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. \(300 \text{K}\)
• B. \(900 \text{K}\)
• C. \(1000 \text{K}\)
• D. \(360 \text{K}\)

Hint:

Efficiency is 1 minus T2/T1. First find the sink temperature from the 50 percent efficiency.

Answer 1

The Correct Option is C

Step 1: Sink first
Half of the heat goes to the sink, so $T_2 = \frac{600}{2} = 300$ K.

Step 2: New ratio
For $\eta = 0.7$, $\frac{T_2}{T_1'} = 0.3$, so $T_1' = \frac{300}{0.3} = 1000$ K. Option (C).

Final Answer:
1000 K. \[ \boxed{\text{(C)}\ 1000\ \text{K}} \]