4^th of July can be quite a bit of fun. There are many outdoor activities. And what goes with outdoor stuff? Drinks. Maybe beer, maybe soda — but either way, cold. So, you have your cooler and you have your drinks. How much ice do you need to get?

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. Why water? This is so I can use the specific heat capacity of water.
• How much water? Well, the standard size is 12 fl. 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.

The Physics

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 (but not the same thermal energy).

Since the ice is at 0^oC, it won’t won’t increase in temperature first. The first thing it will do is to change phase from a solid to a liquid. This also takes energy. 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 really want you want — but it would do the job.

How much energy is associated with a change in temperature?

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

If something is changing phase, then this is the amount of energy needed to make that change.

The l_f is the latent heat of fusion. How much energy per mass you need to make that phase change.

The Estimation

So I have 1 can of soda or beer. How much ice do I need to cool that off? Well, how cool do you want it? Oh, you can’t decide — well that is ok. I will make you a nice plot of final temperature of drink vs. 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. I can write this as:

Just to be clear:

• the i subscript is for the ice
• d subscript is the for the drink
• w is for the specific heat of water
• c is for the can

The is one problem. The above assumes that all of the ice melts. If you proceed with this equation, you will find that the final drink temperature will be colder than the initial ice (for significantly large amounts of starting ice). This works with the energy expression, but it is unrealistic. So, I need to fix my expression. But first, let me proceed with the above expression. If I solve for the final temperature, I get:

Before I make a graph of the final temperature, let me state some of my constants.

• The specific heat of water is 4.18 J/(g °C)
• The specific heat of aluminum is 0.904 J/(g °C)
• The latent heat of fusion for water is 334 J/g

Now for the plot. Remember this is just for one can of your preferred drink and all the ice melts.

Once I get 100 grams of ice, all of it will melt and the drink will be 0 °C. That seems crazy, but it assumes you haven’t lost any energy to the surroundings and all of the energy it took to melt the ice came from the drink to cool it off.

So, if I have a 6 pack of drinks, I would need 600 grams of ice, a 12 pack would need 1.2 kg of ice. Yes, that seems small. Remember this is the ideal case. What if I suppose that I lose 60% of the energy to the surroundings? I would just need to multiply the energy from the melting ice by 0.4 (if I lose 60%, I only get to use 40% of that). This is that same plot with this added in.

So, from this I need about 250 grams of ice per drink to cool it down to 0 °C. That still seems small. If I get a 10 pound bag of ice, how many drinks could this cool off? Really that is the whole point. This would be a starting mass of around 4500 grams. If I need 250 grams per drink, then:

What does this tell me? This says that you need about 1 ten pound bag of ice for a 12 pack. Remember, my calculation was for the case where all the ice melted. You probably don’t want that.