Question

A Carnot engine, whose efficiency is 40% takes heat from a source maintained at temperature 600K. To have an efficiency 60%, the intake temperature for the same exhaust temperature should be

• A. \(1800\) K
• B. \(900\) K
• C. \(720\) K
• D. \(360\) K

Hint:

Use \(\eta=1-\dfrac{T_2}{T_1}\) with the same exhaust temperature.

Answer 1

The Correct Option is B

Step 1: Understanding the Concept
The efficiency of a Carnot engine is \(\eta=1-\dfrac{T_{2}}{T_1}\), where \(T_2\) is the sink (exhaust) temperature.

Step 2: Key Formula or Approach
First find \(T_2\) from the 40% case, then the new source temperature for 60%.

Step 3: Detailed Explanation
\(0.4=1-\dfrac{T_2}{600}\), so \(T_2=360\) K.
For 60%: \(0.6=1-\dfrac{360}{T_1'}\), so \(\dfrac{360}{T_1'}=0.4\).
\[ T_1'=\frac{360}{0.4}=900\ \text{K} \]

Final Answer:
The source temperature must be 900 K, option (B). \[ \boxed{900\ \text{K (B)}} \]