Question:
The temperature of the sink of a Carnot engine is \(250\;K\). In order to increase the efficiency of the Carnot engine from \(25\%\) to \(50\%\), the temperature of the source should be increased by
The temperature of the sink of a Carnot engine is \(250\;K\). In order to increase the efficiency of the Carnot engine from \(25\%\) to \(50\%\), the temperature of the source should be increased by
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For a Carnot engine,
\[
\eta=1-\frac{T_{\text{sink}}}{T_{\text{source}}}
\]
Efficiency increases when the source temperature increases or sink temperature decreases.
Updated On: Jun 22, 2026
- \(\dfrac{1}{3}\times10^3\;K\)
- \(\dfrac{1}{2}\times10^3\;K\)
- \(200\;K\)
- \(\dfrac{1}{6}\times10^3\;K\)
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The Correct Option is D
Solution and Explanation
Step 1: Write the efficiency formula for a Carnot engine.
Efficiency of a Carnot engine is \[ \eta=1-\frac{T_2}{T_1} \] where \[ T_1=\text{source temperature} \] and \[ T_2=\text{sink temperature} \] Given, \[ T_2=250\;K \]
Step 2: Find the initial source temperature.
Initially, efficiency is \[ \eta_1=25\%=0.25 \] Using Carnot efficiency formula, \[ 0.25=1-\frac{250}{T_1} \] \[ \frac{250}{T_1}=0.75 \] \[ T_1=\frac{250}{0.75} \] \[ T_1=\frac{1000}{3}\;K \]
Step 3: Find the new source temperature.
Final efficiency is \[ \eta_2=50\%=0.5 \] So, \[ 0.5=1-\frac{250}{T_1'} \] \[ \frac{250}{T_1'}=0.5 \] \[ T_1'=\frac{250}{0.5} \] \[ T_1'=500\;K \]
Step 4: Find the increase in source temperature.
Required increase is \[ \Delta T=T_1'-T_1 \] \[ =500-\frac{1000}{3} \] \[ =\frac{1500-1000}{3} \] \[ =\frac{500}{3}\;K \] \[ =\frac{1}{6}\times10^3\;K \]
Step 5: Final conclusion.
Hence, the source temperature should be increased by \[ \boxed{\frac{1}{6}\times10^3\;K} \]
Efficiency of a Carnot engine is \[ \eta=1-\frac{T_2}{T_1} \] where \[ T_1=\text{source temperature} \] and \[ T_2=\text{sink temperature} \] Given, \[ T_2=250\;K \]
Step 2: Find the initial source temperature.
Initially, efficiency is \[ \eta_1=25\%=0.25 \] Using Carnot efficiency formula, \[ 0.25=1-\frac{250}{T_1} \] \[ \frac{250}{T_1}=0.75 \] \[ T_1=\frac{250}{0.75} \] \[ T_1=\frac{1000}{3}\;K \]
Step 3: Find the new source temperature.
Final efficiency is \[ \eta_2=50\%=0.5 \] So, \[ 0.5=1-\frac{250}{T_1'} \] \[ \frac{250}{T_1'}=0.5 \] \[ T_1'=\frac{250}{0.5} \] \[ T_1'=500\;K \]
Step 4: Find the increase in source temperature.
Required increase is \[ \Delta T=T_1'-T_1 \] \[ =500-\frac{1000}{3} \] \[ =\frac{1500-1000}{3} \] \[ =\frac{500}{3}\;K \] \[ =\frac{1}{6}\times10^3\;K \]
Step 5: Final conclusion.
Hence, the source temperature should be increased by \[ \boxed{\frac{1}{6}\times10^3\;K} \]
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