Explanation:
The given data is as follows.
mass of ice = 8.8 g, mass of water = 250 g
It is known that latent heat of fusion of ice is 333 J/g. Then, heat energy of ice will be calculated as follows.
Q = [tex]m \times L_{f}[/tex]
= [tex]8.8 g \times 333 J/g[/tex]
= 2930.4 J
As heat absorbed by water = heat released by ice
[tex]250 g \times 4.18 J/g^{o}C \times \Delta T[/tex] = 2930.4 J
[tex]\Delta T[/tex] = 2.8°C
thus, we can conclude that the temperature change in the water upon the complete melting of the ice is 2.8°C.
