Thermodynamics, please help?

Question

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

A container with thermically insulating walls has 2 parts, part A which contains a m(A)=1230 g of water at a temperature of 15.6 degrees Celsius, and part B which contains a m(B)=830 g of ice at a temp. of 0 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)

PS. I think I managed to do it by bringing the water to 0 degrees, but I do not understand why I should do that. Though, it's the only way I managed 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!

Answer 1

You need to add the latent heat and then the rise in temperature above 0 C for the resultant mixture. Here is how it is done:

1230 g of water at 15.6 C has a Heat content of 1230 X 15.6 = 19188 Calories (sp. heat is 1). Q = m.s.t is the formula.

830 grams of water at 0 C needs 830 X 79.7 = 66151 Calories to become water at 0 C. It is much more than what is available (19188 Cals) and so it is clear that all the ice will not melt. And if some ice is left, the mixture has to be at the equilibrium temperature of 0 C for water and ice. How much ice melts can be calculated. Approx. 240 g of ice melts.

If you want to put it as a formula, we can write:

m(A).s.(t1 - t2) = [m(A) + m(B)] .s.(t2 - 0) + m(B) X (L) where t1 is the initial temperature of A and t2 is the final temperature of the mixture and L is the latent heat of ice melting.

Answer 2

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m(A) x T(A) = Heat energy in cals
m(B) x Lheat = caloruies needed to melt ice

830 grams of ice will absorb
830g x 79.7cal/g = 66,151cal of heat , needed to melt the ice.
Change of state of matter for water from as solid to a liquid, is endothermic (absorbs heat)
66,151 cal needed to melt 830 g of ice
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1 cal = 1g x 1 degree celsius
The system has 1230g x 15.6cal/g = 19,188 calories of heat in it
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There is not enough heat to melt all of the ice.
Only 19,188cal / 79.7cal/g = 240 g of ice will melt

At equilibrium
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The water temperature will be a 0 degrees,
There wil be 1230 gm + 240 g = 1470g of water
There will be 830g - 240g =580g ice
19,188 cals of heat energy was used to melt 240 g of ice
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