Can someone explain how this happens?
What is the temperature of the ice?
I am not sure, if that would be a part of the answer then please explain it.
Let me word it this way: if I were making an alcoholic drink and took ice out of my freezer, put a few cubes in the drink and stirred, what factors could cool the drink to below zero?
Second question (if the first doesn't already answer it): would the drink be able to drop below 0°C if the drink was initially at room temperature, and the ice was initially at 0°C?
I meant to mark this as high school competency.
What is the temperature of your freezer? If it is below 0°C, where would be the problem? You'll need a lot of ice, however.
The ice cools the drink; the drink warms the ice. The will meet at a temperature in the middle somewhere. Exactly where depends on the starting temperatures and the relative masses of ice and drink. Also there'll be heat coming in from the room, but we could do this experiment in a good thermos flask to minimse that.
If the temperatures meet in the middle somewhere, what must be true about the ice to make the drink end up below zero? Is this plausible for ice from a home freezer?
Google "freezing point depression."
Everything that has been said so far I understand already. So there's no secret here? It's simply that the ice has to be cold enough to cool the drink below 0°C before reaching the equilibrium temperature?
I had read that it had something to do with the process of ice melting making the ice itself colder?
Is the temperature of the mixture of ice and alcohol always going to meet between the initial temperatures of the alcohol and the ice? Or could the melting ice cooling itself cause the mixture to drop below the ice's initial temperature?
I don't think that's possible.
Regardless of the amount of ice and specific heat capacity of the ice, the equilibrium temperature(final temperature of both substances) reached will be between the initial temperature of the ice and initial temperature of the alcohol, but you could get the alcohol to reach a temperature very close to that of the ice.
As soon as the ice and alcohol reach the same temperature the heat exchange will stop. What you are suggesting seems to go against the laws of thermodynamics. Heat energy always moves towards the colder body. Therefore the alcohol would never get colder than the ice melting, never get colder than the initial ice's temperature. When ice starts melting at zero, it is gaining energy from the alcohol which must have temperatures greater than zero for this to happen, meanwhile it gets cooler but as it gets to zero the heat exchange will stop :)
Hope this helps
Yes and no...
By definition it's the warmest parts of the ice that are melting. The heat to do the actual solid-to-liquid change has to come from somewhere warmer, which has to be the liquid. Heat cannot come from the (colder) ice. So no part of the ice gets colder.
However, we just took away the warmest part of the ice and left the cooler parts. So the temperature of the remaining ice is lower than the temperature of the ice before the warmest part went away. This is a cheat, in a sense. It's analogous to this: the mean of -5, -4, -3, -2, -1, 0 is -2.5. The average of -5, -4, -3, -2, -1 is -3, which is lower. It's only lower because I took away the highest member, not because I lowered any values.
Pressure? Dry ice?
I repeat, and add emphasis,
Heat is given up by the solid phase (ice) in the process of dissolving/diluting alcohol.
Actually, that isn't true(you aren't the only one who said it...): they meet at the freezing temperature of the liquid, regardless of either's starting temperature.
So that means, counter-intuitively, that they can end up colder than either started.
So the only thing that matters here is the freezing temp of the drink (as long as you add enough ice to reach it).
This is exactly what I was trying to fish out. Can you give an "explain like I'm five" explanation on how this happens?
So - it takes energy to mix the water and the alcohol, which I forgot about. When the booze is at 0C then any further melting requires work from somewhere, which means that either the temperature must drop or the ice must stop melting. Obviously dilute booze is a higher entropy state than booze plus ice. So any local melting will end up being a one-way street, where a pure water-ice mixture would be two-way.
Is that about right? Or am I missing something again?
Not necessarily. Put a cube of ice from the freezer in liquid nitrogen. Do you expect the nitrogen to freeze?
It is an interesting situation if the two phases are not the same material. We can have several different options, if we neglect mixing of the substances:
If the freezing point of the liquid is below the freezing point of the solid:
- everything freezes, the final temperature is below the freezing point of the liquid
- nothing of the solid melts, some of the liquid freezes, the final temperature is the freezing point of the liquid.
- nothing melts, nothing freezes, the final temperature is between the freezing/melting points
- nothing freezes, something melts, the final temperature is at the melting point of the solid
- everything melts, the final temperature is above the melting point of the solid (typical result with alcoholic drinks and ice)
If the freezing point of the liquid is above the freezing point of the solid:
- everything freezes
- the liquid freezes completely, the solid partially melts, final temperature at solid melting temperature
- liquid freezes completely, solid melts completely, final temperature between freezing/melting points
- liquid freezes partially, solid melts completely, final temperature at liquid freezing point
- everything melts
It gets more complicated if the two substances mix, e. g. melting ice changes the freezing point of the alcoholic drink.
That can happen below 0, if the ice starts cold enough.
A five year-old doesn't have much chance here, but I'll try to be basic...
This isn't like mixing two liquids together - the concept of an equilibrium temperature somewhere in between doesn't apply. A liquid freezes or a solid melts at one and only one temperature (per a given substance and pressure). If you add heat to ice (put it in a warmer liquid) it will melt at its melting temperature, period. If you remove heat from water, it will freeze at its freezing temperature, period. If those are different temperatures, the lower temperature wins because while the ice can exist as ice at a lower temperature than its freezing point, you can't remove heat from a liquid (by making it melt ice) without dropping its temperature.
This may seem like a bit of a contradiction, but it is actually somewhat similar to the simpler case of a liquid at its boiling point. There is only one boiling point at a given pressure and no matter how much heat you add, that won't change. Conversely, if you lower the pressure above a liquid (say, put it in a vacuum chamber), it will start to boil and that will lower the temperature to get it to the new boiling point.
No, that isn't what I said and is a totally different and much simpler situation because the nitrogen is boiling, not freezing and the temperature is lower, not higher than the ice temp.
You drop ice in LN2, all you get is colder ice.
Iàm sorry, but I donàt see how that can be a general statement. What assumptions are you making here?
I'm pretty sure both of those are impossible (unstable). Could you please detail the conditions that could cause them?
If you have solid steel and liquid water, the steel stays solid and the water stays liquid. Nothing freezes. Nothing melts.
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