Question

A Carnot engine has sink at 250 K. If efficiency increases from 25% to 50%, by how much must sink’s temperature increase?

• A. \( \frac{1}{3} \times 10^2 \text{K} \)
• B. \( \frac{1}{2} \times 10^2 \text{K} \)
• C. 200 K
• D. \( \frac{1}{6} \times 10^2 \text{K} \)

Hint:

Use Carnot efficiency formula: \( \eta = 1 - \frac{T_C}{T_H} \), solve for unknown temp.

Answer 1

The Correct Option is D

Efficiency: \( \eta = 1 - \frac{T_C}{T_H} \) At 25%: \[ 0.25 = 1 - \frac{250}{T_H} \Rightarrow T_H = \frac{250}{0.75} = 333.33\ \text{K} \] At 50%: \[ 0.5 = 1 - \frac{T_C'}{T_H} \Rightarrow T_C' = 0.5 \cdot 333.33 = 166.67\ \text{K} \] Increase in temp = \( 166.67 - 250 = -83.33 \), but correction: sink is cold end, so actually high temp end must be increased. Use: \[ T_H = \frac{250}{1 - 0.25} = 333.33,\quad T_H' = \frac{250}{1 - 0.5} = 500 \Rightarrow \Delta T = 500 - 333.33 = 166.67\ \text{K} = \frac{1}{6} \times 10^3 = \frac{1}{6} \cdot 1000 = 166.67 \]