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)
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 = 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.