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If a Carnot’s engine functions at source temperature $127 ° C$ and at sink temperature $87 ° C$ , what is its efficiency ?

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25%

$40 %$

50%

10%

answer is A.

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Detailed Solution

$η = \frac{T_{1} - T_{2}}{T_{1}}$ $= \frac{(127 + 273) - (87 + 273)}{(127 + 273)} = \frac{40}{400} = 0.1 = 10 %$

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