Question

A Carnot engine has an efficiency of \(\frac{1}{6}\). When the temperature of sink is reduced by 62 K , its efficiency is doubled. The temperature of the source is

• A. \(37^{\circ}\mathrm{C}\)
• B. \(62^{\circ}\mathrm{C}\)
• C. \(99^{\circ}\mathrm{C}\)
• D. \(124^{\circ}\mathrm{C}\)

Hint:

Write efficiency as 1 minus sink over source temperature for both cases and use the 62 K drop in sink temperature.

Answer 1

The Correct Option is C

Step 1: Understanding the Concept
Carnot efficiency is \(\eta = 1 - \dfrac{T_2}{T_1}\).

Step 2: First Case
\(\eta = \dfrac16\), so \(\dfrac{T_2}{T_1} = \dfrac56\), i.e. \(T_2 = \dfrac{5T_1}{6}\).

Step 3: Second Case
Sink becomes \(T_2 - 62\) and \(\eta = \dfrac13\), so \(\dfrac{T_2 - 62}{T_1} = \dfrac23\).

Step 4: Solve
Subtract: \(\dfrac{62}{T_1} = \dfrac56 - \dfrac23 = \dfrac16\), so \(T_1 = 372\ \text{K}\).
In Celsius: \(372 - 273 = 99^{\circ}\text{C}\), option C.

Final Answer:
The source temperature is 99 degrees Celsius. \[ \boxed{\text{(C) }99^{\circ}\text{C}} \]