Question

(a) If 15 kW of power from a heat reservoir at 500 K is input into a heat engine with an efficiency of 37%, what is the power output? (b) What is the temperature of the cold reservoir?

Answer 1

Answer:

(A) Power output will be 5.55 KW (b) lower temperature will be 315 K

Explanation:

We have given efficiency of heat engine $\eta =37%$ = 0.37

Input power = 15 KW

Temperature of heat reservoir $T_H=500K$

(A) We know that $\eta =\frac{output}{input}$

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 $T_L$  is lower temperature and $T_H$ is higher temperature

So $0.37=1-\frac{T_L}{T_H}$

$0.37=1-\frac{T_L}{500}$

$T_L=315K$