Published by:
CGP EDU Academic Team
Published on: September 12, 2026
The efficiency of Carnot engine when source temperature is T 1 and sink temperature is T 2 will be
$(a) \frac{T_1 - T_2}{T_1} \quad (b) \frac{T_2 - T_1}{T_2} \quad (c) \frac{T_1 - T_2}{T_2} \quad (d) \frac{T_1}{T_2}$
Text Solution
Verified by ExpertsThe correct answer is:
A
The efficiency \( \eta \) of a Carnot engine is given by the formula:
\( \eta = 1 - \frac{T_2}{T_1} \)
Step 1: The efficiency can also be expressed as:
\( \eta = \frac{T_1 - T_2}{T_1} \)
Step 2: Rearranging this, we get:
\( \eta = \frac{T_1 - T_2}{T_1} = \frac{T_1 - T_2}{T_1} \)
Thus, the correct answer is option (a): \( \frac{T_1 - T_2}{T_1} \). Therefore, A.
\( \eta = 1 - \frac{T_2}{T_1} \)
Step 1: The efficiency can also be expressed as:
\( \eta = \frac{T_1 - T_2}{T_1} \)
Step 2: Rearranging this, we get:
\( \eta = \frac{T_1 - T_2}{T_1} = \frac{T_1 - T_2}{T_1} \)
Thus, the correct answer is option (a): \( \frac{T_1 - T_2}{T_1} \). Therefore, A.
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