Question

A heat engine operates between two reservoirs at 300K. During each cycle, it absorbs 200 calories from the high temperature reservoir. Calculate:
(a) The maximum efficiency of the system
(b) The maximum works the engine can perform in each cycle

Answer 1

(a) Zero

The maximum efficiency (Carnot efficiency) of a heat engine is given by

$\eta=1-\frac{T_C}{T_H}$

where

$T_C$ is the low-temperature reservoir

$T_H$ is the high-temperature reservoir

For the heat engine in the problem, we have:

$T_C = 300K$

$T_H = 300K$

Therefore, the maximum efficiency is

$\eta=1-\frac{300}{300}=0$

(b) Zero

The efficiency of a heat engine can also be rewritten as

$\eta = \frac{W}{Q_H}$

where

W is the work performed by the engine

$Q_H$ is the heat absorbed from the high-temperature reservoir

In this problem, we know

$\eta=0$

Therefore, since the term $Q_H$ cannot be equal to infinity, the numerator of the fraction must be zero as well, which means

W = 0

So the engine cannot perform any work.