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The temperature inside and outside of re...

The temperature inside and outside of refrigator are `260 K` and `315 K` respectively . Assuming that the refrigerator cycle is reversible, calculate the heat delivered to surroundings for every joule of work done .

Text Solution

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`T_(2) = 260 K , T_(1) = 315 K, W= 1 "joule"`
Coefficient of performance of Carnot refrigerator `beta = (Q_(2))/W = (T_(2))/(T_(1)-T_(2))`
`therefore (Q_(2))/(1) = (260)/(315-260) = (260)/(42) implies Q_(2) = (260)/(42) = 6.19J`
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