What will be the final temperature of the water in an insulated container?

Question

What will be the final temperature of the water in an insulated container as the result of passing 4.76 g of steam [H2O(g)] at 100.0°C into 110.5 g of water at 25.0°C? (Hvap° = 40.6 kJ/mol H2O)

Answer 1

Hvap° = 40.6 kJ/mol H2O = 2.25kJ/g H2O
Heat capacity of water: 4.181 J/g·K
Let the final temperature of the water to be T°C.
(4.76 g)*(2250J/g + (4.181 J/g·K)*(100.0°C - T°C)) = (110.5 g)*(4.181 J/g·K)*(T°C - 25.0°C)
Cancel the units, we have:
4.76*2250 + 4.76*4.181*100.0 - 4.76*4.181*T = 110.5*4.181*T - 110.5*4.181*25.0
T = (4.76*2250+4.76*4.181*100.0+110.5*4.181*25.0)/ (4.76*4.181+110.5*4.181)
= 50.3
So the final temperature of the water to be 50.3°C.

Answer 2

First, see if the ice melts. the warmth of fusion of water is 333 J/g 60g ice x 333J/1g ice = 20,000 J = 20 kJ So sure: The ice would soften, and there is 22 kJ warmth left over to warmth the water. yet now there's a entire of 130g water at 0C. the warmth potential of water is 4.184J/g-C. 22,000J/130gH2O x 1g-C/4.184J = 40degC So the water would upward thrust via 40degC from 0 C to 40C.