The heat exchange must be calculated in several steps:

1. The heat lost by the steam to the ice when the steam condenses

2. The heat gained by the ice when it melts

3. The heat lost by the condensed steam to the original water as the temperature of the condensed steam equalizes with that of the water.

Part 1: The heat released by the steam as it condenses:

ΔH1 = (mass of steam)(latent heat of vaporization) = (0.010kg)(2260kJ/kg)

= 22.60 kJ

Part 2: Heat needed to melt ice:

ΔH2 = (mass of ice)(latent heat of fusion)=(0.050kg)(333 kJ/kg)

= 16.65 kJ

Since heat lost by steam must equal heat gained by water, an additional amount of (22.60-16.65)=5.95 kJ is absorbed by the melted ice and raises the temperature of the water, but is not enough to bring the water to 100° so it won’t boil. [Because (0.050kg)( (4.186 kJ/kg-K)(100°C) > 16.65 kJ]

Heat lost by the 100°C water = heat gained by the 0° water,

Or ΔHcold = -ΔHhot

(0.050kg)(333 kJ/kg)+ (0.050kg)(4.186 kJ/kg-°C)(Tf-0°C) =

[(0.010kg)(2260kJ/kg) + (0.010kg)(4.186 kJ/kg-°C)( 100°C - Tf)]

(16.65kJ + 0.2093 Tf – 0kJ) =( 22.60kJ + 4.186kJ - 0.04186 Tf)

0.25116 Tf = 10.136, Tf = 40.36°C (final temperature)

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