Answer:
Wnet, in, = 133.33J
Explanation:
Given that
Pump heat QH = 1000J
Warm temperature TH= 300K
Cold temperature TL= 260K
Since the heat pump is completely reversible, the combination of coefficient of performance expression is given as,
From first law of thermodynamics,
COP(HP, rev) = 1/(1-TL/TH)
COP(HP, rev) = 1/(1-260/300)
COP(HP, rev) = 1/(1-0.867)
COP(HP, rev) = 1/0.133
COP(HP, rev) = 7.5
The power required to drive the the heat pump is given as
Wnet, in= QH/COP(HP, rev)
Wnet, in = 1000/7.5
Wnet, in = 133.333J. QED
So the 133.33J was the amount heat that was originally work consumed in the transfer.
Extra....
According to the first law, the rate at which heat is removed from the low temperature reservoir is given as
QL=QH-Wnet, in
QL=1000-133.333
QL=866.67J
