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]
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
1 clarification reqd : 1 gm of water is initiallly in wat temperature ?
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]