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The efficiency of a heat engine is given by

n = Wu / QH

Where

Wu - is the useful work (7190 J)

QH - Heat taken from the hot reservoir (13,8 x 4184 J)

1 cal = 4,184 J => 1 kcal = 1000 cal = 4184J

So n = 7190 / (13,8 x 4,184 x 1000) J

and since you want that in percent multiply the result by 100.

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14y ago
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14y ago

5017 J = 5000 J (2 significant figures)

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Q: A heat engine does 7190 J of work in each cycle while absorbing 13.8 kcal of heat from a high-temperature reservoir What is the efficiency in percent of this engine?
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