Answer:
a) [tex]\dot Q_{L} = 4987.776\,\frac{kJ}{min}[/tex], b) [tex]\dot Q = 5787.776\,\frac{kJ}{min}[/tex]
Explanation:
a) The efficiency of the Carnot heat engine is:
[tex]\eta_{th} = \left(1-\frac{T_{L}}{T_{H}} \right)\times 100\,\%[/tex]
[tex]\eta_{th} = \left(1-\frac{300.15\,K}{1173.15\,K} \right)\times 100\,\%[/tex]
[tex]\eta_{th} = 74.415\,\%[/tex]
The power output used to drive the refrigerator is:
[tex]\dot W = (0.744)\cdot \left(800\,\frac{kJ}{min}\right)\cdot \left(\frac{1}{60}\,\frac{min}{s} \right)[/tex]
[tex]\dot W = 9.92\,kW[/tex]
The coefficient of performance of the refrigerator is:
[tex]COP_{R} = \frac{T_{L}}{T_{H}-T_{L}}[/tex]
[tex]COP_{R} = \frac{268.15\,K}{300.15\,K-268.15\,K}[/tex]
[tex]COP_{R} = 8.380[/tex]
The heat removal from the refrigerated space is:
[tex]\dot Q_{L} = \dot W\cdot COP_{R}[/tex]
[tex]\dot Q_{L} = (9.92\,kW)\cdot \left(60\,\frac{s}{min} \right)\cdot (8.380)[/tex]
[tex]\dot Q_{L} = 4987.776\,\frac{kJ}{min}[/tex]
b) The heat rejected from the Carnot heat engine to the ambient air is:
[tex]\dot Q_{L} = (1-0.744)\cdot \left(800\,\frac{kJ}{min} \right)[/tex]
[tex]\dot Q_{L} = 204.8\,\frac{kJ}{min}[/tex]
Now, the heat rejected from the refrigerator to the ambient air is:
[tex]\dot Q_{H} = \dot W + \dot Q_{L}[/tex]
[tex]\dot Q_{H} = (9.92\,kW)\cdot (60\,\frac{s}{min} ) + 4987.776\,\frac{kJ}{min}[/tex]
[tex]\dot Q_{H} = 5582.976\,\frac{kJ}{min}[/tex]
Finally, the total rate of heat rejection to the ambient air is:
[tex]\dot Q = 5582.976\,\frac{kJ}{min} + 204.8\,\frac{kJ}{min}[/tex]
[tex]\dot Q = 5787.776\,\frac{kJ}{min}[/tex]
