Ans : 8000/9Kcal?
1] caculate the least amount of work that must be done to freeze one gram of water at 0 degrees by a refrigerator. temperature of surroundings is 27 degrees. how much heat is passed to surroundings in this process? latent heat of fusion = 80 k cal/gram
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10 Answers
skygirl
·2009-03-14 19:33:37
1 clarification reqd : 1 gm of water is initiallly in wat temperature ?
tapanmast Vora
·2009-03-15 05:35:56
OH NO!!! [2]
wat I did was
w/Q = (T2-T1)/T2
80Kcal/Q = (300-273)/300
sum1 chk my working!!
**edit** OH K [1]