Answer:
No.
Explanation:
The Coefficient of Performance of the reversible heat pump is determined by the Carnot's cycle:
[tex]COP_{HP} = \frac{T_{H}}{T_{H}-T_{L}}[/tex]
[tex]COP_{HP} = \frac{297.15\,K}{297.15\,K-280.15\,K}[/tex]
[tex]COP_{HP} = 3.339[/tex]
The power required to make the heat pump working is:
[tex]\dot W = \frac{300\,kW}{3.339}[/tex]
[tex]\dot W = 89.847\,kW[/tex]
The heat absorbed from the exterior air is:
[tex]\dot Q_{L} = 300\,kW - 89.847\,kW[/tex]
[tex]\dot Q_{L} = 210.153\,kW[/tex]
According to the Second Law of Thermodynamics, the entropy generation rate in a reversible cycle must be zero. The formula for the heat pump is:
[tex]\frac{\dot Q_{L}}{T_{L}} - \frac{\dot Q_{H}}{T_{H}} + \dot S_{gen} = 0[/tex]
[tex]\dot S_{gen} = \frac{\dot Q_{H}}{T_{H}} - \frac{\dot Q_{L}}{T_{L}}[/tex]
[tex]\dot S_{gen} = \frac{300\,kW}{297.15\,K}-\frac{210.153\,kW}{280.15\,K}[/tex]
[tex]\dot S_{gen} = 0.259\,\frac{kW}{K}[/tex]
Which contradicts the reversibility criterion according to the Second Law of Thermodynamics.
