Question

The efficiency of a Carnot engine operating between a reservoir and an sink is \( \eta \). If the temperature of the reservoir is increased by 50% and that of the sink is decreased by 50%, the new efficiency will be:

• A. \( \frac{\eta}{2} \)
• B. \( 4\eta \)
• C. \( \frac{200\eta}{3} \)
• D. \( \frac{\eta}{4} \)

Hint:

In problems involving the efficiency of a Carnot engine, remember that efficiency depends on the ratio of the temperatures of the hot and cold bodies. When the temperatures of either body are changed, adjust the equation accordingly and solve for the new efficiency.

Answer 1

The Correct Option is C

The efficiency of a Carnot engine is given by the formula: \[ \eta = 1 - \frac{T_2}{T_1} \] Where: - \( \eta \) is the efficiency, - \( T_1 \) is the temperature of the reservoir (hot body), - \( T_2 \) is the temperature of the sink (cold body). Now, if the temperature of the reservoir is increased by 50%, the new temperature \( T_1' \) will be: \[ T_1' = 1.5 \times T_1 \] If the temperature of the sink is decreased by 50%, the new temperature \( T_2' \) will be: \[ T_2' = 0.5 \times T_2 \] Thus, the new efficiency \( \eta' \) will be: \[ \eta' = 1 - \frac{T_2'}{T_1'} = 1 - \frac{0.5 \times T_2}{1.5 \times T_1} \] Simplifying: \[ \eta' = 1 - \frac{T_2}{3T_1} = \frac{3T_1 - T_2}{3T_1} \] Since \( \eta = 1 - \frac{T_2}{T_1} \), the original efficiency is \( \eta = \frac{T_1 - T_2}{T_1} \). Thus, the new efficiency \( \eta' \) becomes: \[ \eta' = \frac{3(T_1 - T_2)}{3T_1} = \frac{200 \eta}{3} \] Hence, the new efficiency is \( \frac{200\eta}{3} \).