Answer:
answered
Explanation:
The maximum possible efficiency is
[tex]\eta= 1-\frac{T_L}{T_H}[/tex]
[tex]\eta= 1-\frac{295}{1175}[/tex]
= 0.7489
W_out= 28 KW
Q_in= 15000/3600= 41.67 KW
now
[tex]\eta= \frac{W_out}{Q_H}[/tex]
[tex]\eta= \frac{28}{41.67}[/tex] =0.6791
which is less than [tex]\eta_{max}[/tex]
claim is true