Answer to Question #300718 in Physics for Lovewilly

Question

A Carnot heat engine with an efficiency of 60% receives heat of 3000kj from a source and reject the waste heat to medium at 300k.Determine the source temperature

Answer 1

Given:

$\eta=0.6$

$T_2=300\:\rm K$

A Carnot heat engine efficiency

$$\eta=1-\frac{T_2}{T_1}$$

$$T_1=T_2/(1-\eta)=300/(1-0.6)=750\:\rm K$$