Question

A Carnot engine has an efficiency of 50%. If the temperature of the sink is reduced by \( 40^\circ C \), its efficiency increases by 30%. The temperature of the source will be:

• A. 166.7K
• B. 255.1K
• C. 266.7K
• D. 367.7K

Hint:

To solve for the temperature of the source in a Carnot engine, use the relationship between efficiency and the temperatures of the source and sink: \( \eta = 1 - \frac{T_{{sink}}}{T_{{source}}} \).

Answer 1

The Correct Option is C

Step 1: The efficiency of a Carnot engine is given by: \[ \eta = 1 - \frac{T_{{sink}}}{T_{{source}}}. \] Step 2: The initial efficiency is \( \eta = 0.5 \), so: \[ 0.5 = 1 - \frac{T_{{sink}}}{T_{{source}}}. \] This gives: \[ \frac{T_{{sink}}}{T_{{source}}} = 0.5 \quad \Rightarrow \quad T_{{sink}} = 0.5 T_{{source}}. \] Step 3: After reducing the sink temperature by \( 40^\circ C \), the efficiency increases by 30\%. The new efficiency is \( 0.65 \), so: \[ 0.65 = 1 - \frac{T_{{sink}}'}{T_{{source}}}. \] Substitute \( T_{{sink}}' = T_{{sink}} - 40 \) into the equation and solve for \( T_{{source}} \). 
Step 4: The temperature of the source is found to be \( T_{{source}} = 266.7 \, {K} \).