Two Carnot engines A and B operate in series such that engine A absorbs heat at T1 and rejects heat

Q: Why does Engine B only take half the heat?

The problem explicitly states a condition where Engine B absorbs half of the heat rejected by Engine A. This could be due to a physical bypass or heat loss in a real-world scenario, though here it's just a mathematical constraint to test your understanding.

Q: Can we use efficiency formulas directly?

Yes! You can use $\eta = W/Q_{in} = 1 - T_{sink}/T_{source}$. For Engine A, $W_A = Q_1(1 - T/T_1)$. For Engine B, $W_B = (Q_2/2)(1 - T_3/T)$. Equating them works perfectly too!

Q: Why did we divide the work equation by $Q_2$?

Dividing by $Q_2$ is a strategic algebraic move. It creates ratios like $Q_1/Q_2$ and $Q_3/Q_2$, which we can directly replace with temperature ratios using the Carnot principle.

Q: What if the engines were not Carnot engines?

If they weren't Carnot (reversible) engines, the relation $Q_{rejected}/Q_{absorbed} = T_{sink}/T_{source}$ would not hold. We would need more information about their specific efficiencies.

Q: Is the intermediate temperature $T$ always an average?

Not necessarily! If $W_A = W_B$ and all heat was transferred ($Q_{in, B} = Q_{out, A}$), $T$ would be the exact arithmetic mean $(T_1 + T_3)/2$. But because Engine B only takes half the heat, the formula shifts to a weighted average.

Question

Two Carnot engines $A$ and $B$ operate in series such that engine $A$ absorbs heat at $T_1$ and rejects heat to a sink at temperature $T$. engine $B$ absorbs half of the heat rejected by engine $A$ and rejects heat to the sink at $T_3$. When work done in both the cases is equal, then the value of $T$ is

Options:

(A) $\frac{2}{3}T_1 + \frac{3}{2}T_3$

(B) $\frac{1}{3}T_1 + \frac{2}{3}T_3$

(C) $\frac{3}{2}T_1 + \frac{1}{3}T_3$

(D) $\frac{2}{3}T_1 + \frac{1}{3}T_3$

Sigma Percentile · Accepted Answer

### The Setup: A Tale of Two Engines

Imagine a factory where two heat engines are working together in a chain, or in **series**. The first engine, Engine $A$, is the heavy lifter. It draws a large amount of heat, let's call it $Q_1$, from a blazing hot furnace at temperature $T_1$.

It uses some of this heat to do useful work, $W_A$, and dumps the leftover heat, $Q_2$, into an intermediate sink at temperature $T$.

Now, here is where it gets interesting. Engine $B$ is sitting right next to this sink, ready to use that rejected heat. However, it doesn't take all of it. The problem explicitly states that Engine $B$ only absorbs **half** of the heat rejected by Engine $A$. So, its heat input is $\frac{Q_2}{2}$.

Engine $B$ does its own share of work, $W_B$, and finally exhausts the remaining heat, $Q_3$, into a cold sink at temperature $T_3$.

### The Golden Rule of Carnot

To solve this, we need to rely on the fundamental property of a reversible Carnot engine. For any Carnot engine, the ratio of the heat rejected to the heat absorbed is perfectly equal to the ratio of their absolute temperatures.

Mathematically, this is written as:

$$ \frac{Q_{rejected}}{Q_{absorbed}} = \frac{T_{sink}}{T_{source}} $$

Let's apply this golden rule to Engine $A$. The heat rejected is $Q_2$ and the heat absorbed is $Q_1$. The temperatures are $T$ and $T_1$.

$$ \frac{Q_2}{Q_1} = \frac{T}{T_1} \implies Q_1 = Q_2 \frac{T_1}{T} $$

Now, let's do the same for Engine $B$. The heat rejected is $Q_3$, but remember, the heat absorbed is only $\frac{Q_2}{2}$. The temperatures are $T_3$ and $T$.

$$ \frac{Q_3}{Q_2 / 2} = \frac{T_3}{T} \implies Q_3 = \frac{Q_2}{2} \frac{T_3}{T} $$

### Bridging the Work Done

The core constraint given in the problem is that both engines perform the **exact same amount of work**.

$$ W_A = W_B $$

We know that the work done by any heat engine is simply the difference between the heat it takes in and the heat it throws away.

For Engine $A$, $W_A = Q_1 - Q_2$. 
For Engine $B$, $W_B = \frac{Q_2}{2} - Q_3$.

Equating them, we get:

$$ Q_1 - Q_2 = \frac{Q_2}{2} - Q_3 $$

This equation looks a bit messy with all these different $Q$ variables. But notice that every term can be related back to $Q_2$. Let's divide the entire equation by $Q_2$ to clean it up.

$$ \frac{Q_1}{Q_2} - 1 = \frac{1}{2} - \frac{Q_3}{Q_2} $$

### The Final Mathematical Symphony

Now, we substitute the temperature ratios we derived earlier from the Carnot principle. We know that $\frac{Q_1}{Q_2} = \frac{T_1}{T}$ and $\frac{Q_3}{Q_2} = \frac{1}{2} \frac{T_3}{T}$.

Substituting these into our simplified work equation:

$$ \frac{T_1}{T} - 1 = \frac{1}{2} - \frac{1}{2} \frac{T_3}{T} $$

Let's group all the terms containing the unknown intermediate temperature $T$ on one side, and the constants on the other.

$$ \frac{T_1}{T} + \frac{T_3}{2T} = \frac{3}{2} $$

To clear the denominators and solve for $T$, we can multiply the entire equation by $2T$.

$$ 2T_1 + T_3 = 3T $$

Finally, isolating $T$, we arrive at our beautiful final answer:

$$ T = \frac{2}{3}T_1 + \frac{1}{3}T_3 $$

This perfectly matches option (d). The intermediate temperature is a weighted average of the source and final sink temperatures, heavily skewed towards the hotter source because Engine $B$ only utilized half of the rejected heat!

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Visualized Solution

$E n g in e A S e t u p$

$E n g in e B S e t u p$

$C a r n o t E f f i c i e n cy P r in c i pl e$

$A ppl y in g t o E n g in e A$

$A ppl y in g t o E n g in e B$

$E q u a t in g W or k D o n e$

$S u b s t i t u t in g T e m p er a t u r es$

$F ina l C a l c u l a t i o n$

The Sigma Insight: Heat Engines and Refrigerators

The Setup

The Golden Rule of Carnot

Bridging the Work Done

The Final Mathematical Symphony

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Visualized Solution

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

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

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

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

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

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

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

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