.15 kg of steam at 100 degree Celcius is mixed with .2kg of water at 50 degree and .2 kg of ice at 0 degree. Determine the final temperature and phase(s) of the final state. Assume that no energy is lost to the surroundings.
The Attempt at a Solution
I am thinking that since energy is conserved, the steam will lose energy while the ice and water will gain.
-Q_steam + Q_water + Q_ice = 0
I know Q = mc*deltaT
The answer key I have says that the end result will be 100 degree with .485 kg water, and .065 kg steam. I would try to test if the all of the steam and ice condense and melt if I didn't know this.
So ice will all melt
Q_ice = .2*2100*(T-0) + .2*L_f
Q_water = .2*4190*(T-50)
Q_steam = .15*c*(T-100) + x*L_v
There are 3 possibilities, 1) none of the water boils 2) some of the water boils 3) all of the water boils. In this case, it is option 2. T must be 100
Q_ice = .2*2100*100 + .2*3.34*10^5 = 108800
Q_water = 41900
Q_steam = x*2.256*10^6
x = .0667 kg of steam turns into water
If this is the case, I have 1.43 kg of steam left and .4667 kg of water after the ice melt. This answer is however incorrect. Where did I go wrong?