refrigerators-2?

This is an easy question on heat pumps. Let me show you the steps

1)
output power of heat pump = heat pumping ability
Hence, output power = 6000Btu/h = 6000 x 0.293 watts = 1758W = 1.758kW

2)
coefficient of performance = output power / input power = 1758 / 1000 = 1.758 (**Note CoP is always larger than one in normal cases)

3)
Assuming density of air is taken as 1.2kg/m^3 and the specific heat capacity of air is 1.027kJ/kg/degC. Also note that 70degF = 21.1degC; 90degF = 32.2degC

sensible heat capacity = air flow rate x 1.232 x (32.2 - 21.1) = 1.758 (i.e. 6000Btu/h)

Hence, air flow rate = 1.758 / (1.232 x 11.1) = 1.758 / 13.6752 = 0.129 m^3/s

ANSWER:
coefficient of performance = 1.758
air flow rate to discharge heat = 0.129 m^3/s

THank you for reading