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KIM [24]
3 years ago
10

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.
Chemistry
1 answer:
Alja [10]3 years ago
3 0

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


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