Question #ae278

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Question #ae278

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Answer 1 · 2015-04-28T02:17:11

$E f f = \frac{W}{Q_{h o t}}$ $Eff = W/Q_(hot)$
where $Q_{h o t}$ $Q_(hot)$ is the heat absorbed from hot reservoir and,
W = 1250 J (Work done by heat engine)

therefore, $Q_{h o t} = \frac{1250}{0.24} = 5208.33$ $Q_(hot) = 1250/0.24 = 5208.33$ J (Answer A)

Now, $W = Q_{h o t} - Q_{c o l d}$ $W = Q_(hot) - Q_(cold)$
where $Q_{c o l d}$ $Q_(cold)$ is the heat transferred to cold reservoir.

Implies,
$Q_{c o l d} = Q_{h o t} - W = 5208.33 - 1250 = 3958.33$ $Q_(cold) = Q_(hot) - W = 5208.33 - 1250 = 3958.33$ J (Answer B)

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