Two ideal Carnot engines operate in cascade (all heat given up by one engine is used by the other en

Q: Why is the constant $k$ the same for both engines?

Because the heat $Q$ rejected by the first engine at temperature $T$ is exactly the heat absorbed by the second engine at the same temperature $T$. Since $Q/T = k$ for the first engine's sink and $Q/T = k'$ for the second engine's source, $k$ must equal $k'$.

Q: What if the engines were not ideal Carnot engines?

If they weren't ideal, the relation $Q/T = \text{constant}$ wouldn't hold, and we couldn't express work purely in terms of temperatures without knowing their specific individual efficiencies.

Q: Is the intermediate temperature always the arithmetic mean?

Only when the work done by both engines in the cascade is equal. If their efficiencies were equal instead, the intermediate temperature would be the geometric mean, $\sqrt{T_1 T_2}$.

Q: Can we solve this using standard efficiency formulas instead of the $k$ constant?

Yes! $W_1 = \eta_1 Q_1 = (1 - T/T_1)Q_1$ and $W_2 = \eta_2 Q = (1 - T_2/T)Q$. Since $Q = Q_1(T/T_1)$, substituting this into $W_2$ and equating to $W_1$ will yield the exact same arithmetic mean result.

Q: Why does the heat flow from $T_1$ to $T$ and then to $T_2$?

By the second law of thermodynamics, heat naturally flows from a higher temperature to a lower temperature. For the engines to produce positive work, the strict inequality $T_1 > T > T_2$ must hold true.

Question

Two ideal Carnot engines operate in cascade (all heat given up by one engine is used by the other engine to produce work) between temperatures $T_1$ and $T_2$. The temperature of the hot reservoir of the first engine is $T_1$ and the temperature of the cold reservoir of the second engine is $T_2$. $T$ is temperature of the sink of first engine which is also the source for the second engine. How is $T$ related to $T_1$ and $T_2$, if both the engines perform equal amount of work?

Options:

(A) $T = 0$

(B) $T = \frac{2T_1T_2}{T_1 + T_2}$

(C) $T = \sqrt{T_1T_2}$

(D) $T = \frac{T_1 + T_2}{2}$

Sigma Percentile · Accepted Answer

### The Cascade of Carnot Engines: Unraveling the Intermediate Temperature

Thermodynamics is often described as the poetry of physics, and the Carnot cycle is its most elegant sonnet. When we talk about a single Carnot engine, we are dealing with the theoretical maximum efficiency a heat engine can achieve between two temperatures. But what happens when we chain them together?

Imagine a majestic waterfall driving a massive turbine. The water that flows out of this first turbine hasn't lost all its gravitational potential; it can still fall further to drive a second turbine below it. This is exactly the physical intuition behind a **cascade of Carnot engines**. The heat rejected by the first engine isn't wasted into the void; it becomes the lifeblood—the source heat—for the second engine.

### Analyzing the Setup

Let's break down the architecture of our cascade system. We have two ideal Carnot engines, $E_1$ and $E_2$, operating in series.

1. **Engine 1 ($E_1$)**: It draws heat $Q_1$ from a hot reservoir at a high temperature $T_1$. It performs some mechanical work $W_1$ and rejects the remaining heat $Q$ into an intermediate reservoir at temperature $T$.
2. **Engine 2 ($E_2$)**: This engine picks up the baton. It draws that exact same heat $Q$ from the intermediate reservoir at temperature $T$, performs mechanical work $W_2$, and finally rejects heat $Q_2$ into a cold sink at temperature $T_2$.

The crucial link here is the intermediate reservoir at temperature $T$. It acts as a perfect thermal bridge, absorbing heat from $E_1$ and immediately supplying it to $E_2$.

### The Master Equation: The Carnot Principle

To solve this, we need to invoke the fundamental property of a reversible Carnot cycle. For any ideal Carnot engine, the ratio of the heat transferred to the absolute temperature of the reservoir is a constant. This is a direct consequence of the fact that the total entropy change for a reversible cycle is zero ($\oint \frac{dQ}{T} = 0$).

For Engine 1, this means:
$$ \frac{Q_1}{T_1} = \frac{Q}{T} $$

For Engine 2, this means:
$$ \frac{Q}{T} = \frac{Q_2}{T_2} $$

Notice the beautiful symmetry? Because the heat $Q$ rejected by $E_1$ is exactly the heat absorbed by $E_2$, the ratio $\frac{Q}{T}$ is the same for both engines! Let's call this universal constant $k$.

$$ \frac{Q_1}{T_1} = \frac{Q}{T} = \frac{Q_2}{T_2} = k $$

This allows us to express all heat transfers purely in terms of temperatures:
- $Q_1 = k T_1$
- $Q = k T$
- $Q_2 = k T_2$

### Formulating the Work Done

By the First Law of Thermodynamics (conservation of energy), the work done by a cyclic heat engine is simply the difference between the heat absorbed and the heat rejected.

For Engine 1:
$$ W_1 = Q_1 - Q = k T_1 - k T = k(T_1 - T) $$

For Engine 2:
$$ W_2 = Q - Q_2 = k T - k T_2 = k(T - T_2) $$

We have now successfully expressed the work done by both engines using a single common constant $k$ and their respective operating temperatures.

### Equating and Solving

The problem provides us with a very specific constraint: **both engines perform an equal amount of work**.

$$ W_1 = W_2 $$

Substituting our derived expressions:
$$ k(T_1 - T) = k(T - T_2) $$

Since $k$ is a non-zero constant (as heat is actually flowing), we can elegantly cancel it out from both sides:
$$ T_1 - T = T - T_2 $$

Now, it's just a matter of simple algebra. Let's group the $T$ terms on one side:
$$ T + T = T_1 + T_2 $$
$$ 2T = T_1 + T_2 $$
$$ T = \frac{T_1 + T_2}{2} $$

The intermediate temperature $T$ is exactly the **arithmetic mean** of the hot and cold reservoir temperatures!

### The Efficiency Variation (A JEE Favorite)

While we have solved the problem, a true master of physics always asks, *"What if?"*

What if the problem stated that the **efficiencies** of both engines were equal, instead of their work outputs?

The efficiency of a Carnot engine is given by $\eta = 1 - \frac{T_{	ext{sink}}}{T_{	ext{source}}}$.

If $\eta_1 = \eta_2$:
$$ 1 - \frac{T}{T_1} = 1 - \frac{T_2}{T} $$

Canceling the 1s and the negative signs:
$$ \frac{T}{T_1} = \frac{T_2}{T} $$

Cross-multiplying yields:
$$ T^2 = T_1 T_2 $$
$$ T = \sqrt{T_1 T_2} $$

In this scenario, the intermediate temperature would be the **geometric mean** of the extreme temperatures.

**Key Takeaway:** Always read the constraints carefully. Equal work leads to the arithmetic mean, while equal efficiency leads to the geometric mean. Mastering these subtle variations is what separates a good student from a great one.

Select Answer:

Visualized Solution

$Cascade Configuration$

$Carnot Engine Principle$

$Work Done by Engine 1$

$Work Done by Engine 2$

$Equating Work Done$

$Intermediate Temperature$

$What if Efficiencies were Equal?$

The Sigma Insight: Heat Engine, Second Law of Thermodynamics and Carnot Engine

The Cascade of Carnot Engines

Analyzing the Setup

The Master Equation

Formulating the Work Done

Equating and Solving

The Efficiency Variation (A JEE Favorite)

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