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: Understanding the Question:
Efficiency of a Carnot engine depends on the temperatures of the source (\( T_1 \)) and sink (\( T_2 \)). We have two conditions to find the source temperature.
Step 2: Key Formula or Approach:
Efficiency \( \eta = 1 - \frac{T_2}{T_1} \).
Step 3: Detailed Explanation:
Case 1: \( \eta_1 = 1/6 \).
\[ \frac{1}{6} = 1 - \frac{T_2}{T_1} \Rightarrow \frac{T_2}{T_1} = \frac{5}{6} \Rightarrow T_2 = \frac{5}{6} T_1 \]
Case 2: Efficiency \( \eta_2 = 1/3 \) when sink is lowered by 57 K. New sink temp is \( T_2 - 57 \).
\[ \frac{1}{3} = 1 - \frac{T_2 - 57}{T_1} \]
\[ \frac{T_2 - 57}{T_1} = 1 - \frac{1}{3} = \frac{2}{3} \]
\[ T_2 - 57 = \frac{2}{3} T_1 \]
Substitute \( T_2 = \frac{5}{6} T_1 \) into this equation:
\[ \frac{5}{6} T_1 - 57 = \frac{2}{3} T_1 \]
\[ \frac{5}{6} T_1 - \frac{4}{6} T_1 = 57 \]
\[ \frac{1}{6} T_1 = 57 \]
\[ T_1 = 57 \times 6 = 342\text{ K} \]
Step 4: Final Answer:
The temperature of the source is \( 342\text{ K} \).