Question:
A Carnot engine operates between a source and a sink. The efficiency of the engine is \(40\%\) and the temperature of the sink is \(27^\circ\text{C}\). If the efficiency is to be increased to \(50\%\), then the temperature of the source must be increased by
A Carnot engine operates between a source and a sink. The efficiency of the engine is \(40\%\) and the temperature of the sink is \(27^\circ\text{C}\). If the efficiency is to be increased to \(50\%\), then the temperature of the source must be increased by
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For a Carnot engine, efficiency depends only on absolute temperatures: \(\eta=1-\frac{T_2}{T_1}\). Always convert Celsius to Kelvin before calculation.
Updated On: Jun 15, 2026
- \(80\ \text{K}\)
- \(120\ \text{K}\)
- \(100\ \text{K}\)
- \(160\ \text{K}\)
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The Correct Option is C
Solution and Explanation
Step 1: Convert sink temperature into Kelvin.
Given sink temperature is
\[ 27^\circ\text{C} \]
So,
\[ T_2=27+273=300\ \text{K} \]
Step 2: Use Carnot efficiency formula.
Efficiency of a Carnot engine is
\[ \eta=1-\frac{T_2}{T_1} \]
Initially,
\[ \eta=40\%=0.4 \]
So,
\[ 0.4=1-\frac{300}{T_1} \]
\[ \frac{300}{T_1}=0.6 \]
\[ T_1=\frac{300}{0.6}=500\ \text{K} \]
Step 3: Find new source temperature for \(50\%\) efficiency.
New efficiency is
\[ \eta'=50\%=0.5 \]
Therefore,
\[ 0.5=1-\frac{300}{T_1'} \]
\[ \frac{300}{T_1'}=0.5 \]
\[ T_1'=\frac{300}{0.5}=600\ \text{K} \]
Step 4: Find increase in source temperature.
\[ \Delta T=T_1'-T_1 \]
\[ =600-500 \]
\[ =100\ \text{K} \]
Step 5: Final conclusion.
Hence, the source temperature must be increased by
\[ \boxed{100\ \text{K}} \]
Given sink temperature is
\[ 27^\circ\text{C} \]
So,
\[ T_2=27+273=300\ \text{K} \]
Step 2: Use Carnot efficiency formula.
Efficiency of a Carnot engine is
\[ \eta=1-\frac{T_2}{T_1} \]
Initially,
\[ \eta=40\%=0.4 \]
So,
\[ 0.4=1-\frac{300}{T_1} \]
\[ \frac{300}{T_1}=0.6 \]
\[ T_1=\frac{300}{0.6}=500\ \text{K} \]
Step 3: Find new source temperature for \(50\%\) efficiency.
New efficiency is
\[ \eta'=50\%=0.5 \]
Therefore,
\[ 0.5=1-\frac{300}{T_1'} \]
\[ \frac{300}{T_1'}=0.5 \]
\[ T_1'=\frac{300}{0.5}=600\ \text{K} \]
Step 4: Find increase in source temperature.
\[ \Delta T=T_1'-T_1 \]
\[ =600-500 \]
\[ =100\ \text{K} \]
Step 5: Final conclusion.
Hence, the source temperature must be increased by
\[ \boxed{100\ \text{K}} \]
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