Sankalp NEET Full Test-2 Question-22 Solution

Question 22: A Carnot engine has an efficiency of 50% when its source is at a temperature 327^{\circ}\mathrm{C}. The temperature of the sink is :

(1) 100^{\circ}\mathrm{C}

(2) 200^{\circ}\mathrm{C}

(3) 27^{\circ}\mathrm{C}

(4) 15^{\circ}\mathrm{C}

Answer: Option (3)

Explanation:

The efficiency of a Carnot engine is given by

\eta = 1 - \frac{T_2}{T_1}

where T_1 is the absolute temperature of the source and

T_2 is the absolute temperature of the sink.

The temperature of the source is

T_1 = 327 + 273 = 600 \, \mathrm{K}

Given efficiency is

\eta = 50\% = 0.5

Substituting the values,

0.5 = 1 - \frac{T_2}{600} \frac{T_2}{600} = 0.5 T_2 = 300 \, \mathrm{K}

Converting the sink temperature into degree Celsius,

T_2 = 300 - 273 = 27^{\circ}\mathrm{C}

Hence, the temperature of the sink is 27^{\circ}\mathrm{C}.

Therefore, the correct answer is Option (3).