Question

A Carnot engine is working between $127^{\circ}C$ and $27^{\circ}C$. Keeping the sink temperature unaltered, the temperature at which the source has to be kept so as to double its efficiency is ________.

• A. $400^{\circ}C$
• B. $273^{\circ}C$
• C. $327^{\circ}C$
• D. $525^{\circ}C$
• E. $600^{\circ}C$

Hint:

Always convert Celsius to Kelvin ($T_{K} = T_{C} + 273$) before using thermodynamics formulas.

Answer 1

The Correct Option is C

Step 1: Concept
Efficiency $\eta = 1 - \frac{T_2}{T_1}$ (temperatures in Kelvin).

Step 2: Meaning
$T_2 = 27 + 273 = 300~K$. Initial $T_1 = 127 + 273 = 400~K$. Initial $\eta = 1 - \frac{300}{400} = 0.25$.

Step 3: Analysis
New efficiency $\eta' = 2 \times 0.25 = 0.5$. $0.5 = 1 - \frac{300}{T'_1} \implies \frac{300}{T'_1} = 0.5 \implies T'_1 = 600~K$.

Step 4: Conclusion
$T'_1 = 600 - 273 = 327^{\circ}C$. Final Answer: (C)