+\dot W[/tex] (Eq. 2)
If we get that [tex]\dot W = 5.10\,kW[/tex] and [tex]\dot Q_{L} = 11.667\,kW[/tex], then the rate of heat transfer to the outside air is:
[tex]\dot Q_{H} = 11.667\,kW+5.10\,kW[/tex]
[tex]\dot Q_{H} = 16.767\,kW[/tex]
The rate of heat transfer to the outside air is 16.767 kilowatts (1006.02 kilojoules per minute).
