Question:
A Carnot engine with efficiency 40% takes heat from a source at 500 K. To increase the efficiency to 60%, keeping temperature of the sink same, the new temperature of the source will be
A Carnot engine with efficiency 40% takes heat from a source at 500 K. To increase the efficiency to 60%, keeping temperature of the sink same, the new temperature of the source will be
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Find the sink temperature from the first efficiency and use it again for the second.
Updated On: Nov 13, 2026
- 900 K
- 750 K
- 1000 K
- 450 K
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The Correct Option is B
Solution and Explanation
Step 1: Understanding the Concept
Carnot efficiency is \(\eta=1-\dfrac{T_2}{T_1}\).
Step 2: Sink temperature
\(0.4=1-\dfrac{T_2}{500}\), so \(T_2=300\ \text{K}\).
Step 3: New source
\(0.6=1-\dfrac{300}{T_1'}\), so \(\dfrac{300}{T_1'}=0.4\).
Step 4: Calculation
\(T_1'=750\ \text{K}\).
Final Answer:
The new source temperature is 750 K. \[ \boxed{\text{(B) }750\text{ K}} \]
Carnot efficiency is \(\eta=1-\dfrac{T_2}{T_1}\).
Step 2: Sink temperature
\(0.4=1-\dfrac{T_2}{500}\), so \(T_2=300\ \text{K}\).
Step 3: New source
\(0.6=1-\dfrac{300}{T_1'}\), so \(\dfrac{300}{T_1'}=0.4\).
Step 4: Calculation
\(T_1'=750\ \text{K}\).
Final Answer:
The new source temperature is 750 K. \[ \boxed{\text{(B) }750\text{ K}} \]
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