A carnot engine takes 3 × 10⁶ cal. of heat from a reservoir at 627° C, and gives it to a sink at 27° C. The work done by the engine is (a) 4.2 × 10⁶ J (b) 8.4 × 10⁶ J (c) 16.8 × 10⁶ J (d) zero

Answer: (b). T₁ = 900 K, T₂ = 300 K: η = 1 - (300)/(900) = 2/3. W = 2/3 × 3 × 10⁶ = 2 × 10⁶ cal W = 2 × 10⁶ × 4.2 = 8.4 × 10⁶ J So the answer is (b) 8.4 × 10⁶ J.

A carnot engine takes 3 × 10⁶ cal. of heat from a reservoir at 627°… | JEE Main 2003 (AIEEE)

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Heat Engines, Refrigerators & Second Law

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

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Answer: (b)

$T_1 = 900\text{ K}$ , $T_2 = 300\text{ K}$ : $\eta = 1 - \frac{300}{900} = \frac{2}{3}$ .
$W = \frac{2}{3} \times 3 \times 10^6 = 2 \times 10^6\text{ cal}$
$W = 2 \times 10^6 \times 4.2 = 8.4 \times 10^6\text{ J}$
So the answer is (b) $8.4 \times 10^6\ J$ .

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