Q203 / 225 · JEE Main 2003 · Physics

A Carnot engine takes $3 \times 10^6 \mathrm{cal}$. of heat from a reservoir at $627^{\circ} \mathrm{C}$, and gives it to a sink at $27^{\circ} \mathrm{C}$. The work done by the engine is

  • A
    $4.2 \times 10^6 \mathrm{~J}$
  • Correct answer
    B
    $8.4 \times 10^6 \mathrm{~J}$
  • C
    $16.8 \times 10^6 \mathrm{~J}$
  • D
    Zero

Solution

$\eta=\frac{(627+273)-(273+27)}{627+273}$ $=\frac{900-300}{900}=\frac{600}{900}=\frac{2}{3}$ work $=(\eta) \times$ Heat $=\frac{2}{3} \times 3 \times 10^6 \times 4.2 \mathrm{~J}=8.4 \times 10^6 \mathrm{~J}$

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

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