Efficiency of Five Carnot Engines in Series

To find the efficiency $\eta$ of each Carnot engine, we analyze the heat transfer through the series of five engines.

For any individual Carnot engine with efficiency $\eta$, the relation between the heat absorbed per cycle ($Q_{\text{in}}$) and the heat released per cycle ($Q_{\text{out}}$) is given by: $\eta = 1 - \frac{Q_{\text{out}}}{Q_{\text{in}}} \implies Q_{\text{out}} = Q_{\text{in}}(1 - \eta)$

Let $Q_0$ be the heat absorbed by the first engine.

1. First Engine:

   • Heat absorbed = $Q_0$
   • Heat released = $Q_1 = Q_0 (1 - \eta)$

2. Second Engine:

   • Heat absorbed = $Q_1$
   • Heat released = $Q_2 = Q_1 (1 - \eta) = Q_0 (1 - \eta)^2$

3. Third Engine:

   • Heat absorbed = $Q_2$
   • Heat released = $Q_3 = Q_0 (1 - \eta)^3$

4. Fourth Engine:

   • Heat absorbed = $Q_3$
   • Heat released = $Q_4 = Q_0 (1 - \eta)^4$

5. Fifth Engine:

   • Heat absorbed = $Q_4$
   • Heat released = $Q_5 = Q_0 (1 - \eta)^5$

The total work done $W$ by all five engines in one cycle is equal to the total heat input to the system minus the final heat output released to the lowest temperature reservoir: $W = Q_0 - Q_5 = Q_0 \left[ 1 - (1 - \eta)^5 \right]$

Thus, the net efficiency $\eta_{\text{net}}$ of the system is: $\eta_{\text{net}} = \frac{W}{Q_0} = 1 - (1 - \eta)^5$

We are given that $\eta_{\text{net}} = \frac{211}{243}$. Equating this to our expression: $1 - (1 - \eta)^5 = \frac{211}{243}$

Rearranging terms: $(1 - \eta)^5 = 1 - \frac{211}{243}$

$(1 - \eta)^5 = \frac{32}{243}$

Expressing $32$ and $243$ in terms of powers of $5$: $32 = 2^5 \quad \text{and} \quad 243 = 3^5$

So, $(1 - \eta)^5 = \left( \frac{2}{3} \right)^5$

Taking the fifth root on both sides: $1 - \eta = \frac{2}{3}$

$\eta = 1 - \frac{2}{3} = \frac{1}{3} \approx 0.33$

Thus, the value of $\eta$ is approximately 0.33 (or $\frac{1}{3}$).