A Carnot engine has an efficiency of 1/6. When the temperature of the sink is reduced by 62°C, its e

Q: Why didn't we convert $62^\circ\text{C}$ to Kelvin by adding 273?

Because 62 is a *change* in temperature ($\Delta T$), not an absolute temperature. The size of one degree Celsius is exactly the same as one Kelvin. So, a drop of $62^\circ\text{C}$ is identical to a drop of $62 \text{ K}$.

Q: Can I just use Celsius values directly in the $\eta = 1 - \frac{T_2}{T_1}$ formula?

Absolutely not! The Carnot efficiency formula is derived using the absolute thermodynamic temperature scale. Using Celsius will give you completely wrong, and sometimes even negative, efficiencies.

Q: How did the efficiency become $\frac{2}{6}$?

The problem explicitly states that when the sink temperature is reduced, the efficiency is *doubled*. Since the initial efficiency was $\frac{1}{6}$, doubling it gives $2 \times \frac{1}{6} = \frac{2}{6}$.

Q: Why does reducing the sink temperature increase the efficiency?

Look at the formula $\eta = 1 - \frac{T_2}{T_1}$. If you make $T_2$ (the sink) smaller, the fraction $\frac{T_2}{T_1}$ becomes smaller. Subtracting a smaller number from 1 leaves a larger efficiency. Physically, a colder sink means the engine can extract more useful work from the heat.

Q: What if the source temperature was increased instead of reducing the sink temperature?

Increasing the source temperature $T_1$ also decreases the fraction $\frac{T_2}{T_1}$, which would also increase the efficiency. The new equation would look like $\eta_{new} = 1 - \frac{T_2}{T_1 + \Delta T}$.

Question

A Carnot engine has an efficiency of 1/6. When the temperature of the sink is reduced by 62°C, its efficiency is doubled. The temperatures of the source and the sink are respectively,

Options:

(A) 62°C, 124°C

(B) 99°C, 37°C

(C) 124°C, 62°C

(D) 37°C, 99°C

Sigma Percentile · Accepted Answer

### The Ideal Engine: Carnot's Masterpiece

Imagine a perfect machine, one that extracts heat from a blazing hot source, converts a portion of it into useful mechanical work, and dumps the rest into a cold sink. This is the Carnot engine, the theoretical limit of thermodynamic efficiency. The efficiency $\eta$ of this ideal engine is governed by a beautifully simple equation:

$$\eta = 1 - \frac{T_2}{T_1}$$

Here, $T_1$ is the absolute temperature of the hot source, and $T_2$ is the absolute temperature of the cold sink. The word **absolute** is critical here—these temperatures must strictly be measured in Kelvin. Using Celsius in this ratio is a guaranteed path to a wrong answer.

### Decoding the First State

The problem presents us with a puzzle in two acts. In the first act, we are told the engine operates with an efficiency of $\frac{1}{6}$. Let's plug this directly into our master equation:

$$\frac{1}{6} = 1 - \frac{T_2}{T_1}$$

By rearranging this equation, we can isolate the ratio of the sink temperature to the source temperature:

$$\frac{T_2}{T_1} = 1 - \frac{1}{6} = \frac{5}{6}$$

We don't know the individual temperatures yet, but we know their ratio is locked at $\frac{5}{6}$. We will hold onto this key piece of information for the next act.

### The Temperature Trap: Celsius vs. Kelvin

Now comes the twist. The problem states: *"When the temperature of the sink is reduced by $62^\circ	ext{C}$, its efficiency is doubled."*

This is where many students fall into a classic trap. They see $62^\circ	ext{C}$ and immediately try to add $273$ to convert it to Kelvin. But wait! This is not an absolute temperature; it is a **change** in temperature ($\Delta T$). Because the size of one degree Celsius is exactly the same as the size of one Kelvin, a temperature drop of $62^\circ	ext{C}$ is physically identical to a temperature drop of $62 	ext{ K}$.

Therefore, our new sink temperature is simply $(T_2 - 62)$. The efficiency is doubled, so the new efficiency is $2 	imes \frac{1}{6} = \frac{2}{6}$.

### The Mathematical Symphony

Let's set up the efficiency equation for this new, modified state:

$$\frac{2}{6} = 1 - \frac{T_2 - 62}{T_1}$$

We can split the fraction on the right side to make it easier to digest:

$$\frac{2}{6} = 1 - \frac{T_2}{T_1} + \frac{62}{T_1}$$

Notice how the double negative turned into a positive $\frac{62}{T_1}$. Now, remember that ratio we found in the first act? We know that $\frac{T_2}{T_1} = \frac{5}{6}$. Let's substitute that right into our new equation:

$$\frac{2}{6} = 1 - \frac{5}{6} + \frac{62}{T_1}$$

Since $1 - \frac{5}{6}$ is just $\frac{1}{6}$, the equation simplifies beautifully:

$$\frac{2}{6} = \frac{1}{6} + \frac{62}{T_1}$$

Subtracting $\frac{1}{6}$ from both sides leaves us with:

$$\frac{1}{6} = \frac{62}{T_1}$$

Cross-multiplying gives us the absolute temperature of the source:

$$T_1 = 62 	imes 6 = 372 	ext{ K}$$

### The Final Reveal

We have the source temperature in Kelvin, but our options are in Celsius. To convert back, we subtract $273$:

$$T_1 = 372 - 273 = 99^\circ	ext{C}$$

Now, what about the sink temperature, $T_2$? We can use our trusty ratio $\frac{T_2}{T_1} = \frac{5}{6}$:

$$T_2 = \frac{5}{6} 	imes 372 = 5 	imes 62 = 310 	ext{ K}$$

Converting this to Celsius:

$$T_2 = 310 - 273 = 37^\circ	ext{C}$$

And there we have it! The temperature of the source is **$99^\circ	ext{C}$** and the temperature of the sink is **$37^\circ	ext{C}$**. This perfectly matches option (b). By carefully navigating the units and trusting the algebra, the solution unfolds naturally.

Animated Solution for Physics - Thermodynamics: A Carnot engine has an efficiency of 1/6. When the temperature of the sink is reduced by 62°C, its efficiency is doubled. The temperatures of the source and the sink are respectively,

Select Answer:

Visualized Solution

The Carnot Engine

Initial State

Modified State

Setting up the New Equation

Substituting the Ratio

Solving for Source Temperature

Solving for Sink Temperature

Final Answer

The Sigma Insight: Heat Engines and Refrigerators

The Ideal Engine

Decoding the First State

The Temperature Trap

The Mathematical Symphony

The Final Reveal

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