Answer
The required Power 0.46watt
Step-by-step explanation:
In the reverse Carnot cycle, work is done to extract heat from one system and expel it into another via four processes, two isothermal and two isentropic.
Given data
Given that the heat transfer from the house
Q=750kj/min
Temperature of the house T1=24°C to Kelvin 24+273=297k
Outdoor temperature T2=35°c
To Kelvin =35+273=308k
We are assuming a steady state operation
Hence the coefficient of performance c.o.p for a reversed Carnot cycle operating between two temperatures is
C.O.P= 1/(T2/T1) - 1
=1(308/297)-1
=1/1.037-1
=1/0.037
COP=27.02
The expression for net work done
Is given as
Wnet=Quantity of heat transfer/COP
Wnet=750/27.02
Work =27.75j/min
To get the power let's us divide the work done by 60min
Power =27.75/60
Power =0.46watt