1.

The inside and outside temperature of a refrigerator are 273 K and 303 K respectively. Assuming that refrigerator cycle is reversible for every joule of work done, the heat delivered to the surroundings will be

Answer»

10 J
20 J
30 J
50 J

Solution :As `beta = (Q_(2))/(W) = (T_(2))/(T_(1) - T_(2))`
`:. Q_(2) = W(T_(2))/(T_(1) - T_(2)) = 1 xx (273)/(303 - 273) = (273)/(30) ~= 9J`
Heat delivered to the SURROUNDINGS,
`Q_(1) = Q_(2) + W = 9 + 1 = 10 J`


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