Answer:
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[tex]_H[/tex] [ W[tex]_{net, in[/tex] ] Q[tex]_L[/tex] ← [ House ]
Rate of heat removed from the house; Q[tex]_L[/tex] = 750 kJ/min = ( 750 kJ/min × ( 1 kW / 60 kJ/min ) ) = 12.5 kW
Net-work input; W[tex]_{net, in[/tex] = 5.25 kW
a) The coefficient of performance of the air conditioner; COP.
COP = Q[tex]_L[/tex] / W[tex]_{net, in[/tex]
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[tex]_H[/tex] = Q[tex]_L[/tex] + W[tex]_{net[/tex]
we substitute
Q[tex]_H[/tex] = 12.5 kW + 5.25 kW
Q[tex]_H[/tex] = 17.75 kW
Q[tex]_H[/tex] = ( 17.75 × 60 ) kJ/min
Q[tex]_H[/tex] = 1065 kJ/min
Therefore, the rate of heat transfer to the outside air is 1065 kJ/min
