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

#### powerof

1. Homework Statement

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. Homework 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.

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#### jtbell

Mentor
We know that ΔT is equal for all the components when the equilibrium point is reached
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?

#### 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:

#### jtbell

Mentor
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].
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).

But this is assuming that both phase transitions are complete, but how do I know that?
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.

#### 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.

#### jtbell

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.

#### 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.

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

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