Here's how much ice you need to cool your beer, according to math

Here is the question: How much ice do you actually need to get your drinks cold?

Let me start with some assumptions.

Suppose you get n drinks and these start at room temperature. Let me say room temperature is 22 °C (about 72 °F).

You start with ice and drinks. The ice is just at 0 °C.

The cans are filled with water. I am actually surprised that canned water isn't more popular. Think about it. Okay, but assume it's filled with water, so I can then use the specific heat capacity of water.

How much water? Well, the standard size is 12 fluid ounces. This would be 355 ml or 355 grams of water.

The can is aluminum and about 15 grams.

The cooler has no mass. Yes, it is one of those massless coolers that you can get from the store. Also, the amount of energy transfer while the drinks are cooling is small.

So, how will this work? Let me start by saying that things have thermal energy. The hotter they are and the bigger they are, the more thermal energy they have. What I want is to transfer thermal energy from the drinks to the ice.

This is one of the cool things about temperature: when you leave stuff in contact for a while, they reach the same temperature. Be careful, don't confuse temperature with thermal energy.

If you put some pizza on aluminum foil and heat it up in the oven, both the foil and the pizza will eventually reach the same temperature. Even though the pizza has the same temperature as the foil, you can easily burn yourself on the pizza since it has much more thermal energy.

Also, the pizza is much tastier than the foil. If you put a drink in zero degree Celsius water, the first thing it will do is change phase from a solid to a liquid. This phase transition requires energy input into the ice.

After that, the water (that was ice) will increase in temperature while the drinks decrease in temperature. At the final point, there will be drinks and water. This probably isn't what you want, but it would do the job.

How much energy is associated with a change in temperature? It turns out that the change in thermal energy for an object depends on the change in temperature, the mass, and the specific heat capacity.

?Ethermal = mC?T

Here, m is the mass of the thing, ?T is the change in temperature, and C is the specific heat capacity of the thing. Also, different things have different specific heat capacities. This is why a hot, foam-based coffee cup does not burn you but the coffee inside (which is about the same temperature) does.

If something is changing phase, like from a solid to a liquid, then this also takes energy. The amount of energy needed for a phase change depends on the mass and the latent heat of fusion. Now for an estimation. Let's say I have one can of soda or beer.

How much ice do I need to cool that off? Well, how cool do you want it? If you can't decide, that is okay. I will make you a nice plot of final temperature of drink versus the amount of starting ice. Remember, I am assuming the drink (and the aluminum can) start at 22 oC.

The key here is that the change in energy of the ice (turning to water) plus the change in energy of the drink must be zero. The problem is with the change in energy of the ice.

If you assume all the ice melts and all of this energy comes from the decrease in thermal energy from the drinks, the drink could end up colder than the starting temperature of the ice. And while this is okay in terms of conservation of energy, it just doesn't happen This is the key. In situations like this, the objects change temperature until they all end at the same temperature.

So, by using the above ideas, I can make a plot of the final temperature of your one drink as a function of the starting mass of ice (at zero degrees Celsius).

The arrow indicates the lowest temperature you could get the drink-it wouldn't get colder than the starting temperature of the ice. This says that with 100 grams of ice, you could end up with a mixture of zero degrees Celsius of water with your drink.

Expanding this to a six-pack (of drinks, not my stomach) you would need 600 grams of ice. Let me make this a little bit more realistic. The above calculation assumes all the thermal energy from the drink goes into the ice water. In reality, some other thermal energy will go into the ice (from the cooler and from the outside of the cooler).

Suppose the ice gets 60 percent of the energy from things other than the drinks. In that case I would need 250 grams of ice per drink, or 1.5 kg for a six-pack and 3 kg for a twelve-pack of drinks.

What if I think of this in a different way? Instead of determining the amount of ice, suppose I purchase a 10-pound (4.5 kg) bag of ice? How many drinks would this cool down? Using the same calculations above, I get eighteen drinks.

So, what is the answer? I think I would recommend one 10-pound bag for every twelve drinks. This way, not all the ice will melt and you can keep your drinks cool for a longer period of time.

Excerpted from Geek Physics: Surprising Answers to the World's Most Interesting Questions by Rhett Allain with permission of Turner Publishing.