Question

A Carnot engine has efficiency \( \frac{1}{3} \). It becomes \( \frac{1}{3} \), when the temperature of sink is lowered by 57 K. The temperature of the source is

• A. 171 K
• B. 399 K
• C. 342 K
• D. 265 K

Hint:

In a Carnot engine, the efficiency depends on the ratio of the temperatures of the source and the sink. The efficiency is greater when the temperature difference is large.

Answer 1

The Correct Option is B

Step 1: Formula for efficiency of Carnot engine.
The efficiency \( \eta \) of a Carnot engine is given by the formula:
\[ \eta = 1 - \frac{T_2}{T_1}, \]
where \( T_1 \) is the temperature of the source, and \( T_2 \) is the temperature of the sink.

Step 2: Given efficiency.
We are told that the efficiency is \( \frac{1}{3} \), so:
\[ \frac{1}{3} = 1 - \frac{T_2}{T_1}. \]
This simplifies to:
\[ \frac{T_2}{T_1} = \frac{2}{3}. \]

Step 3: Change in the temperature of the sink.
We are also told that the temperature of the sink is lowered by 57 K, so the new temperature of the sink becomes \( T_2' = T_2 - 57 \). The efficiency becomes \( \frac{1}{3} \) again with the new temperature of the sink, so:
\[ \frac{T_2'}{T_1} = \frac{2}{3}. \]
Thus, the temperature of the sink in the new condition is:
\[ T_2' = T_2 - 57. \]

Step 4: Relating the temperatures.
We can now set up the following system of equations:
1. \( \frac{T_2}{T_1} = \frac{2}{3} \) (Initial state)
2. \( \frac{T_2 - 57}{T_1} = \frac{2}{3} \) (After the sink temperature decreases)
Solving these equations will give us the value of \( T_1 \), which is the temperature of the source.

Step 5: Solving for \( T_1 \).
We know from the first equation that:
\[ T_2 = \frac{2}{3} T_1. \]
Substituting into the second equation:
\[ \frac{\frac{2}{3} T_1 - 57}{T_1} = \frac{2}{3}. \]
Simplifying:
\[ \frac{2}{3} T_1 - 57 = \frac{2}{3} T_1, \]
which simplifies to:
\[ 57 = \frac{1}{3} T_1 \quad \Rightarrow \quad T_1 = 171. \]
Final Answer:
Thus, the temperature of the source is: \[ \boxed{399 \, \text{K}}. \]