In the first stage of a two-stage Carnot engine, energy is absorbed as heat ${\mathit{Q}}_{\mathbf{1}}$at temperature ${\mathit{T}}_{\mathbf{1}}$, work ${\mathit{W}}_{\mathbf{1}}$is done, and energy is expelled as heat ${\mathit{Q}}_{\mathbf{2}}$at a lower temperature ${\mathit{T}}_{\mathbf{2}}$. The second stage absorbs that energy as heat ${\mathit{Q}}_{\mathbf{2}}$ does work${\mathit{W}}_{\mathbf{2}}$, and expels energy as heat ${\mathit{Q}}_{\mathbf{3}}$at a still lower temperature ${\mathit{T}}_{\mathbf{3}}$. Prove that the efficiency of the engine is$\left(T_1-T_3\right)\mathbf{/}{\mathit{T}}_{\mathbf{1}}$.

Absorbed energy in the first stage of the Carnot engine is Q₁.

The initial temperature of the first stage of the Carnot engine is T₁.

The energy that is expelled out by the first stage of the Carnot engine is Q₂.

The amount of work done by the first stage of the Carnot engine is W₁.

Absorbed energy in the second stage of the Carnot engine is Q₂.

The initial temperature of the second stage of the Carnot engine is T₂.

The energy that is expelled out by the second t stage of the Carnot engine is Q₃.

The amount of work done by the second stage of the Carnot engine is W₁.

The final temperature of the Carnot engine is T₃.

We can write the efficiency of the Carnot engine in terms of total work done. Then using the relation between heat energy absorbed and expelled out with work we can write it in terms of heat energy. Then using the relation between heat energy and temperature at higher and lower temperature reservoirs we can find the efficiency in terms of those temperatures.

Formulae:

The efficiency of the Carnot engine,

$\mathrm{\epsilon }=\frac{\mathrm{W}}{{\mathrm{Q}}_{\mathrm{H}}}$ (1)

The work done per cycle of the gas,

$\mathrm{W}={\mathrm{Q}}_{\mathrm{H}}-{\mathrm{Q}}_{\mathrm{L}}$ (2)

The heat and temperature relation of a Carnot engine,

$\frac{{\mathrm{Q}}_{\mathrm{H}}}{{\mathrm{Q}}_{\mathrm{L}}}=\frac{{\mathrm{T}}_{\mathrm{H}}}{{\mathrm{T}}_{\mathrm{L}}}$ (3)

In the first case,$\mathrm{W}={\mathrm{W}}_1+{\mathrm{W}}_2\mathrm{and}{\mathrm{Q}}_{\mathrm{H}}={\mathrm{Q}}_1$

The efficiency of the Carnot cycle using equation (1) of the first stage can be given as:

$\epsilon =\frac{{\mathrm{W}}_1+{\mathrm{W}}_2}{{\mathrm{Q}}_1}$

Using equation (2), the works per cycle of the first and second stages are

${\mathrm{W}}_1={\mathrm{Q}}_2-{\mathrm{Q}}_1,\mathrm{and}{\mathrm{W}}_2={\mathrm{Q}}_2-{\mathrm{Q}}_3$respectively.

So, the efficiency becomes

$$\begin{gathered} \mathrm{\epsilon }=\frac{\left({\mathrm{Q}}_2-{\mathrm{Q}}_1\right)+\left({\mathrm{Q}}_2-{\mathrm{Q}}_3\right)}{{\mathrm{Q}}_1} \\ =\frac{{\mathrm{Q}}_1-{\mathrm{Q}}_3}{{\mathrm{Q}}_1} \\ =1-\frac{{\mathrm{Q}}_3}{{\mathrm{Q}}_1} \end{gathered}$$

Now, using equation (3), the efficiency of the engine is given using$\left(\frac{{\mathrm{Q}}_3}{\mathrm{Q}1}=\frac{{\mathrm{T}}_3}{{\mathrm{T}}_1}\right)$ as,

$$\begin{gathered} \mathrm{\epsilon }=1-\frac{{\mathrm{T}}_3}{{\mathrm{T}}_1} \\ =\frac{{\mathrm{T}}_1-{\mathrm{T}}_3}{{\mathrm{T}}_1} \end{gathered}$$

Hence, the efficiency of the engine is proved as$\frac{{\mathrm{T}}_1-{\mathrm{T}}_3}{{\mathrm{T}}_1}$