Answer:
[tex]Q'_H=1065kJ/min[/tex]
Explanation:
From the question we are told that
Heat Remove rate Q_o= 750 kJ/min
Power Draw P=5.25
a)Generally the coefficient of performance (COP) of an air conditioner is mathematically given as
[tex]COP=\frac{Q_o}{P}[/tex]
[tex]COP=\frac{750kJ/min}{5.25kw}[/tex]
Where
[tex]750kJ/min=12.5kJ/sec[/tex]
[tex]COP=\frac{12.5kJ/sec}{5.25kw}[/tex]
[tex]COP=2.38[/tex]
b)Generally the equation for power is mathematically represented as
[tex]P=Q_H-Q_o[/tex]
[tex]P+Q_o=Q_H[/tex]
[tex]Q_H=P+Q_o[/tex]
[tex]Q_H=5.25+12.5[/tex]
[tex]Q_H=17.75kW[/tex]
Generally rate of heat transfer to the outside air is mathematically given as
[tex]Q'_H=17.75*60[/tex]
[tex]Q'_H=1065kJ/min[/tex]
