00:01
So in this problem in part a, the amount of heat needed to change 2 kilogram of ice as 0 degrees celsius into water at 0 degrees celsius is going to be given by q equals ml.
00:19
And this l is actually lf, which is leaden heat of fusion of ice.
00:24
And laden heat of fusion of ice is actually 3 .35 times.
00:31
10 to the 5 jule per kg per kg so from here we can find q equals 2 kilogram times 3 .35 times 10 to 5 jule per kg and this is going to be equal to 6 .7 times 10 to the 5 jule now this is the additional energy that the water has then the ice right so so this is the water has this much of the energy more than the ice itself.
01:13
But according to the mass energy relation, this energy is stored as additional mass of water.
01:19
And that mass can be found from this relation here, e .n .e .n.
01:26
Equals q equals delta m .c squared.
01:32
So additional energy here, heat energy is the additional rest mass energy, which is equal to additional mass time.
01:39
X squared and from here delta m equals q over c squared which is 6 .7 times 10 to the 6 .7 times 10 to the 5 jule over 3 times 10 to the 8 meter per second squared and this is going to be equal to 7 .44 times 10 to the negative 12 kilogram.
02:15
All right.
02:16
So we have the additional mass that the water has at zero degrees celsius compared to the ice at zero degrees celsius...