Question:
The efficiency of a Carnot engine is \(50\%\) and the temperature of the sink is \(500\,\text{K}\). Keeping the source temperature constant, the required sink temperature to raise the efficiency of the engine to \(60\%\) is
The efficiency of a Carnot engine is \(50\%\) and the temperature of the sink is \(500\,\text{K}\). Keeping the source temperature constant, the required sink temperature to raise the efficiency of the engine to \(60\%\) is
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For a Carnot engine,
\[
\eta=1-\frac{T_2}{T_1}.
\]
To increase efficiency while keeping the source temperature fixed, the sink temperature must be decreased.
Updated On: Jul 9, 2026
- \(100\,\text{K}\)
- \(400\,\text{K}\)
- \(500\,\text{K}\)
- \(600\,\text{K}\) \bigskip
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The Correct Option is B
Solution and Explanation
Concept:
The efficiency of a Carnot engine is
\[
\eta
=
1-\frac{T_2}{T_1},
\]
where
\[
T_1=\text{source temperature},
\qquad
T_2=\text{sink temperature}.
\]
Step 1: Find the source temperature. Initially, \[ \eta=50\%=0.5, \qquad T_2=500\,\text{K}. \] Using \[ 0.5 = 1-\frac{500}{T_1}, \] \[ \frac{500}{T_1} = 0.5. \] \[ T_1 = 1000\,\text{K}. \]
Step 2: Calculate the new sink temperature for \(60\%\) efficiency. Given, \[ \eta=60\%=0.6. \] Using \[ 0.6 = 1-\frac{T_2'}{1000}, \] \[ \frac{T_2'}{1000} = 0.4. \] \[ T_2' = 400\,\text{K}. \]
Step 3: Write the final answer. \[ \boxed{T_2'=400\,\text{K}} \] \[ \boxed{\text{Answer = (B)}} \]
Step 1: Find the source temperature. Initially, \[ \eta=50\%=0.5, \qquad T_2=500\,\text{K}. \] Using \[ 0.5 = 1-\frac{500}{T_1}, \] \[ \frac{500}{T_1} = 0.5. \] \[ T_1 = 1000\,\text{K}. \]
Step 2: Calculate the new sink temperature for \(60\%\) efficiency. Given, \[ \eta=60\%=0.6. \] Using \[ 0.6 = 1-\frac{T_2'}{1000}, \] \[ \frac{T_2'}{1000} = 0.4. \] \[ T_2' = 400\,\text{K}. \]
Step 3: Write the final answer. \[ \boxed{T_2'=400\,\text{K}} \] \[ \boxed{\text{Answer = (B)}} \]
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