Question

What is the efficiency (w/qh) of an ideal carnot heat engine operating between a hot region at t= 400 k and a cold one at t= 300 k?

Answer 1

The efficiency of an ideal Carnot heat engine can be written as:

`\eta = 1- (T_(cold))/(T_(hot))`
where

`T_(cold)` is the temperature of the cold region

`T_(hot)` is the temperature of the hot region

For the engine in our problem, we have
`T_(cold)=300 K` and
`T_(hot)=400 K`, so the efficiency is

`\eta= 1 - (300 K)/(400 K)=0.25`