Medium MCQ +4 / -1 PYQ · JEE Mains 2020

A carnot engine having an efficiency of ${1 \over {10}}$ is being used as a refrigerator. If the work done on the refrigerator is 10 J, the amount of heat absorbed from the reservoir at lower temperature is :

  1. A 90 J Correct answer
  2. B 99 J
  3. C 1 J
  4. D 100 J

Solution

n = $1 - {{{T_C}} \over {{T_H}}}$ <br><br>$\Rightarrow$ ${1 \over {10}}$ = $1 - {{{T_C}} \over {{T_H}}}$ <br><br>$\Rightarrow$ ${{{T_C}} \over {{T_H}}} = {9 \over {10}}$ <br><br>As for carnot engine T $\propto$ Q <br><br>$\therefore$ ${{{Q_C}} \over {{Q_H}}} = {9 \over {10}}$ <br><br>$\Rightarrow$ ${{Q_H} = {{10} \over 9}{Q_C}}$ <br><br>As W = Q<subh> - Q<sub>C</sub> <br><br>$\Rightarrow$ 10 = ${{{10} \over 9}{Q_C}}$ - Q<sub>C</sub> <br><br>$\Rightarrow$ Q<sub>C</sub> = 90 J</subh>

About this question

Subject: Physics · Chapter: Thermodynamics · Topic: Zeroth and First Law

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