Question:
A Carnot engine with efficiency 50% takes heat from a source at 600 K. To increase the efficiency by 20%, keeping temperature of the sink same, the new temperature of the source will be
A Carnot engine with efficiency 50% takes heat from a source at 600 K. To increase the efficiency by 20%, keeping temperature of the sink same, the new temperature of the source will be
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Efficiency is 1 minus T2/T1. First find the sink temperature from the 50 percent efficiency.
Updated On: Oct 1, 2026
- \(300 \text{K}\)
- \(900 \text{K}\)
- \(1000 \text{K}\)
- \(360 \text{K}\)
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The Correct Option is C
Solution and Explanation
Step 1: Find sink temperature
\(\eta = 1 - \frac{T_2}{T_1}\), so \(0.5 = 1 - \frac{T_2}{600}\) gives \(T_2 = 300\ \text{K}\).
Step 2: New efficiency
The efficiency increases by 20 percentage points, from \(50\%\) to \(70\%\).
Step 3: New source temperature
\(0.7 = 1 - \frac{300}{T_1'}\) gives \(\frac{300}{T_1'} = 0.3\), so \(T_1' = 1000\ \text{K}\). Option (C).
Final Answer:
The new source temperature is 1000 K. \[ \boxed{\text{(C)}\ 1000\ \text{K}} \]
\(\eta = 1 - \frac{T_2}{T_1}\), so \(0.5 = 1 - \frac{T_2}{600}\) gives \(T_2 = 300\ \text{K}\).
Step 2: New efficiency
The efficiency increases by 20 percentage points, from \(50\%\) to \(70\%\).
Step 3: New source temperature
\(0.7 = 1 - \frac{300}{T_1'}\) gives \(\frac{300}{T_1'} = 0.3\), so \(T_1' = 1000\ \text{K}\). Option (C).
Final Answer:
The new source temperature is 1000 K. \[ \boxed{\text{(C)}\ 1000\ \text{K}} \]
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