The structure of a house is such that it loses heat at a rate of 4500 kJ/h per C difference between the indoors and outdoors. A heat pump that requires a power input of 5.000 kW is used to maintain this house at 24C. Determine the lowest outdoor temperature for which the heat pump can meet the heating requirements of this house. (Include a minus sign if required.) The lowest outdoor temperature for which the heat pump can meet the heating requirements of this house is C. \fSol'n ? Find the coefficient of Performany : COP : Q H Win 1:25 KW COP = 5. 00 kxWv Cop = Find the Lowest temperature using cop : TH Cops TH - temperature of TH- TL higher energy TL - temperative of THE 240 +273.15 lower energy. TA = 297- 15 k must be absolute TH TH- TL COP(TA- TL ) - TH COP(TA) - COP (TL) = TH