Thermodynamics Homework: Solving for Final Temperature and Heat Exchange

[fluvly]

Homework Statement

It's a direct translation from italian, forgive the mistakes:

A container with thermically insulating walls has 2 parts, part A which contains a quantity m(A)=1230 g of water at a temperature of 15.6 degrees Celsius, and part B which contains a quantity m(B)=830 g of ice at a temperature of zero degrees Celsius. The two parts are divided by an insulating plate.

If the insulating plate is removed, what is the final temperature of the system?

What is the heat exchanged between water and ice?

(latent heat of fusion for ice: r=79.7 cal/g, specific heat of water: 1cal/gK, temperature of fusion of ice: 0degrees celsius)

Homework Equations

Q=mc(Tf-Ti)
Q=r(m1-m2)

The Attempt at a Solution

It's just an attempt :)

Q to bring water to zero degrees = 1230 * 1 * (0-15.6) = -19188

19188 = r *(m1-m2)
19188= 79.7*(830-x)
19188 = 66151 - 79.7x
x= 589 g

So 589 g of ice are still there at the temperature of equilibrium, which then is 0 degrees Celsius.

It should be correct, but I do not understand why I should bring the water to 0 degrees. Though, it's the only way I manage to do it.

I tried to solve it considering the final temperature as an unknown quantity, but then the unknown quantities were two, with only one equation.

Please give me a method to understand what the final temperature is without me guessing it at first.

Thanks!

 

[denverdoc]

I hate late night thoughts as too many lately have been way off, but before a phase change, all of the water gets to zero C, if possible. I'll tell you why I suspect so--my friend brought over these odd plastic beverage containers that contain some water or fluid in the hollow beverage container itsself. I poured some soda into one, and after three hours it was shockingly cold. Short of slosh I have never had a beverage that cold.

 

[Andrew Mason]

If the insulating plate is removed, what is the final temperature of the system?

What is the heat exchanged between water and ice?

(latent heat of fusion for ice: r=79.7 cal/g, specific heat of water: 1cal/gK, temperature of fusion of ice: 0degrees celsius)


...

The Attempt at a Solution

It's just an attempt :)

Q to bring water to zero degrees = 1230 * 1 * (0-15.6) = -19188

19188 = r *(m1-m2)
19188= 79.7*(830-x)
19188 = 66151 - 79.7x
x= 589 g

So 589 g of ice are still there at the temperature of equilibrium, which then is 0 degrees Celsius.

It should be correct, but I do not understand why I should bring the water to 0 degrees. Though, it's the only way I manage to do it.

Your method is correct. The reason you bring it to 0 is evident in your answer. Is there enough heat transferred from the water to melt all the ice? What is the temperature of ice?

AM