Question

The efficiency of a Carnot's engine is 25%, when the temperature of sink is 300 K. The increase in the temperature of source required for the efficiency to become 50% is

• A. \( 225 \, {K} \)
• B. \( 400 \, {K} \)
• C. \( 200 \, {K} \)
• D. \( 100 \, {K} \)

Hint:

When solving problems involving Carnot engines, always remember that the efficiency is determined by the temperatures of the hot and cold reservoirs.

Answer 1

The Correct Option is C

The efficiency \(\eta\) of a Carnot engine is given by: \[ \eta = 1 - \frac{T_c}{T_h} \] where \(T_c\) is the temperature of the cold reservoir (sink) and \(T_h\) is the temperature of the hot reservoir (source). Initially, the efficiency is 25%, with \(T_c = 300 \, {K}\), so: \[ 0.25 = 1 - \frac{300}{T_h} \Rightarrow T_h = \frac{300}{0.75} = 400 \, {K} \] To achieve 50% efficiency: \[ 0.50 = 1 - \frac{300}{T_h'} \Rightarrow T_h' = \frac{300}{0.50} = 600 \, {K} \] The increase in the temperature of the source is: \[ \Delta T = T_h' - T_h = 600 \, {K} - 400 \, {K} = 200 \, {K} \]