Two Carnot engines A and B are operated in series. The first one, A receives heat at T1(= 600 K) and

Q: Why is the heat rejected by Engine A exactly equal to the heat absorbed by Engine B?

Because they are connected in series. The intermediate reservoir at temperature $T_2$ acts as a perfect thermal conductor between the two engines. All the heat $Q_2$ dumped by Engine A is picked up by Engine B.

Q: Can we use Celsius for these calculations instead of Kelvin?

Absolutely not! Thermodynamic temperature ratios (like $\frac{Q_1}{Q_2} = \frac{T_1}{T_2}$) are derived from the absolute thermodynamic temperature scale. You must always convert temperatures to Kelvin before taking ratios.

Q: What if the engines were not Carnot engines?

If they were irreversible engines, the relation $\frac{Q_1}{Q_2} = \frac{T_1}{T_2}$ would not hold. We would only know that $\frac{Q_1}{Q_2} > \frac{T_1}{T_2}$, and we couldn't solve for an exact temperature without more information about their specific efficiencies.

Q: Is $T_2$ always the arithmetic mean of $T_1$ and $T_3$ for engines in series?

No, it is only the arithmetic mean when the **work outputs** are equal. If the problem stated that their **efficiencies** were equal, $T_2$ would be the geometric mean ($\sqrt{T_1 T_3}$).

Q: Does the intermediate reservoir do any work?

No, the reservoir itself does zero work. It merely acts as a thermal buffer, absorbing heat from Engine A and supplying it to Engine B.

Question

Two Carnot engines $A$ and $B$ are operated in series. The first one, $A$ receives heat at $T_1(= 600	ext{ K})$ and rejects to a reservoir at temperature $T_2$. The second engine $B$ receives heat rejected by the first engine and in turn rejects to a heat reservoir at $T_3(= 400	ext{ K})$. Calculate the temperature $T_2$ if the work outputs of the two engines are equal.

Options:

(A) 600 K

(B) 500 K

(C) 400 K

(D) 300 K

Sigma Percentile · Accepted Answer

### Visualizing the Series Setup

Imagine we have two Carnot engines, $A$ and $B$, connected in series. This means the heat rejected by the first engine becomes the heat input for the second engine. Engine $A$ takes heat $Q_1$ from a source at $T_1 = 600	ext{ K}$, does work $W_1$, and rejects heat $Q_2$ to an intermediate reservoir at temperature $T_2$.

Then, Engine $B$ takes this exact heat $Q_2$, does work $W_2$, and rejects heat $Q_3$ to a sink at $T_3 = 400	ext{ K}$. The intermediate reservoir acts merely as a thermal buffer, transferring the heat without doing any work itself.

### The Master Equation: Equating Work

According to the first law of thermodynamics, the work done by any heat engine is simply the difference between the heat absorbed and the heat rejected.

For Engine $A$, the work is:
$$W_1 = Q_1 - Q_2$$

And for Engine $B$, the work is:
$$W_2 = Q_2 - Q_3$$

The problem gives us a crucial condition: the work outputs of both engines are equal. This means $W_1 = W_2$. Let's equate our expressions:
$$Q_1 - Q_2 = Q_2 - Q_3$$

Now, let's rearrange this equation to group our terms. If we move the $Q_2$ from the left to the right, and the $Q_3$ from the right to the left, we get:
$$Q_1 + Q_3 = 2Q_2$$

### The Magic of the Carnot Cycle

To connect this heat equation to temperatures, let's divide the entire equation by $Q_2$. This gives us:
$$\frac{Q_1}{Q_2} + \frac{Q_3}{Q_2} = 2$$

Here is where the magic of the Carnot cycle comes in. For a reversible Carnot engine, the ratio of heats is equal to the ratio of their absolute temperatures. So, $\frac{Q_1}{Q_2}$ is exactly $\frac{T_1}{T_2}$. Similarly, $\frac{Q_3}{Q_2}$ is $\frac{T_3}{T_2}$.

Let's substitute these temperature ratios back into our equation. Look at that! We now have an equation entirely in terms of temperatures:
$$\frac{T_1}{T_2} + \frac{T_3}{T_2} = 2$$

### Final Calculation

Since the denominators are the same, we can combine the fractions to get:
$$\frac{T_1 + T_3}{T_2} = 2$$

Rearranging this to solve for $T_2$, we find that $T_2$ is simply the arithmetic mean of $T_1$ and $T_3$:
$$T_2 = \frac{T_1 + T_3}{2}$$

Finally, let's substitute the given values. $T_1$ is $600	ext{ K}$ and $T_3$ is $400	ext{ K}$. 
$$T_2 = \frac{600 + 400}{2} = \frac{1000}{2} = 500	ext{ K}$$

So, the temperature of the intermediate reservoir must be **$500	ext{ K}$** for the engines to produce equal work.

**A Pro Tip:** This is a favorite concept for JEE. What if the question asked for equal efficiencies instead of equal work? In that case, you would equate $1 - \frac{T_2}{T_1}$ to $1 - \frac{T_3}{T_2}$. Solving that would give $T_2$ as the geometric mean, $T_2 = \sqrt{T_1 T_3}$. Always read carefully whether it's equal work or equal efficiency!

Select Answer:

Visualized Solution

Visualizing the Series Setup

Work Done by Each Engine

Equating the Work Outputs

Rearranging the Heat Equation

Dividing by $Q_{2}$

Carnot Principle

Substituting Temperatures

Solving for $T_{2}$

Final Calculation

The Way Forward: Equal Efficiency

The Sigma Insight: Heat Engines and Refrigerators

Visualizing the Series Setup

The Master Equation

The Magic of the Carnot Cycle

Final Calculation

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Select Answer:

Visualized Solution

Visualizing the Series Setup

Work Done by Each Engine

Equating the Work Outputs

Rearranging the Heat Equation

Dividing by .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Q2​

Carnot Principle

Substituting Temperatures

Solving for .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } T2​

Final Calculation

The Way Forward: Equal Efficiency

The Sigma Insight: Heat Engines and Refrigerators

Visualizing the Series Setup

The Master Equation

The Magic of the Carnot Cycle

Final Calculation

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