fluid dynamics - What cools a drink?

When you stick ice in a drink, AFAICT (the last physics I took was in high school) two things cool the drink:

• The ice, being cooler than the drink, gets heat transferred to it from the drink (Newton's law of cooling). (This continues after it melts, too.)


• The ice melts, and the cool liquid mixes with the drink, which makes the mixture feel cooler.

My first question is, to what extent does each of these cool the drink: which has a greater effect, and how much greater?

Secondly, how much of the cooling by the first method (heat transfer) is without melting the ice? That is, is there any significant amount of heat transfer to each speck of ice before it's melted, and how much does that cumulatively affect the drink's temperature?

I suppose all this will depend on a bunch of variables, like the shape and number of ice cubes and various temperatures and volumes. But any light that can be shed, let's say for "typical" situations, would be appreciated.

Answer

There are three processes to take into account:

1. The warming of ice towards the melting point if it was originally below $0^{\circ} C$.


2. The melting of ice itself


3. The warming of the resulting water

The 1. and 3. part is addressed by heat capacity of ice and water respectively and the amount of heat will be directly proportional to temperature difference and weight of the water/ice. The proportionality constant (actually it also depends on the temperature but not very strongly so let's just ignore that) is called specific heat. For water it is about twice as large as that of ice at temperatures around $0^{\circ} C$.

As for the 2. part, this has to do with latent heat. Simply put, this is an amount of heat you need to change phases without changing temperature. Less simply put, when warming you are just converting the heat into greater wiggling of water molecules around their stable positions in the crystal thereby increasing their temperature. But at the melting point that heat will instead go into breaking chemical bonds between molecules in the ice lattice.

Now, latent heat is really big (you need lots of energy to break those bonds). To get a hang on it: you would need the same amount of heat to warm water from $0^{\circ} C$ to $80^{\circ} C$ as you would need to melt the same amount of ice.

Now, presumably you want your drink cold in the end so that temperature for 3. will be close to $0^{\circ} C$ and also the ice cubes should be pretty warm (no use in producing ice cubes of e.g. $-50^{\circ} C$, right?). This means that these processes won't contribute much cooling. It's fair to say that melting of the ice takes care of everything.

Note: we can also quickly estimate how much ice you need by neglecting the processes 1. and 3. Say you are starting with a warm drink of $25^{\circ} C$ and you want to get it to $5^{\circ} C$. So, reusing the argument about the $80^{\circ} C$ difference being equivalent to a latent heat of the same mass, we see that you need four times less ice than water to get the job done.