Question

As shown in the figure, five Carnot engines, each with efficiency \(\eta\) and same number of cycles per unit time, are operating between six heat reservoirs. The amount of heat released per cycle by one engine is completely absorbed by the next engine. Consider \(Q_0\) to be the amount of heat absorbed per cycle by the first engine and \(W\) as the amount of total work done by all the engines per cycle, then the net efficiency of the system is found to be \[ \eta_{\mathrm{net}} = \frac{W}{Q_0} = \frac{211}{243} \] The value of \(\eta\) is _______.

Hint:

Rather than adding up each engine's work one by one, think about the chain as a single combined system: heat rejected by one engine is fully picked up by the next, so those transfers cancel out inside the system. Focus only on what heat enters the whole chain and what heat finally leaves it, then connect that difference to the total work done.

Answer 1

Step 1: Write efficiency relation for each engine.
For each Carnot engine: \[ \eta=\frac{W_i}{Q_{i-1}} \] Hence: \[ W_i=\eta Q_{i-1} \] Heat rejected by first engine: \[ Q_1=Q_0-W_1 \] \[ Q_1=Q_0(1-\eta) \] Similarly: \[ Q_2=Q_1(1-\eta) \] Thus: \[ Q_n=Q_0(1-\eta)^n \]

Step 2: Find total work done.
Work done by: \[ i^{\text{th}} \] engine: \[ W_i=\eta Q_{i-1} \] Thus: \[ W = \eta Q_0 \left[ 1+(1-\eta)+(1-\eta)^2+(1-\eta)^3+(1-\eta)^4 \right] \] Using geometric series: \[ W = \eta Q_0 \cdot \frac{1-(1-\eta)^5}{1-(1-\eta)} \] \[ W = Q_0\left[1-(1-\eta)^5\right] \] Hence: \[ \eta_{\mathrm{net}} = \frac{W}{Q_0} = 1-(1-\eta)^5 \]

Step 3: Use given net efficiency.
Given: \[ 1-(1-\eta)^5 = \frac{211}{243} \] Thus: \[ (1-\eta)^5 = 1-\frac{211}{243} \] \[ = \frac{32}{243} \] \[ = \left(\frac23\right)^5 \] Therefore: \[ 1-\eta=\frac23 \] \[ \eta=\frac13 \]

Step 4: Identify the final answer.
Therefore: \[ \boxed{\frac13} \]

Answer 2

Concept:

• Every unit of heat rejected by one engine is fully absorbed by the next, so these intermediate heat transfers cancel out when the five engines are treated as a single combined system. The only heat entering this combined system from outside is $Q_0$ (into the first engine), and the only heat leaving it is $Q_5$ (rejected by the fifth engine). By overall energy conservation, the total work is simply $W = Q_0 - Q_5$, with no need to add up each engine's work separately.
• Since every engine has the same efficiency $\eta$, the heat passed down the chain shrinks by the same factor $(1-\eta)$ at each stage.

Step 1: Track the heat as it passes down the chain
Each Carnot engine rejects heat equal to $(1-\eta)$ times what it absorbed. So after engine 1, the heat handed to engine 2 is $Q_1 = Q_0(1-\eta)$. After engine 2, $Q_2 = Q_1(1-\eta) = Q_0(1-\eta)^2$. Continuing this pattern through all five engines:
$Q_5 = Q_0(1-\eta)^5$

Step 2: Apply overall energy conservation to the whole chain
$W = Q_0 - Q_5 = Q_0 - Q_0(1-\eta)^5 = Q_0\left[1-(1-\eta)^5\right]$

Step 3: Form the net efficiency and substitute the given value
$\eta_{\text{net}} = \dfrac{W}{Q_0} = 1-(1-\eta)^5 = \dfrac{211}{243}$
$(1-\eta)^5 = 1-\dfrac{211}{243} = \dfrac{32}{243}$

Step 4: Solve for $\eta$
Recognize $32 = 2^5$ and $243 = 3^5$, so $(1-\eta)^5 = \left(\dfrac{2}{3}\right)^5$.
Taking the fifth root of both sides: $1-\eta = \dfrac{2}{3}$, so $\eta = \dfrac{1}{3}$.

Final Answer: $\eta = \dfrac{1}{3}$