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Question

A Carnot engine takes in $$3000\ kcal$$ 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


A
4.2×106J
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B
8.4×106J
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C
16.8×106J
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D
Zero
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Solution

The correct option is B $$8.4\times 10^{6}J$$
Given : $$Q_{1} = 3000 \times 10^{3} cal = 3\times 10^{6} cal$$
$$T_{1} = 627^{\circ}C = 627 + 273 = 900\ K$$
$$T_{2} = 27^{\circ}C = 27 + 273 = 300\ K$$
Since $$\dfrac {Q_{1}}{Q_{2}} = \dfrac {T_{1}}{T_{2}}\Rightarrow Q_{2} = \dfrac {T_{2}}{T_{1}} \times Q_{1}$$
$$Q_2= \dfrac {300}{900} \times 3\times 10^{6} = 10^{6} cal$$
work done $$W = Q_{1} - Q_{2}$$
$$= 3\times 10^{6} - 10^{6} = 2\times 10^{6} cal$$
$$= 2\times 10^{6} \times 4.2\ J = 8.4\times 10^{6} J$$.

Physics

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