The temperature of inside and outside of a refrigerator (working based on carnot cycle) are 273K and 303K respectively. Assuming that the refrigerator cycle is reversible, for every joule of work done, the heat delivered to the surrounding will be nearly :

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10J

20J

30J

50J

answer is A.

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Detailed Solution

GivenTemperature of inside T2=273 KTemperature of outside T1=303 Kk=Q2W=T2T1−T2=27330Heat delivered to surrounding∴Q1=Q2+W=k+1W=9+1 J=10 J

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