Question

A 500 mL bottle at room temperature (25˚C) of water is poured over 120 g of ice that is (-8˚C). What will be the final temperature of the water be when all the ice has melted? You can ignore loss of heat to the room.

Answer 1

q1 = heat to raise T of solid ice from -8 to zero C.
q1 = mass ice x specific ice x (Tf - Ti) where Tf is 0 C and Ti is -8 C.

q2 = heat to melt ice at zero C to liquid H2O at zero C.
q2 = mass ice x heat fusion

q3 = heat to raise water from zero C to final T.
q3 = mass H2O x specific heat H2O x (Tf - Ti). Ti is zero.

q4 = heat lost by the 500 mL H2O at 25 C.
q4 = mass H2O x specific heat H2O x (Tf - Ti). Ti = 25 C.

q1 + q2 + q3 + q4 = 0

Plug all of that into q1 through q4 and solve for T. Tf is close to 18 C.

Answer 2

Dr. Bob,

When I use the values that I have for latent heat of fusion and vaporization, I do not get the right answer.

I have that latent heat of vaporization is 597.3 - 0.564T, and latent heat of fusion is 79.7 cal/gH2O.

Answer 3

Heat fusion is about 80 cal/g so your 79.7 cal/g (probably a better number than my memory) should be ok. However, I remember the heat vaporization as about 540 cal/g. But hold on ! You don't have heat vap anywhere in the equation.
You need specific heat ice
You need heat fusion water/ice.
You need specific heat liquid water.
Nowhere do you need heat vaporization for water. Somewhere you are substituting incorrectly. Allow me to use round numbers.
q1 = [120 x 0.500 x (8)] [Note: I think I remember specific heat ice = about 0.5 cal/g but check that out.]
q2 = [120 x 80]
q3 = [120 x 1 x (Tf-0)]
q4 = [500 x 1 x (Tf - 25)]
I think the answer is in the neighborhood of 15-20 C.