Question:

A Carnot engine operates between 600 K and 300 K. Efficiency is:

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To maximize the efficiency of a Carnot engine, you must either increase the temperature of the source (\( T_H \)) or decrease the temperature of the sink (\( T_L \)). A 100% efficient engine is theoretically impossible as it would require a sink at absolute zero (0 K).
Updated On: Jun 3, 2026
  • 25%
  • 50%
  • 75%
  • 100%
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The Correct Option is B

Solution and Explanation

Concept: The efficiency (\( \eta \)) of a Carnot engine, which is a theoretical ideal thermodynamic cycle, depends solely on the absolute temperatures of the hot reservoir (source) and the cold reservoir (sink).
• Formula: \( \eta = 1 - \frac{T_L}{T_H} \)
• Efficiency in percentage: \( \eta\% = \left( 1 - \frac{T_L}{T_H} \right) \times 100 \)
• \( T_H \): Temperature of the hot reservoir (Source).
• \( T_L \): Temperature of the cold reservoir (Sink).

Step 1:
Identifying the given temperatures.
Temperature of the source (\( T_H \)) = 600 K
Temperature of the sink (\( T_L \)) = 300 K

Step 2:
Substituting values into the efficiency formula.
\[ \eta = 1 - \frac{300}{600} \] \[ \eta = 1 - \frac{1}{2} = 0.5 \]

Step 3:
Converting to percentage.
\[ \eta\% = 0.5 \times 100 = 50\% \]
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