Question

A Carnot engine operating between temperatures \(T_1\) and \(T_2\) has efficiency 0.2. When \(T_2\) is reduced by 50 K, its efficiency increases to 0.4. Then \(T_1\) and \(T_2\) are respectively

• A. 200 K, 150 K
• B. 250 K, 200 K
• C. 300 K, 250 K
• D. 300 K, 200 K
• E. 300 K, 150 K

Hint:

Always use Absolute Temperature (Kelvin) in thermodynamic efficiency formulas.

Answer 1

The Correct Option is B

Concept: Efficiency (\(\eta\)) of a Carnot engine: \(\eta = 1 - \frac{T_2}{T_1}\).

Step 1: Set up equations.
Case 1: \(0.2 = 1 - \frac{T_2}{T_1} \implies \frac{T_2}{T_1} = 0.8\)
Case 2: \(0.4 = 1 - \frac{T_2 - 50}{T_1} \implies \frac{T_2 - 50}{T_1} = 0.6\)

Step 2: Solve for temperatures.
Subtracting equations: \(\frac{50}{T_1} = 0.2 \implies T_1 = 250 \text{ K}\).
Using Case 1: \(T_2 = 0.8 \times 250 = 200 \text{ K}\).