Answer:
(A) Power output will be 5.55 KW (b) lower temperature will be 315 K
Step-by-step explanation:
We have given efficiency of heat engine
= 0.37
Input power = 15 KW
Temperature of heat reservoir

(A) We know that
![\eta =(output)/(input)]()
So [text]0.37=\frac{output}{15}[text]
Output = 5.55 KW
(B) We also know that [text]\eta =1-\frac{T_L}{T_H}0.37=\frac{output}{15}[text], here
is lower temperature and
is higher temperature
So


