# Calorimetry: mix of ice, water vapor inside a copper container

1. Jun 18, 2014

### powerof

1. The problem statement, all variables and given/known data

We insert into a copper container (weighing 1.5 Kg) 3 Kg of water vapour at 100 ºC. Inside the container there are 10 Kg of ice at -10ºC. Find the ΔT when the system reaches the equilibrium.

Known data: the specific heats of water, copper and ice and the latent heat of ice and water vapour.

2. Relevant equations

Q=mcΔT

3. The attempt at a solution

We know that ΔT is equal for all the components when the equilibrium point is reached, therefore:

$\left\{\begin{matrix}Q_{absCopp}+Q_{absIce}=Q_{relsVapour} \\ \Delta T=\frac{Q_{abs(Copp)}}{m_{Cu}c_{Cu}}=\frac{Q_{abs(Ice)}}{m_{ice}c_{ice}}=\frac{Q_{rels(Vapour)}}{m_{vapor}c_{vapor}} \end{matrix}\right.$

I don't know how to proceed. The changes of state confuse me. Please give me some clues.

Have a nice day.

2. Jun 18, 2014

### Staff: Mentor

All the components have the same Tfinal at equilibrium, by definition. They have different Tinitial's. What does this tell you about their ΔT's?

3. Jun 18, 2014

### powerof

It would mean that ΔT=Tfinal-(-10ºC) for copper and for ice (assuming the copper and ice have the same temperatures) and ΔT=100-Tfinal for water vapour.

The amount of heat Q released by the water vapour is Q1+Q2=Q(vapour) where Q1=mcΔT(vapour) [the usual formula] and Q2=mL(vapour) [the total heat released after the complete phase transition, ie, condensation].

Similarly for the ice we would have the heat absorbed when going from -10ºC to 0ºC and then the heat absorbed when going from solid to liquid.

But this is assuming that both phase transitions are complete, but how do I know that? Also, how is the heat released from the condensation distributed? I mean how much of that heat is used to warm and melt the ice and how much to warm the container?

Last edited: Jun 18, 2014
4. Jun 18, 2014

### Staff: Mentor

Keep in mind that the specific heat of water is different before condensation (when it's still vapor) and after condensation (when it's now liquid).

Similarly for the ice: the specific heat is different before melting (when it's still ice) and after melting (when it's now liquid).

I don't know a way to decide this in advance. Maybe someone else has an idea. However, if this assumption is incorrect, you should get a nonsensical value for Tfinal. (Hint: this assumption means that all the water ends up as liquid.) Then you can change your assumption and try again.

5. Jun 18, 2014

### powerof

Writing in LateX is somewhat tiring, so I scanned the paper on which I did my calculations for you to read, if you don't mind:

http://imgur.com/a/eUIyp#0

The result appears to make sense (~90ºC) so that's reassuring.

6. Jun 18, 2014

### Staff: Mentor

It looks like you've accounted for all the different Q's, and you're fairly close. However, for heat of vaporization of water, I have 2260 J/g = 540 cal/g, not 500; and for specific heat of copper, I have 389 J/kg-K not 398.

As a check on the arithmetic, you can calculate all the Q's separately, using your final temperature, and verify that they add up correctly.

7. Jun 18, 2014

### powerof

Well, the accurate values of the constants are the least of my concerns. I'm not dealing with a real world problem so I don't need precision. As long as I understood it conceptually, I'm happy. Thank you for your help.

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