Answer:
a) [tex]\eta = 42\,\%[/tex], b) [tex]COP_{R} = 29[/tex]
Explanation:
a) The thermal efficiency is:
[tex]\eta = \frac{\dot Q_{in} - \dot Q_{out}}{\dot Q_{in}}\times 100\,\%[/tex]
[tex]\eta = \frac{1\,MW-0.58\,MW}{1\,MW} \,\times 100\,\%[/tex]
[tex]\eta = 42\,\%[/tex]
b) The coefficient of performance is:
[tex]COP_{R} = \frac{\dot Q_{L}}{\dot W}[/tex]
[tex]COP_{R} = \frac{0.58\,MW}{0.02\,MW}[/tex]
[tex]COP_{R} = 29[/tex]
