Question:
The efficiency of a Carnot engine working between temperatures \(T_1\) and \(T_2\) (where \(T_1 > T_2\)) is given by:
The efficiency of a Carnot engine working between temperatures \(T_1\) and \(T_2\) (where \(T_1 > T_2\)) is given by:
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Always remember: efficiency depends only on temperature ratio for Carnot engines, not on working substance.
Updated On: Jun 10, 2026
- \( \eta = 1 - \frac{T_1}{T_2} \)
- \( \eta = 1 - \frac{T_2}{T_1} \)
- \( \eta = \frac{T_1}{T_2} - 1 \)
- \( \eta = \frac{T_2}{T_1} \)
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The Correct Option is B
Solution and Explanation
Concept:
A Carnot engine is an ideal reversible heat engine operating between two reservoirs at temperatures \(T_1\) (source) and \(T_2\) (sink). It gives the maximum possible efficiency for any heat engine.
Efficiency is defined as:
\[
\eta = \frac{W}{Q_1}
\]
Step 1: Work done in one cycle From first law of thermodynamics: \[ W = Q_1 - Q_2 \] So: \[ \eta = \frac{Q_1 - Q_2}{Q_1} = 1 - \frac{Q_2}{Q_1} \]
Step 2: Carnot condition For a reversible Carnot engine: \[ \frac{Q_2}{Q_1} = \frac{T_2}{T_1} \]
Step 3: Final efficiency Substituting: \[ \eta = 1 - \frac{T_2}{T_1} \] Thus: \[ \boxed{\eta = 1 - \frac{T_2}{T_1}} \]
Step 1: Work done in one cycle From first law of thermodynamics: \[ W = Q_1 - Q_2 \] So: \[ \eta = \frac{Q_1 - Q_2}{Q_1} = 1 - \frac{Q_2}{Q_1} \]
Step 2: Carnot condition For a reversible Carnot engine: \[ \frac{Q_2}{Q_1} = \frac{T_2}{T_1} \]
Step 3: Final efficiency Substituting: \[ \eta = 1 - \frac{T_2}{T_1} \] Thus: \[ \boxed{\eta = 1 - \frac{T_2}{T_1}} \]
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