Mass of ice m = 63 gms
Temperature of Ice T ice = -18C
Temperature of Water T water = 25C
Specific heat of Ice C ice= 2.09 J/g C
Specific heat of water C water = 4.184 J/g C
Delta H fusion for water = 334 J/g
Now there are three stages
1. Heat energy while converting -18C Ice to 0C Ice
Q1 = m x C ice x delta T = 63 x 2.09 x (0 - (-18)) = 2370.06 J
2. Heat energy while converting 0C ice to 0C water
Q2 = m x Delta H fusion for water = 63 x 334 = 21042 J
3. Heat energy while converting 0C water to 25C water
Q3 = m x C water x delta T = 63 x 4.184 x (25 - 0) = 6589.8 J
Q = Q1 + Q2 + Q3 = 30001.86 = 30 KJ
Heat Energy required = 30 KJ
