Question

A Carnot engine has efficiency \(\frac{1}{6}\). It becomes \(\frac{1}{3}\), when the temperature of \(\operatorname{sink}\) is lowered by

57 K . The temperature of the source is

• A. 171 K
• B. 399 K
• C. 342 K
• D. 285 K

Hint:

Write the efficiency before and after lowering the sink temperature and solve for the source temperature.

Answer 1

The Correct Option is C

Step 1: Understanding the Concept
Carnot efficiency is \(\eta=1-\dfrac{T_2}{T_1}\), where \(T_1\) is the source and \(T_2\) the sink temperature.

Step 2: First case
\(\dfrac{1}{6}=1-\dfrac{T_2}{T_1}\), so \(T_2=\dfrac{5}{6}T_1\).

Step 3: Second case
\(\dfrac{1}{3}=1-\dfrac{T_2-57}{T_1}\), so \(T_2-57=\dfrac{2}{3}T_1\).

Step 4: Calculation
Substituting: \(\dfrac{5}{6}T_1-57=\dfrac{2}{3}T_1\), so \(\dfrac{T_1}{6}=57\) and \(T_1=342\text{ K}\).

Final Answer:
The source temperature is 342 K. \[ \boxed{\text{(C) }342\text{ K}} \]