How can ice cool an alcoholic drink below 0°C?

[stinsonbr]

Can someone explain how this happens?

 

[DrClaude]

What is the temperature of the ice?

 

[stinsonbr]

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.

 

[mfb]

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.

 

[Ibix]

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?

 

[Bystander]

Google "freezing point depression."

 

[stinsonbr]

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?

 

[NihalRi]

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

 

[Ibix]

I had read that it had something to do with the process of ice melting making the ice itself colder?

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.

 

[MarkJW]

Pressure? Dry ice?

 

[Bystander]

I repeat, and add emphasis,

Google "freezing point depression."

Heat is given up by the solid phase (ice) in the process of dissolving/diluting alcohol.

 

[russ_watters]

The ice cools the drink; the drink warms the ice. The will meet at a temperature in the middle somewhere.

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).

 

[stinsonbr]

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?

 

[Ibix]

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?

 

[mfb]

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.

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.

As soon as the ice and alcohol reach the same temperature the heat exchange will stop.

That can happen below 0, if the ice starts cold enough.

 

[russ_watters]

This is exactly what I was trying to fish out. Can you give an "explain like I'm five" explanation on how this happens?

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.

 

[russ_watters]

Not necessarily. Put a cube of ice from the freezer in liquid nitrogen. Do you expect the nitrogen to freeze?

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.

 

[DrClaude]

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.

Iàm sorry, but I donàt see how that can be a general statement. What assumptions are you making here?

 

[russ_watters]

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:
...
- 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

I'm pretty sure both of those are impossible (unstable). Could you please detail the conditions that could cause them?

 

[jbriggs444]

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.

 

[russ_watters]

Iàm sorry, but I donàt see how that can be a general statement. What assumptions are you making here?

This thread is about a water/alcohol mixture (perhaps also with sugar) in ice. If there are more complicated/different situations that have different outcomes, I don't thing it is wise to muddle the thread with them, particularly when the OP specifically asked for simplicity.

 

[Bystander]

No one's made ice cream?!

 

[DrClaude]

This thread is about a water/alcohol mixture (perhaps also with sugar) in ice. If there are more complicated/different situations that have different outcomes, I don't thing it is wise to muddle the thread with them, particularly when the OP specifically asked for simplicity.

Even then, I don't see why you say temperature is not a factor. If the initial temperature of the ice is low enough, you can end up with the liquid freezing completely and the final mixture ending up below the freezing point of the liquid.

 

[russ_watters]

No one's made ice cream?!

Jeez, seriously!

For those that haven't, here's how you set it up:

Take a large bucket and put some very salty water in it.

Add ice.

Now you have extra-cold water.

 

[russ_watters]

Even then, I don't see why you say temperature is not a factor. If the initial temperature of the ice is low enough, you can end up with the liquid freezing completely and the final mixture ending up below the freezing point of the liquid.

Only if you have a really bad bartender.

[Edit] Actually, even with a really bad bartender, that would be very difficult to achieve.

 

[russ_watters]

If you have solid steel and liquid water, the steel stays solid and the water stays liquid. Nothing freezes. Nothing melts.

Ok, fair enough. And that's assumptions/a situation far outside the scope of the OP's question. I submit that by expanding the issue so much before he has an understanding of the situation he's asking about (particularly when simplicity was requested), that makes reaching an understanding of that situation less likely.

 

[mfb]

Ok, fair enough. And that's assumptions/a situation far outside the scope of the OP's question. I submit that by expanding the issue so much before he has an understanding of the situation he's asking about (particularly when simplicity was requested), that makes reaching an understanding of that situation less likely.

The steel/water example is just an extreme one, the ice/drink setup has the same properties. If ice and drink are at -2C for example, they are in thermal equilibrium (by definition of temperatures). Nothing melts or freezes, assuming the alcohol content is sufficient.

 

[ToddSformo]

How about this: freezing is exothermic reaction, latent heat is given off. Melting is an endothermic reaction. So when ice is added to alcohol, it takes energy to melt the ice, and the energy comes from the liquid, in the form of heat, which, in turn, lowers the temperature of the ice-alcohol system?

 

[mfb]

If the temperature is below zero, no melting happens.

You cannot reach an equilibrium temperature below zero if ice melts (neglecting transient effects from finite heat conductivity and so on)

 

[russ_watters]

If ice and drink are at -2C for example, they are in thermal equilibrium (by definition of temperatures). Nothing melts or freezes, assuming the alcohol content is sufficient.

