A 10.0 g ice cube is placed into 250 g of water with an initial temperature of 20.0 C. If the water drops to a temperature of 16.8 C, has a specific heat of 4.18 J/g*K, what is the enthalpy of fusion of the ice. Ignore the fact that the ice, once melted, has to be heated again.

the mass of ice taken = 10 g

the mass of water = 250 g

initial temperature of water = 20 C

the final temperature of water = 16. 8 C

specific heat of water = 4.18 J/g*K

the heat absorbed by ice to melt = heat loss by water

heat loss by water = mass X specific heat of water X change in temperature

heat loss by water = 250 X 4.18 X (20-16.8) = 3344 Joules

heat gained by ice = 3344 J

heat gained by ice = enthalpy of fusion X moles of ice

moles of ice = mass / molar mass = 10 / 18 = 0.56 moles

enthalpy of fusion = 3344 / 0.56 = 5971.43 J / mole