Thermodynamics: Final temp of Ice + steam?

[Tabatron]

Homework Statement

In an insulated container, .50kg of steam at 140C is mixed with 2.0kg of ice, at -20C. What is the final temp inside the container if heat exchanges with the container are ignored?

This is a multiple choice question and the answer is 60C.

Homework Equations

mC(Tf-Ti)
Lf
Lv

The Attempt at a Solution

I'm stuck on how to figure out what mass of the ice and steam will turn into water, do I assume that all turns to water? I'm also confused on where to go after that.

Qsteam = .5(2010)(100-140) + m(22.6 x 10^5) + m(4186)(Tf-100)
Qice = 2(2090)(0-(-20)) + m(33.5 x 10^4) + m(4186)(Tf-0)

As for the weird Lf and Lv values, their in J/Kg and I'm just using the numbers given to me from my teachers formula chart.

EDIT: Ok, so I assumed that all turns to water and got an answer of Tf = -163, which I remember my teacher saying that this means not all of the ice melts. Is this correct?

Any help is appreciated.

 

[Mindscrape]

Okay, good job on factoring in the changes of state. You seem to be having trouble with the units of Lf and Lv? Anyway, if you do have those correct, you are basically there. You just have to realize that the two heats must be equal because the container is thermally isolated, the heat can only exchange within the system.

 

[Tabatron]

The Lf and Lv values I obtained of my formula chart.

I'm still confused though, how do I solve for how much mass of ice will melt?

 

[Mindscrape]

Since you've already assumed that the two states are going to both turn into water, which is smart of course, then just use the respective masses and assume they all turn to water (that no mass went into energy).