[cseil]

Hello, I've got this problem.

You can create water droplets at -40°C but they're so unstable that they become ice. Assuming that every drop is isolated (adiabatic transformation) calculate the final temperature of the drops and the ΔS and ΔH for this process.

Every droplet is 1g, Cp of liquid water is 1 cal/g*K and Cp of ice is 0.5 cal/g*K. The ΔH of fusion for the water is 80 cal/g at 0°C.

I can calculate the ΔH and ΔS dividing the process into three steps:
H2O (liquid) from -40°C to 0°C,
H2O (liquid) to H2O (ice) at 0°C,
H2O (ice) at 0°C to H2O (ice) at -X°C.

I tried to do something like this:

q=0
[tex]Cp(l) (273-233) - \frac{\Delta H}{273} + Cp(s)(T_x-273) = 0[/tex]

I get Tx=193.18K (-80°C). Do you think it's right?

[mjc123]

If it releases latent heat of crystallisation, it will  warm up.
What is "ΔH/273"?

[cseil]

Sorry, it's just ΔH.
I think I mixed up the entropy and the enthalpy 

But if I use just ΔH I get Tx=353°K=80°C!!
I don't think it makes sense.

[mjc123]

Do you get ice at 80°C?

Work out ΔH for water  ice at -40°C. What does that heat do?

[Enthalpy]

Maybe I saw 1g drops in Guyana where rain is seriously loud, but I wouldn't call them "droplets".

Anyway, you should check whether the amount of cold from -40°C suffices to freeze all the water.

More generally, changes of state ususally mean much heat, as compared to temperature changes. This is even more true for vaporization. It's good to have in mind the equivalents of the melting and evaporation heats in terms of liquid ΔT and realize the implications.

[cseil]

I can't get it, I don't know 

If I calculate the ΔH(fus) at -40°C I get 100cal (because it's 80+ΔCpdT = 80-0.5(233-273)).
So the heat released will be -100cal for the crystallisation.

I don't know what to do now.

[mjc123]

So what happens if you put 100 cal into 1g of ice at -40°C?

[Enthalpy]

Or try a different route, referring to liquid water at 0°C:
How much less heat does liquid water at -40°C contain?
How much less heat does solid water at 0°C contain?
Can all the water freeze?
If no, what is the temperature?
If yes, how much could is available, and to what temperature does it bring the ice, starting from 0°C?

[cseil]

So what happens if you put 100 cal into 1g of ice at -40°C?

Well, the temperature goes up.
100 cal is the quantity CpdT where Cp is 0.5 cal/g*K.

So I can calculate the final temperature:

[tex]100 = Cp(T_x - 233)[/tex]

and the result is Tx=433K.

Now, even if this is right, I get the problem that now I don't have ice and I don't have the liquid neither.. at this temperature I have vapor!

[Enthalpy]

the result is Tx=433K. Now, even if this is right [...]

It is not, as you guessed.

Did you already check how much water can freeze?

By the way, the equilibrium state of water does not obey an analytical equation like Cp*ΔT. Instead, it's like "under 0°C it's this way BUT at 0°C it's that AND between 0°C and the boiling point still an other", and this has no good differentiation properties, so attempts to solve it by one single equation like Cp*ΔT, which is analytic, are futile. You do need an "if then" approach.

[cseil]

the result is Tx=433K. Now, even if this is right [...]

It is not, as you guessed.

Did you already check how much water can freeze?

By the way, the equilibrium state of water does not obey an analytical equation like Cp*ΔT. Instead, it's like "under 0°C it's this way BUT at 0°C it's that AND between 0°C and the boiling point still an other", and this has no good differentiation properties, so attempts to solve it by one single equation like Cp*ΔT, which is analytic, are futile. You do need an "if then" approach.

I understand that, but I have to put it into an equation, and I don't understand how to.
What do you mean "how much water can freeze?".

I have drops at -40°C, every drop becomes ice. When it happens, 100 cal are released.
This quantity of heat goes to the ice, because every drop is thermically isolated.

Now, I have ice at -40°C and 100 cal into it. The analytical way to calculate how it does change temperature is the one I wrote, but it doesn't work, because as you said I have to consider the system in three different parts.

The only thing I can think is:
if 100 cal go to the ice, I have an increase of temperature till it goes to 0°C.

[tex]Cp_{ice}(273-233) = 20 cal[/tex]

So when the drop is at -40°C and absorb 20 calories, it goes to 0°C.
So the 20% of the heat of crystallisation makes the ice go to 0°C.

When it goes to 0°C there's fusion, the fusion absorbs 80 cal for every drop. So we have 20 cal for the change of temperature from 233 to 273K and 80 cal for the fusion. It makes 100cal in total.

Now I don't understand what happens. I have liquid water at 0°C, if it crystallises again there'll be 80 cal released and 80 cal adsorbed from the ice for the fusion and so so on.

Does it make sense?

[cseil]

Anyone, guys?
Do you think it's correct or not? 

Thanks

[Enthalpy]

[...] I have to put it into an equation [...]

You can't put it in a single equation. You must first check what happens, then decide which equation applies. Not having done it, you will continue to fail.

[...] every drop becomes ice. When it happens, 100 cal are released. []

No. And neither.

Good luck!

[cseil]

I understood that but I don't know how to.
So... ok!