# Calorimetry Problem (Isolated System)

This problem has me scratching my head, I was hoping you guys could help me understand what's going on.

A 0.50 kg chunk of ice at -10 °C is placed in 3.0 kg of water at 20 °C. At what temperature and what phase will the final mixture be? How much ice melts?

Here's my attempt at the solution:

Let Q1 be the heat required to raise the temperature of ice to 0 °C.

Q1 = miceCiceΔTice ≈ 1.05 × 104 J

Let Q2 be the heat required to change phase of ice to water.

Q2 = miceLf ice ≈ 1.67 × 105 J

Let Q3 be the heat required to lower the temperature of water to 0 °C.

Q3 = mwCwΔTw ≈ -2.55 × 105 J

Let Q4 be the heat required to change phase of water to ice.

Q4 = - mwLf ice ≈ 9.99 × 105 J

Since we're dealing with an isolated system, ∑Q = 0;

so Q ice -> water = Q water -> ice

The max heat ice can take before it all melts is Q1 + Q2

And the max heat water can lose before it all freezes is Q3 + Q4

So am I correct to say that if Q1 + Q2 ≥ Q3 + Q4 then, all water would freeze because that would happen before all ice melts? What about cases where possibly only some of the ice melts?

Any help is appreciated, thanks!

I've found that the best procedure is to ignore (temporarily) the fact that the system is isolated, and guess at an answer. The guess should be a simple situation: one phase, at a transition temperature. In this case, I'd probably guess all liquid water at 0 °C. But the guess doesn't matter; the result will be the same in any case>

Then use your individual calculations (in the questions) to find out the individual heat flows for the steps needed to bring the two parts of the original system to that "guess" point.

In this case you'd need to warm the ice to freezing, melt the ice, and cool the warm water to freezing. You've already calculated all three.

Only now do we remember that there is zero net heat flow in/out of the system.

In a perfect world, these three calculated heat-in and heat-out values would add to zero, and you would be finished. (Not likely)

In some cases, you'll need to remove some heat to reach balance; freeze some of the water, and in an extreme case, freeze it all and cool the ice down.

In the rest of the cases, you'll need to add some heat to reach zero net heat flow, warming the single mass of water.

You could even throw in some steam above 100 Celsius; the process would be the same.

It's an unusual example of Hess's Law. https://en.wikipedia.org/wiki/Hess%27s_law

So am I correct to say that if $Q_1 + Q_2 \geq Q_3 + Q_4$ then, all water would freeze because that would happen before all ice melts? What about cases where possibly only some of the ice melts?

No, because you would use energy released from melting the ice to freeze some of the water.

The methodic way to approach this problem is to first figure out which phase the result will be in. In order to freeze all matter, $Q_1 \geq Q_3 + Q_4$ must hold, so that the ice is still frozen and all the original water is too.

If this turns out to be false, see if everything will end up liquid, for which $Q_1 + Q_2 \leq Q_3$ must be true.

If both turn out false, you will end up with two phases, of which we know both will be at $0°C$. Therefore the difference of $Q_1$ and $Q_3$ is available for freezing or melting (depending on which is bigger), as the rest is used to bring both to $0°C$. Then it is just a matter of figuring out how much mass you can freeze/melt with the excess energy.

• Could you briefly elaborate on why the conditions are Q1≥Q3+Q4 for a solid final mixture and Q1+Q2≤Q3 for a liquid one? I know it should be easy to follow but for some reason I can't make the connection. Commented Nov 3, 2017 at 5:26
• For a final liquid you need to be able to warm all the ice to $0°C$ and then melt it with the energy stored in the liquid phase to begin with. You can not use $Q_4$ for that, because then you would be generating ice, not giving you a completely liquid result. The argument for all solid is analogous.
– noah
Commented Nov 3, 2017 at 14:34
• Oh I just realized the $Q_3$ I used is the negative of the $Q_3$ you defined. Sorry about that.
– noah
Commented Nov 3, 2017 at 14:36