Question

If we had 2kg of ice at 0°C, and 3 kg of water vapor at 100°C, what will be the final temperature, and what state will the water be in. If the water is in a mix of states, how much water is in each state? Answer should be roughly 400 K

Answer 1

energy released as vapor converts into water at 100 deg ,

Q1 = m Lv = (3)(2260 x 10^3) = 6780 x 10^3 J

Energy absorbed by ice as it converts into water.

Q2 = m Lf = (2 x 334 x 10^3) = 668 x 10^3 J

to raise temp of water to 100 deg C,

Q3 = 2 x 4186 x (100 -0) = 837.2 x 10^3 J

now we have 5 kg water at 100 deg C.

And energy left = Q! - Q2 - Q3 = 5602.8 x 10^3

water can be turned into vapor,

Q = m Lv

5602.8 x 10^3 = m (2258x 10^3)

m = 2.5 kg

hence final temp = 100 deg C ,

water = 5- 2.5 = 2.5 kg

vapor = 2.5 kg