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: Subtract the two relations
The sink temperature falls by 57 K, so the ratio $\dfrac{T_2}{T_1}$ changes from $\dfrac{5}{6}$ to $\dfrac{2}{3}$.

Step 2: Change in ratio
$\dfrac{57}{T_1}=\dfrac{5}{6}-\dfrac{2}{3}=\dfrac{1}{6}$.

Step 3: Solve
$T_1=6\times57=342\text{ K}$.

Final Answer:
Hence the source is at 342 K. \[ \boxed{\text{(C) }342\text{ K}} \]