chemistry (heat of fusion/vaporization)

i need help setting this problem up.

You have an insulated container of water that has a temperature of 83.2 C, you continue to add ice until the temperature reaches .6 C and you remove the excess ice, your volume increased by 79.8 mL. What amount of hot water did you have in the container to start with?

If we assume the ice is at zero degrees C (which may or may not be so) AND that 79.8 g ice melts to produce 79.8 g water (at zero C, which isn't so because the density of ice is not 1.00 g/mL. Then
massH2O x specificheatwater x (Tf - Ti) + waterfromice x specificheat x (Tf - Ti) + massice to melt x heat fusion ice = 0
massH2O x 4.184 x (83.2-6.0) + 79.8 x 4.184 x (6.0-0.0) + 79.8 x 334 = 0

Solve for massH2O. Check my thinking. Check the numbers.

Ignore thermal expansion.

Assume the volume increase is due to melted ice. I don't think this problem can be done unless an initial temperature of the ice is known.

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asked by abby

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