Agreed, and in the real world that can happen since sometimes people keep their alcohol in the same freezer as their ice. The wording of the question in the title implies that isn't the starting condition.

Keeping within the narrow question, the interesting/difficult/counterintuitive behavior is that both the ice and drink can cool down when you combine them if both start above the freezing point of the drink.

Looking back at the first page, it still looks to me like most of the first 10 or so posts weren't just overly general, they were just plain wrong.

 

[Zachary Smith]

In the real world, the glass is empty and being refilled long before that.

 

[russ_watters]

How about this: freezing is exothermic reaction, latent heat is given off. Melting is an endothermic reaction. So when ice is added to alcohol, it takes energy to melt the ice, and the energy comes from the liquid, in the form of heat, which, in turn, lowers the temperature of the ice-alcohol system?

Correct.

 

[russ_watters]

If the temperature is below zero, no melting happens.

If both start below zero, yes.

You cannot reach an equilibrium temperature below zero if ice melts (neglecting transient effects from finite heat conductivity and so on)

That remains the main issue of the thread and it is not correct.

That, again, is why I think it is a bad idea to expand the scope - I didn't even realize we were still in disagreement about the main issue!

I'll be clear:
If you add ice at 0C to a liquid which is at 10C, but has a freezing point of -5C, the final mixture will be both at about -5C (ignoring dilution).

Heck (the ice cream making example), you can even start with a water/ice mixture at 0C, dump a bunch of salt in, stir, and the temperature will drop.

http://www.thekitchn.com/freezing-science-the-role-of-s-124357

 

[votingmachine]

The general assumption that needs to be made is that the masses of the ice and the drink, and the temperatures of the ice and drink, are not so extreme, so that the final system is in an equilibrium between solid and liquid states. People are making the obvious statements that a kilogram of ice, cooled to absolute zero, will freeze a small drink to some very low temperature, or a small ice cube at exactly the freezing point, in a large warm drink, will end up with a large, slightly less warm drink. Let's ignore those situations ... it is important to note that an adequate amount of ice is needed, and an adequate amount of ordinary drink ... but the ordinary drink preparation sort of assumes those.

So in that ordinary drink, the liquid phase is some solution of alcohol, solutes and water. The freezing point of a solution is below 0-degrees-C. Freezing point depression, as pointed out repeatedly. The ice is at a freezer temperature, about -20-degrees-C. The ice mass, and the phase change of melting ice provide a heat sink for some of the heat in the drink-solution. The ice cools the drink, warming the ice, and melting some ice, turning it into water which dilutes the drink. At equilibrium, the drink is at the freezing point for that particular solution, which is below 0-degrees-C. The mass of ice is also at that same temperature. The temperature will not increase, while there is ice mass remaining that can be warmed, absorbing heat in the warming process. The phase change of ice to water is also a potent heat sink. And the solution around that ice is acting just like adding salt onto a snowy driveway ...

At the equilibrium temperature, the surface of the ice cube is both melting and freezing at the same rate. The interior of the ice cube, which is protected by the surface, still contains pure water, which is below the freezing point for that water. The drink-solution will not decrease further in temperature, as at its freezing point it remains at constant temperature, while the heat of fusion is removed. The ice cube will not increase in temperature, as it is in temperature equilibrium with the surrounding drink-solution.

If it helps, think of dropping cold, steel balls at -20-degrees-C into a drink. The balls will warm and the drink-solution will cool. The drink-solution can only cool to the freezing point of the solution. At that point frozen solution begins to accumulate on the balls.

There are some drinks that contain so much sugar or alcohol that they will not freeze at -20-degrees-C. In those cases, the temperature will simply be the result of the heat capacities of the solution and the solid.

 

[russ_watters]

Salt on a snowy road is another example: the salt does not warm the snow up to 0C, it makes the plain snow melt at below 0C.

The mechanism may be confusing: how does the snowy road know it is salt and not gravel sitting on it if they are separate? Just like with evaporation, there is a distribution of energy, where molecules spontaneously exit and re-enter the solid. When they exit, they mix with the salt and can't re-freeze. This lowers the average kinetic energy (temperature) of the molecules in the ice. Repeat until equilibrium is reached or all of the ice is gone.