Question

When the temperature of the source of a Carnot engine is at 400 K, its efficiency is 25%. The required increase in temperature of the source to increase the efficiency to 50% is

• A. 800 K
• B. 600K
• C. 100 K
• D. 400 K
• E. 200 K

Answer 1

The Correct Option is D

In a Carnot engine, the efficiency \( \eta \) is given by the formula: 

\[ \eta = 1 - \frac{T_C}{T_H} \] Where: - \( \eta \) is the efficiency of the engine, - \( T_C \) is the temperature of the cold reservoir (in Kelvin), - \( T_H \) is the temperature of the hot reservoir (in Kelvin). Given that the initial temperature of the source (hot reservoir) is \( T_H = 400 \, \text{K} \), and the efficiency is 25%, we can write: \[ 0.25 = 1 - \frac{T_C}{400} \] Solving for \( T_C \): \[ \frac{T_C}{400} = 0.75 \quad \Rightarrow \quad T_C = 0.75 \times 400 = 300 \, \text{K} \] Now, we are asked to find the increase in temperature of the source when the efficiency increases to 50%. Using the efficiency formula again for \( \eta = 0.5 \): \[ 0.5 = 1 - \frac{T_C}{T_H'} \] Where \( T_H' \) is the new temperature of the source. Solving for \( T_H' \): \[ \frac{T_C}{T_H'} = 0.5 \quad \Rightarrow \quad T_H' = \frac{T_C}{0.5} = \frac{300}{0.5} = 600 \, \text{K} \] The required increase in the temperature of the source is: \[ \Delta T = T_H' - T_H = 600 - 400 = 200 \, \text{K} \]

Correct Answer:

Correct Answer: (D) 400 K