Q95 / 120 · JEE Main 2007 · Physics

A Carnot engine, having an efficiency of $\eta=1 / 10$ as heat engine, is used as a refrigerator. If the work done on the system is $10 \mathrm{~J}$, the amount of energy absorbed from the reservoir at lower temperature is

  • A
    $99 \mathrm{~J}$
  • Correct answer
    B
    $90 \mathrm{~J}$
  • C
    $1 \mathrm{~J}$
  • D
    $100 \mathrm{~J}$

Solution

$\begin{aligned} \mathrm{W} & =\mathrm{Q}_2\left(\frac{\mathrm{T}_1}{\mathrm{~T}_2}-1\right) \quad \eta=1-\frac{\mathrm{T}_2}{\mathrm{~T}_1} \\ 10 & =\mathrm{Q}_2\left(\frac{10}{9}-1\right) \quad \frac{1}{10}=1-\frac{\mathrm{T}_2}{\mathrm{~T}_1} \Rightarrow \frac{\mathrm{T}_2}{\mathrm{~T}_1}=1-\frac{1}{10}=\frac{9}{10} \\ 10 & =\mathrm{Q}_2\left(\frac{1}{9}\right) \Rightarrow \frac{\mathrm{T}_1}{\mathrm{~T}_2}=\frac{10}{9} \\ \mathrm{Q}_2 & =90 \mathrm{~J}\end{aligned}$

Hence, the correct answer is option: 'B'.

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