Question

The efficiency of a Carnot engine is found to increase from 25% to 40% on increasing the temperature (\(T_1\)) of the source alone through 100 K. The temperature (\(T_2\)) of the sink is given by:

• A. \( 300 \, K \)
• B. \( 250 \, K \)
• C. \( 325 \, K \)
• D. \( 125 \, K \)

Hint:

For Carnot engine problems, always start with the efficiency formula. If efficiency changes, set up equations for both initial and final states and solve simultaneously.

Answer 1

The Correct Option is A

Step 1: Define Efficiency Formula The efficiency of a Carnot engine is given by: \[ \eta = 1 - \frac{T_2}{T_1} \] where: \( T_1 \) = Temperature of the heat source \( T_2 \) = Temperature of the heat sink Step 2: Define Given Conditions Initially, the efficiency is given as 25%, i.e., \[ \frac{T_2}{T_1} = 1 - 0.25 = 0.75 \] which gives: \[ T_2 = 0.75 T_1 \] After increasing \( T_1 \) by 100 K, the new efficiency is 40%, i.e., \[ \frac{T_2}{T_1 + 100} = 1 - 0.40 = 0.60 \] which gives: \[ T_2 = 0.60 (T_1 + 100) \] Step 3: Solve for \( T_2 \) Equating both expressions for \( T_2 \): \[ 0.75 T_1 = 0.60 (T_1 + 100) \] Expanding and solving: \[ 0.75 T_1 = 0.60 T_1 + 60 \] \[ 0.75 T_1 - 0.60 T_1 = 60 \] \[ 0.15 T_1 = 60 \] \[ T_1 = \frac{60}{0.15} = 400 \, K \] Now, substituting \( T_1 = 400 K \) into \( T_2 = 0.75 T_1 \): \[ T_2 = 0.75 \times 400 = 300 \, K \] Thus, the correct answer is \( 300 K \).