Questions
The temperature of inside and outside of a refrigerator 273 k and 303 k respectively. Assuming that the refrigerator cycle is reversible, for every Joule of work done, the heat delivered to the surrounding will be nearly
detailed solution
Correct option is A
Cop=Heat removed from inside(Q2)Work done=273303−273⇒Q2=27330×1 Joule=9.1 Joule∴ Q1=Q2+Work done=(9.1+1)J=10.1 JouleTalk to our academic expert!
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