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

(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
`\eta =37%` = 0.37

Input power = 15 KW

Temperature of heat reservoir
`T_H=500K`

(A) We know that

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-(T_L)/(T_H)`

`0.37=1-(T_L)/(500)`

`T_L=315K`