Thermal equilibrium for ice being added to liquid water

[Mangoes]

Homework Statement

0.05625 m^3 of H2O(s) is added to a thermally isolated system of 20 L of H2O (l). If the ice was initially at 0 C and the liquid water was initially at 95 C, what is the final state and temperature?

The Attempt at a Solution

I missed class for this so I'm going by what I remember as my book doesn't exactly cover this from what I can tell.

I looked up the density for solid water at 0 C and converted the volume to mass, there are 51.4 kg H2O (s)

I looked up the density for liquid water at 95 C and converted the volume to mass, there are 19.2 kg H2O (l)

Ice is going to gain energy as it goes from a phase change of solid to liquid. The liquid mass that was once ice is then going to gain energy as it reaches thermal equilibrium with the liquid water originally at 95 C. The source of the heat is going to be the liquid water. Since heat in = heat out,

q (phase change) + q (temp. raise) = q (temp. lower)

I calculated q (phase change) by looking up the heat of fusion of H2O which is 334 J/g.
q (phase change) = (334 J/g)(51,400g)

The specific heat of water is (4.184 J/g),
q (temp. raise) = (4.184 J/g)(51,400g) Δt

Similarly,
q (temp. lower) = (4.184 J/g)(19,200g) Δt

Putting it all together:

(334 J/g)(51,400g) + (4.184 J/g)(51,400g) Δt = (4.184 J/g)(19,200g) Δt

But this gives a Δt of 127.4 K which doesn't make sense considering the initial temperatures. It's been a while since I've covered this topic and can't really tell what I'm doing wrong. Would anyone be able to point me in the right direction?

 

[SteamKing]

Not all of the ice (solid water) can melt; there's not enough heat which can be extracted from the liquid water to do so. You basically have a big glass of ice water here.

 

[Mangoes]

That's a first for any of these problems I've ever seen.

How would I go about finding out the maximum amount of energy that can be extracted from the liquid water?

I'd assume that if not all of the ice can be converted to liquid water, then only enough can be converted so that which has been converted to liquid reaches thermal equilibrium with the liquid water... but then the remaining ice remains there without any change in temperature? Seems unintuitive...

 

[SteamKing]

Remember, once the liquid water and the ice have reached the same temperature, heat transfer stops. There's nothing that says ice and water can't coexist at 0 degrees C.

Since the container is 'thermally isolated', no additional heat can enter from the surroundings to melt the ice. It's one reason why a thermos bottle or a dewar's flask is a handy device to have on a warm day, or even a tall glass with a lot of ice cubes.

 

[HalfLostAndSearching]

From the information given, it appears that the final state and temperature will depend on the specific heat capacities and densities of both ice and liquid water. It is also important to consider the total mass of the system, as the addition of 0.05625 m^3 of ice will change the overall mass and therefore the specific heat capacity of the system. Additionally, the thermally isolated system will prevent any heat exchange with the surroundings, meaning the energy for the phase change and temperature changes must come from within the system itself. To accurately solve for the final state and temperature, it may be helpful to set up an equation with all of the relevant variables and solve for the unknowns. It may also be useful to double check the values used for the specific heat capacities and heat of fusion, as these can vary slightly depending on the source.