- #1

Ian Baughman

- 36

- 2

## Homework Statement

In a container of negligible mass, equal amounts (in weight) of ice at 0°C and steam at 100°C are mixed at atmospheric pressure. Assuming no heat exchange with the surroundings, what is the temperature when the system reaches equilibrium? What are the fractions of weights of ice, water, and steam?

## Homework Equations

Q = mCΔT

Q = mL

_{f}

Q = mL

_{v}

## The Attempt at a Solution

1) First I began by calling the mass M.

2) Then I calculated the heat required to melt the ice since it is already at 0°C:

Q = ML

_{f}= M(3.34×10^{5})Q = ML

_{v}= M(2.256×10^{6})5) This is where I run into trouble. I know we have all the ice (mass M) now in the form of water plus some of the steam that was condensed to water. I assume that the equilibrium temperature is under 100°C therefore the solution would be 100% water but I don't know how to show that or find the equilibrium temperature.