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: $$\Delta T = T_h' - T_h = 600 \, {K} - 400 \, {K} = 200 \, {K}$$