Three Carnot engines operate in series between a heat source at a temperature T1 and a heat sink at

Q: What does it mean for engines to operate "in series"?

When engines operate in series, the heat rejected by the first engine acts as the heat input for the second engine, and so on. The sink of the first engine is the source for the second.

Q: Why did we equate $1 - T_2/T_1$ to $1 - T_3/T_2$?

The problem explicitly states that all three engines are "equally efficient". The efficiency of the first engine is $1 - T_2/T_1$ and the second is $1 - T_3/T_2$. Equating them directly translates the problem's condition into math.

Q: How did you realize it was a Geometric Progression (GP)?

After canceling the '1's and the negative signs from the equated efficiencies, we got $T_2/T_1 = T_3/T_2 = T_4/T_3$. A sequence where the ratio of consecutive terms is constant is the exact definition of a Geometric Progression!

Q: Is there a shortcut if the question asked for equal work done instead of equal efficiency?

Yes! If the work done by engines in series is equal, their operating temperatures form an Arithmetic Progression (AP) instead of a Geometric Progression (GP). This is a very handy trick for competitive exams.

Q: Can a Carnot engine ever have 100% efficiency?

No, unless the sink temperature is absolute zero (0 Kelvin), which is practically impossible according to the Third Law of Thermodynamics.

Question

Three Carnot engines operate in series between a heat source at a temperature $T_1$ and a heat sink at temperature $T_4$ (see figure). There are two other reservoirs at temperatures $T_2$ and $T_3$, as shown with $T_1 > T_2 > T_3 > T_4$. The three engines are equally efficient if

Options:

(A) $T_2 = (T_1^3 T_4)^{1/4}; T_3 = (T_1 T_4^3)^{1/4}$

(B) $T_2 = (T_1^2 T_4)^{1/3}; T_3 = (T_1 T_4^2)^{1/3}$

(C) $T_2 = (T_1 T_4)^{1/2}; T_3 = (T_1^2 T_4)^{1/3}$

(D) $T_2 = (T_1 T_4^2)^{1/3}; T_3 = (T_1^2 T_4)^{1/3}$

Sigma Percentile · Accepted Answer

### The Setup: Engines in Series

Imagine a waterfall cascading down a mountain, hitting multiple turbines one after another. This is exactly what is happening in our problem, but with heat instead of water!

We have three Carnot engines connected in series. The first engine takes heat from the hottest source at $T_1$, does some work, and rejects the remaining heat to a reservoir at $T_2$.

This rejected heat doesn't go to waste. It becomes the input for the second engine, which operates between $T_2$ and $T_3$. Finally, the third engine takes the heat rejected at $T_3$ and dumps its exhaust into the final cold sink at $T_4$.

### The Master Equation: Carnot Efficiency

The problem gives us a beautiful constraint: **all three engines are equally efficient**.

To use this, we need the formula for the efficiency of a Carnot engine. The efficiency $\eta$ is given by:
$$ \eta = 1 - \frac{T_{	ext{sink}}}{T_{	ext{source}}} $$

Let's apply this to our three engines. For the first engine, the source is $T_1$ and the sink is $T_2$. For the second, it's $T_2$ and $T_3$. For the third, it's $T_3$ and $T_4$.

Equating their efficiencies, we get:
$$ 1 - \frac{T_2}{T_1} = 1 - \frac{T_3}{T_2} = 1 - \frac{T_4}{T_3} $$

### The Mathematical Symphony: Emergence of a GP

Now, let's simplify this equation. We can subtract $1$ from all sides and cancel the negative signs. This leaves us with a very elegant relationship:
$$ \frac{T_2}{T_1} = \frac{T_3}{T_2} = \frac{T_4}{T_3} $$

Look closely at this result. The ratio of any temperature to the previous one is constant! In mathematics, a sequence of numbers with a constant ratio is called a **Geometric Progression (GP)**.

Let's call this common ratio $r$. So, we have:
$$ r = \frac{T_2}{T_1} = \frac{T_3}{T_2} = \frac{T_4}{T_3} $$

### Solving the Sequence

Since the temperatures form a GP, we can express any temperature in the sequence using the first term $T_1$ and the common ratio $r$.

The final temperature $T_4$ is the fourth term in the sequence, so:
$$ T_4 = T_1 \cdot r^3 $$

From this, we can isolate our common ratio $r$:
$$ r = \left( \frac{T_4}{T_1} ight)^{\frac{1}{3}} $$

Now, finding the intermediate temperatures $T_2$ and $T_3$ is just a matter of substitution. $T_2$ is the second term, so:
$$ T_2 = T_1 \cdot r = T_1 \left( \frac{T_4}{T_1} ight)^{\frac{1}{3}} $$

By bringing $T_1$ inside the bracket (which becomes $T_1^3$), we get:
$$ T_2 = \left( T_1^3 \cdot \frac{T_4}{T_1} ight)^{\frac{1}{3}} = \left( T_1^2 T_4 ight)^{\frac{1}{3}} $$

Similarly, $T_3$ is the third term:
$$ T_3 = T_1 \cdot r^2 = T_1 \left( \frac{T_4}{T_1} ight)^{\frac{2}{3}} $$

Bringing $T_1$ inside the bracket gives:
$$ T_3 = \left( T_1^3 \cdot \frac{T_4^2}{T_1^2} ight)^{\frac{1}{3}} = \left( T_1 T_4^2 ight)^{\frac{1}{3}} $$

### The Grand Finale

We have successfully found the intermediate temperatures! 
$$ T_2 = \left( T_1^2 T_4 ight)^{\frac{1}{3}} $$
$$ T_3 = \left( T_1 T_4^2 ight)^{\frac{1}{3}} $$

Comparing this with our options, we can confidently select **Option (b)** as the correct answer.

### A Pro Tip for Exams

Here is a golden nugget for your competitive exams. In this problem, the **efficiencies** were equal, which led to the temperatures forming a **Geometric Progression (GP)**.

However, if a problem states that the **work done** by engines in series is equal, the temperatures will form an **Arithmetic Progression (AP)**! Remembering this duality can save you precious minutes during the exam.

Select Answer:

Visualized Solution

$Visualizing the Setup$

$Carnot Efficiency Formula$

$Equating Efficiencies$

$The Geometric Progression$

$Finding the Common Ratio r$

$Solving for T_{2}$

$Solving for T_{3}$

$Conclusion$

$Bonus Concept$

The Sigma Insight: Heat Engines and Refrigerators

The Setup

The Master Equation

The Mathematical Symphony

Solving the Sequence

The Grand Finale

A Pro Tip for Exams

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Select Answer:

Visualized Solution

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

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

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

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

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Finding the Common Ratio r

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

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Solving for T3​

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

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