Answer:
(A) Power output will be 5.55 KW (b) lower temperature will be 315 K
Explanation:
We have given efficiency of heat engine [tex]\eta =37%[/tex] = 0.37
Input power = 15 KW
Temperature of heat reservoir [tex]T_H=500K[/tex]
(A) We know that [tex]\eta =\frac{output}{input}[/tex]
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 [tex]T_L[/tex] is lower temperature and [tex]T_H[/tex] is higher temperature
So [tex]0.37=1-\frac{T_L}{T_H}[/tex]
[tex]0.37=1-\frac{T_L}{500}[/tex]
[tex]T_L=315K[/tex]
