How fast would a bucket of water cool down?

Could you not put the brew into a container that allowed more area to be cooled? Rather than a bucket how about somethng more shallow, or more than one container? You know what a yard glass looks like? Could you use something (actually several somethings) like that? It sems like the problemis the ara being cooled is a lot saller than the total volume of liqid, so if you had more area being cooled theprocess would go faster.
What about a reverse of the process in Clan of the Cave Bear? They heated rocks in the fire, and then dropped the rocks into the liquid. They would take out those rocks and add more. Could you apply the same principle here? There are all sorts of freezer thingies that you can buy now that you freeze then put next to food. The water does not escape, so you would not be altering the chemial composition of your brew.
Just a suggestion. I don’t know how or if that would affect the taste, or if the liquid is not to be disturbed…
I know this does not answer your original question. My mind took off when you said how hard it was to cool the liquid quckly. Sorry.

@Trillian: Thanks for the suggestions, the question wasn’t really concerning the cooling process though :).. Im thinking about buying a cooling coil to that end..

I was mostly annoyed because i had no idea how to work out the maths on how fast the bucket would cool, and i was wondering if anyone had any pointers.

Sorry, I am completely boned in the math department.

I think this is a problem in blackbody radiation, assuming there’s no appreciable heat transfer from conduction (into the surface on which the bucket rests) or evaporation from the liquid surface. I recall that the rate of heat transfer is proportional to the 4th power of the temperature difference (initially 15 deg K here).

You’d also have to assume spherical geometry to work this out, since clearly the cooling rate depends on surface area-to-volume ratio. Cooling would be much faster if the bucket were shaped like a large, shallow pan or long, coiled tubing.

Once you work out the rate per (degK)^4, then I guess you can set up a diff eq and determine how long until the temperature falls to 5 deg. Sorry can’t help further.

@gasman Not bad! Now lets see what I can remember…

While eb (emissive power) is directly proportional to the T^4 (where T = temperature) the actual rate of cooling, (or q) is dependent on area (A)

q = eb * A

However, if you just have a pot sitting outside, then it’s more of a convection issue

q = h * A * delta-T

where delta-T and the area is basically the surface area of a cylinder the size of whatever the water is in.

In truth there are two basic ways to cool things quickly that do not involve cryogenics. One is to increase the surface area immensely. That is why we have radiators; I’ve seen may a still use them to cool stuff off quickly.

I’ve also seem misting, as a mist has far more surface area than a pot of liquid. For something less drastic, aeration works. Make a little waterfall or something. To see how well that works, take a got cup of something and an empty cup, then pour the liquid back and forth a few times. If you have the cups far enough apart to allow it just a little time to get some good contact with the air, you will see how that you can cool stuff off fairly quickly without refrigerant. I can have water come out of my showerhead hotter than Hell and it’s barely warm by the time it hits the tub six feet later.

The other way to cool it rapidly involves shitloads of coolant, and apparently you’ve already tried that route without success.

Mr. Laureth uses an immersion chiller for his homebrewing. He says it improved the cooling time “from hours to about 15 minutes.”

@jerv—What if there’s no wind—then not much convection? Also I forgot about conduction into air surrounding the bucket, but I have a feeling the dominant mechanism of heat loss is radiative. I know it is in the case of human bodies (37 C = 98.6 F) at room temperature (22 C = 72 F) in still air, where it so happens by cosmic coincidence that again delta-T = 15 deg K.

@gasman Arguing convection versus radiation is almost a matter of semantics. Hot air rises, cooler air takes it’s place, so heat transfer leads to some degree of air circulation anyways.
Still, more wind will speed the process, but I’ve never been outside when the wind is dead calm; everywhere I’ve ever been, the wind is always at least enough to move a cloud of smoke or make the flame of a Bic lighter flicker.

Also bear in mind that the human body actually generates heat while a 10 liter bucket of hot water merely stores heat. I guarantee that the bucket will cool to ambient far quicker than a human. Well, unless you kill them, in which case the human’s greater surface area will cause it to cool quicker than the bucket.

Some of it also depends on what material the bucket is made of. There is a big difference between a copper stock-pot and a cast iron cauldron, and a plastic tub is a different beast altogether. In fact, the plastic tub will insulate it and effectively reduce the amount of surface area that is radiating heat as heat will only really be able to go out the top; since plastic doesn’t have great thermal conductivity, the heat will not be able to go through the walls of the bucket very well. Exactly how much of a difference it makes depends on the geometric proportions, but it could mean the difference between an hour and an entire evening.

@laureth How much water do you pump through there to get the job done? Some people have water bills to consider while others have limits to the output of their wells, so that could matter quite a bit.

Okay, nice answers, especially @gasman and @jerv, but so far no definite one to this. Lets try to make some simplifications.. Say we have 1L of water instead of 10L. Say we store it in a cube instead, this would give us a 10×10 cm cube. A cube has 6 sides and each side is 10cm^2, this gives us a total surface area of 60cm^2.

q = h * A * delta-T

Let’s see if i got this right

q=rate of cooling (degrees/sec?)
h=some constant?
A=surface area = 60 cm^2
delta-T=Temperature difference (degrees, starts at -15 and shrinks as time passes)

Judging by the units i guess h is some coefficient for how fast the convection of the material is? (1/sec*cm^2)? If we for the sake of argument assume that the cube of water has no walls, what is this coefficient for water?

Say the side and bottom walls are made of plastic, how would that complicate things? Is the insulation from the plastic so big that the convection from those areas are negligeble compared to the top (water) surface?

Is this the equation you guys are talking about? Convection–diffusion equation

I mean, this is a really common question, that should be answerable with high school maths. “How high is that building”, and “how fast does that cool” seems equally compilicated questions. I can answer the first question easily by trigonometry, but not the second question, and i would love some intuiton on how to do it.

@jerv – Mr. Laureth says,“I turn the faucet on cold, all the way, and I stop when the wort is cooled.” I would assume that anyone seriously looking into chilling wort with a water-heavy mechanism such as that would take into consideration their own water situation before making any buying decisions. Water is relatively cheap where I live. As always, there’s a trade-off between cheap/convenient/fast, pick two. ;)

There are also people who run the water through the copper tubing and then capture it for use in some other way, such as flushing the toilet or watering the garden.