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A Carnot engine takes 3 x $10^{6}$ cal of heat from a reservoir at $627^{0} C$ and gives it to a sink at $27^{0} C$ . The work done by the engine is

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zero

${16.8 × 10}^{6} J$

$4.2 \times 10^{6} J$

$8.4 \times 10^{6} J$

answer is B.

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Detailed Solution

$η = 1 - \frac{T_{2}}{T_{1}} = \frac{W}{Q}$

$\Rightarrow W = (1 - \frac{T_{1}}{T_{2}}) Q = {1 - \frac{(273 + 27)}{(273 + 627)}} Q$

$\Rightarrow W = (1 - \frac{300}{900}) \times 3 \times 10^{6} c a l$

$= 2 \times 10^{6} \times 4.2 J = 8.4 \times 10^{6} J$

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