91Ó°ÊÓ

Three Carnot engines operate between the temperature limits of

1. 400 and 500K
2. 500 and 600K
3. 400 and 600K

We can use the concept of entropy for Carnot, real, and perfect engines. Using the efficiency of the Carnot cycle we can get the magnitudes of the work done per cycle between each case of the temperature limits.

Formulae:

The efficiency of the Carnot cycle, ${\mathrm{Î\mu }}_c=1-\frac{T_L}{T_H}or{\mathrm{Î\mu }}_c=\frac{\left|W\right|}{\left|Q_H\right|}$ …(¾±)

An engine is a device that, operating in a cycle, extracts energy as heat$\left|Q_H\right|$from a high temperature reservoir and does certain amount of work $\left|W\right|$.

Thus, using equation (i), we can get the magnitude of work done per cycle as follows

$\left|W\right|=\left(1-\frac{T_L}{T_H}\right)\left|Q_H\right|$ …(¾±¾±)

The amount of work done is directly proportional to the efficiency of the engine.

For case (a): Given temperature limits are 400 and 500K

Thus, the efficiency of the cycle for these limits is given using equation (ii) as follows:

${\mathrm{Î\mu }}_c=1-\frac{400\mathrm{K}}{500\mathrm{K}}$

For case (b): Given temperature limits are 500 and 600K

Thus, the efficiency of the cycle for these limits is given using equation (ii) as follows:

role="math" localid="1661318902333" $$\begin{gathered} {\mathrm{Î\mu }}_c=1-\frac{500\mathrm{K}}{600\mathrm{K}} \\ =0.17 \end{gathered}$$

For case (c): Given temperature limits are 400 and 600K

Thus, the efficiency of the cycle for these limits is given using equation (ii) as follows:

$$\begin{gathered} {\mathrm{Î\mu }}_c=1-\frac{400\mathrm{K}}{600\mathrm{K}} \\ =0.33 \end{gathered}$$

Hence, the ranking according to the magnitudes of the work done by the engines per cycle, greatest first is c) > a) > b).