- #71
justaman0000
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It doesn't. Unless you could get the ice below the freezing point of pure alcohol. Water freezes at 32 degrees F. Alcohol is much lower.stinsonbr said:Can someone explain how this happens?
It doesn't. Unless you could get the ice below the freezing point of pure alcohol. Water freezes at 32 degrees F. Alcohol is much lower.stinsonbr said:Can someone explain how this happens?
What about the temperature of the ice? I'm sure it wasn't 0 C. The freezer in my frig at home is set at 0 F, which is about -20 C. Any idea what temperature for the ice?russ_watters said:Let's stick with the OP's main question and merge it with my testing:I used 100 mL of 50% by volume, but let's go with 100g, and 44% by mass.
Room temp: 20C
The goal is excess ice, but I'll have to guess: 50g.
BTW the freezing point of alcohol is under -170 degreez F. So, there is probably no way you are going to get ice that cold.justaman0000 said:It doesn't. Unless you could get the ice below the freezing point of pure alcohol. Water freezes at 32 degrees F. Alcohol is much lower.
0C. It was specified by the OP in what I quoted and is easily achieved with an ice bucket, per my experiment.Chestermiller said:What about the temperature of the ice? I'm sure it wasn't 0 C. The freezer in my frig at home is set at 0 F, which is about -20 C. Any idea what temperature for the ice?
No problem.russ_watters said:0C. It was specified by the OP in what I quoted and is easily achieved with an ice bucket, per my experiment.
I greatly prefer 0C to -20C here because of the persuasive power of both constituents starting warmer than the final mix. It also makes the calculations slightly easier.
Bystander said:No one's made ice cream?!
Okay, then I agree with the conclusion.DrStupid said:In case of water/ethanol mixtures the "frozen liquid" is pure water ice.
I haven't included that in my calculations (yet), but, if anything, they would add heat to the mixture. The calculations I have done so far were just to see what the maximum amount of cooling could be.Kevin McHugh said:Has anybody taken into account the heat capacity of the calorimeters? Don't forget, they are in contact with the solutions and are removing heat to come to thermal equilibrium.
I don't mean to be a pain, but I want to make sure I understand the analysis, so I worked it backwards and here is what I got:Chestermiller said:Hi guys. I completed some preliminary calculations for the case that Russ specified:
100 gm alcohol/water solution, 44% alcohol mass percent at 20 C
50 gm ice at 0 C
I ran the calculation for the case of no heat of mixing. This is a pretty easy calculation to do. Here's what I found:
At final equilibrium, essentially all the ice will have melted, and the final temperature would be about -18 C. This would be close to (although slightly above) the freezing point of the final solution, which would contain about 29% ethanol...
If anyone is interested in the model development, I will be glad to flesh out the analysis.
Agreed.mfb said:I'm not sure about that. The stopping condition is an equal rate of freezing and melting. At the freezing point of the liquid, freezing of the liquid and melting of frozen liquid (!) is in equilibrium.
Why? Throughout the dropping temperature, the ice preferentially melts because it can go into solution with the liquid and can't easily come out of solution. What changes when the freezing point of the liquid is reached is that the liquid can no longer hold any more "melted ice". The mixture is saturated and any more dilution would raise the freezing temp and require some to freeze.But the frozen liquid melts easier/faster than the ice, and we are looking for an equilibrium of freezing of water molecules onto ice and melting of ice.
Agreed.DrStupid said:Me too, but you are referring to the average concentration and I also to the local concentration. That means without dilution (as I mean it) the local concentration of water would increase around the melting ice or even result in a layer of pure water (a practical example would be ice in oil). In this sense you are assume infinite fast dilution of the melting water. Thus it seems we were just talking cross purposes about the same situation...
The main effect is a shift of then equilibrium between melting and freezing to lower temperatures (or with other words: the freezing-point depression). Ice and pure water are in an equilibrium at 0 °C. The addition of alcohol reduces the freezing-point with the result that the ice is still melting at 0 °C but the solution do not longer freeze at this temperature. The endothermic melting without the exothermic freezing decreases the temperature until there is no ice left or the freezing point of the solution is reached (whatever happens first).
Combining these two and analyzing my experiments last night, we detect a flaw that I may not have explicitly detailed but at least was thinking: I hadn't checked the freezing temperature of alcohol (-32C at 50% BV) and so didn't realize just how cold it is. That has (or manifests) two effects on my experiment:justaman0000 said:BTW the freezing point of alcohol is under -170 degreez F. So, there is probably no way you are going to get ice that cold.
What about the heat released by the 56 g of water in cooling from 20 C to 0 C?russ_watters said:I don't mean to be a pain, but I want to make sure I understand the analysis, so I worked it backwards and here is what I got:
50g ice absorbs 100*334 = 17,200J when it melts
44g of ethanol releases 44*2.44*38=4,079J in dropping from 20C to -18C
106g of water releases 106*4.186*18=7,987J in dropping from 0C to -18C
Yeah, duh. Let's try that again:Chestermiller said:What about the heat released by the 56 g of water in cooling from 20 C to 0 C?
You are right. And it does not even matter how exactly the liquid freezes.russ_watters said:Why? Throughout the dropping temperature, the ice preferentially melts because it can go into solution with the liquid and can't easily come out of solution. What changes when the freezing point of the liquid is reached is that the liquid can no longer hold any more "melted ice". The mixture is saturated and any more dilution would raise the freezing temp and require some to freeze.
This can all be precisely calculated.gjonesy said:I find this discussion very interesting, yet counter intuitive in certain scenarios. Considering the op didn't give any particular circumstance, would these same results hold under different conditions.
Example let's say I have a bottle of brandy 20% alc per volume in the trunk of my car and its sat there all day (on a hot day) and I plan to drink it at the bond fire later that afternoon. It comes out of my trunk at 50c (122 f) and I have just a few 8 oz cups to share with my friends at the bond fire. Could you put enough ice in that size container to bring the alcohol down to 0c? Or would the heat from the alcohol win out given the limited space, melting all the ice before the solution could cold down to 0c?
Just wondering.
Chestermiller said:This can all be precisely calculated.
You have to be a wizard to master Oz?mfb said:Oz are a weird unit.
I bought a better mug and verified that insulation matters a lot. The melting rate of the ice (and therefore the internal heat transfer) drops as the temperature drops, which means you need really good insulation to approximate the adiabatic case. In reality, I only achieved -22C in a -18C freezer, despite a freezing point that should have been about -32C.russ_watters said:2. With the large temperature differences and (I think) poorly insulated mug, heat transfer with the surroundings is occurring at enough of a rate to matter. Hence, my mixture got nowhere close to its freezing point.
I don't understand your description at all. There is no time-evolution that has to be taken into account with perfect isolation, room temperature is not relevant at all, and no melting points (apart from ice at 0 C) are relevant.gjonesy said:I assumed the whole of 50c and a steady drop of the same rate, (38 degrees c) based on the earlier calculations. I didn't account for the heat exchange of the solution. I did assume total ice melt and at a higher rate, so unless the liquid that lowers the melting point of water(salt solution/alcohol/ ethylene glycol) is at a steady room temp and not already heated, the heat energy transfer wins and the solution total temp will be higher. That is why I asked the question. Its counter intuitive to think that added heat energy wouldn't factor in.
I did a different conversion and got a different calculation 31.1 c so right around 25c.