As shown in the figure, five Carnot engines, each with efficiency eta and same number of cycles per

Q: Why is the total work simply $Q_0 - Q_5$? Shouldn't we add $W_1 + W_2 + W_3 + W_4 + W_5$?

You absolutely can add them up, and you will get the exact same result! However, using the First Law of Thermodynamics on the entire system as a single black box is much faster. The only energy entering the system is $Q_0$, and the only energy leaving as heat is $Q_5$. The difference must be the total work done.

Q: Can we just add the efficiencies of the five engines to get the net efficiency?

No, efficiency is not additive. Each subsequent engine operates on a smaller amount of heat input. If you added them, you might get an efficiency greater than 100%, which violates the Second Law of Thermodynamics.

Q: What if the engines had different efficiencies, say $\eta_1, \eta_2, \dots$?

The logic remains exactly the same! The final heat rejected would be $Q_5 = Q_0(1-\eta_1)(1-\eta_2)(1-\eta_3)(1-\eta_4)(1-\eta_5)$. The net efficiency would then be $1 - (1-\eta_1)(1-\eta_2)\dots(1-\eta_5)$.

Q: How am I supposed to calculate the 5th root of $32/243$ without a calculator?

In JEE Advanced, numbers are rarely random. Whenever you see a high power, look for small prime bases. $32$ is a classic power of $2$ ($2^5$), and $243$ is a classic power of $3$ ($3^5$). Recognizing these powers is a crucial skill.

Q: Is this cascading setup used in real life?

Yes! While real engines aren't perfect Carnot engines, the principle of cascading is used in Combined Cycle Power Plants. The exhaust heat from a high-temperature gas turbine is used to boil water and run a secondary steam turbine, significantly boosting the overall plant efficiency.

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 $\eta_{	ext{net}} = \frac{W}{Q_0} = \frac{211}{243}$. The value of $\eta$ is:

Sigma Percentile · Accepted Answer

### The Cascade of Carnot Engines

Imagine a beautifully orchestrated cascade of five Carnot engines. The first engine takes in a massive chunk of heat, performs some useful work, and rejects the rest. But in this setup, that rejected heat isn't wasted into the environment—it becomes the exact input for the second engine. This chain reaction continues all the way down to the fifth engine.

Our goal is to find the efficiency $\eta$ of a single engine, given the net efficiency of the entire five-engine system.

### The Geometric Progression of Heat

Let's break down the thermodynamics of a single engine. For any engine with efficiency $\eta$, the work done is $W = \eta Q_{	ext{in}}$. By the first law of thermodynamics, the heat rejected is:

$$Q_{	ext{out}} = Q_{	ext{in}} - W = Q_{	ext{in}}(1 - \eta)$$

Now, let's trace the heat through our cascade. The first engine takes in $Q_0$ and rejects $Q_1$:

$$Q_1 = Q_0(1 - \eta)$$

This $Q_1$ is fed into the second engine, which then rejects $Q_2$:

$$Q_2 = Q_1(1 - \eta) = Q_0(1 - \eta)^2$$

Do you see the pattern? Each time the heat passes through an engine, it gets multiplied by a factor of $(1 - \eta)$. Following this geometric progression, by the time the heat leaves the fifth and final engine, it has been reduced to:

$$Q_5 = Q_0(1 - \eta)^5$$

### The Master Equation for Net Efficiency

What is the net efficiency of this entire system? Net efficiency is defined as the total work done by all engines combined, divided by the initial heat input $Q_0$.

Instead of calculating the work done by each engine and adding them up, we can use a much more elegant approach: **Conservation of Energy**. The total work done by the system must equal the total heat that entered the system minus the final heat that left the system.

$$W_{	ext{total}} = Q_0 - Q_5$$

Therefore, the net efficiency is:

$$\eta_{	ext{net}} = \frac{W_{	ext{total}}}{Q_0} = \frac{Q_0 - Q_5}{Q_0} = 1 - \frac{Q_5}{Q_0}$$

Substituting our expression for $Q_5$, we get a beautifully simple master equation:

$$\eta_{	ext{net}} = 1 - (1 - \eta)^5$$

### Final Calculation

The problem states that the net efficiency is $\frac{211}{243}$. Let's equate this to our master equation:

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

Rearranging the terms to isolate the $\eta$ term:

$$(1 - \eta)^5 = 1 - \frac{211}{243} = \frac{32}{243}$$

At first glance, taking a fifth root might seem daunting. But look closely at the numbers. $32$ is exactly $2^5$, and $243$ is exactly $3^5$. The problem was designed to resolve perfectly!

Taking the fifth root of both sides:

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

Solving for $\eta$:

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

And there we have it. Each individual Carnot engine operates at an efficiency of $33.33\%$.

Enter Numerical Value:

Visualized Solution

$Visualizing the Cascade$

$Efficiency of a Single Engine$

$First Engine Analysis$

$Second Engine Analysis$

$Generalizing to the Fifth Engine$

$Total Work Done$

$Net Efficiency Formula$

$Equating to Given Value$

$Algebraic Manipulation$

$Final Calculation$

The Sigma Insight: Heat Engines and Refrigerators

The Cascade of Carnot Engines

The Geometric Progression of Heat

The Master Equation for Net Efficiency

Final Calculation

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Enter Numerical Value:

Visualized Solution

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Visualizing the Cascade

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Efficiency of a Single Engine

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } First Engine Analysis

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Second Engine Analysis

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Generalizing to the Fifth Engine

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Total Work Done

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Net Efficiency Formula

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Equating to Given Value

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Algebraic Manipulation

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Final Calculation