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Cooling a cup of coffee with help of a spoon

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Re: Cooling a cup of coffee with help of a spoon

#41
post #8

Earlier quoted context omitted.

First, remove your sugar crystals from the thermos of liquid nitrogen...

Should you crush the sugar crystals (to promote rapid cooling) or should you use the largest possible crystals so there are more bonds to be broken? I'm going to guess that there's a sweet spot in there somewhere; a priori it seems you want the maximum size that will dissolve completely.

If I remember the answer correctly, it went something like this. Dump all sugar into the tea at once and wait for the temperature to settle. Now split sugar in two halves, dissolve first half, wait for the temperature to stabilize, dissolve second half. Through math it worked out that the second method cooled the tea more. So the answer was to go with the infinitely small pieces of sugar :)

Re: Cooling a cup of coffee with help of a spoon

#42

Feynman would get a cup of coffee, a spoon and a thermometer.

Why the downvotes? I usually don't like oneliners and reference to Feynman/Einstein, but I think that this comment has an important point. I will try to expand it.

Some topics in physics are easy to model and get a closed formula but some are not. In particular, fluids flow and heat flow are very difficult (if the system is not very symmetric).

To make an industrial heat exchanger, the chemical engineers use "books" that have a lot of examples for different chemical products and different geometrical configurations and try to find the more similar to get some experimental constants. These constant can be placed in formulas to get the required size of the new heat exchanger, but are only reliable if the new one is similar to the original example.

A few yeas ago, I worked at a school and we need an experiment for a physic olympiad. We measure the cooling of a plastic cup of water in different conditions. One of the ideas was to measure the effect of a metal spoon, but in the pretesting we found out that the effect is negligible. We got the bigger difference using a thin plastic lid, the cooling time was much longer.

So, to find out the best method to cool a cup of coffee you can't use a pencil and a peace of paper. You should measure it experimentally wit a thermometer. (Perhaps a computer model simulator might work.)

And in one of the Feynman books, he was trying to see if jelly gets hard if you stir it while cooling. He waited for a freezing day, got a pan with hot jelly, something to stir and a jacket. In another book someone told him that a piece of rubber from the Challenger shuttle would become hard when cold. He got some water, ice cubes and somthing to hold the rubber to test it. So probably Feynman Feynman would get a cup of coffee, a spoon and a thermometer.

Re: Cooling a cup of coffee with help of a spoon

#43
I think that the explanation is wrong (or at least it is a oversimplification.)

The main problem is that the explanation ignores completely the air. Static air is a good isolator, therefore much of the heat must be transported by thermal convection or forced convection (blowing). If the air doesn't move, it wouldn't be useful to make each part of the coffee visit the surface.

The second problem is that the it says that most of the heat is lost by evaporation of the water molecules (latent evaporation heat). I think that this is true when the temperature is over 195 F/90 C. But when the water is cooler I think that thermal conduction is more important, and it is not related to the "high-kinetic-energy outlier water molecules". And also some heat is lost by the floor and walls of the cup, but I think that most of the heat is lost by the surface. [More details in "experiment" bellow.]

The third problem is more an oversimplification. It is incorrect imagine that some of the molecules of the water are "high-kinetic-energy outlier". Let imagine that we mentally "paint" the fast molecules at initial time. Very soon these molecules will bounce with other molecules and change their velocity, so a few instants later the fast molecules will be other molecules. To estimate the time that we have to wait until the fast molecules bounce we can use the "mean free time". This time is useful in a gas where the molecules are far apart, and for air in usual conditions it is ~5E-10seg. Obviously water is not a gas, but the molecules are more close together, so this the time we should wait is much smaller. But the time that a drop of water needs to go from the bottom to the top of the cup is ~.1 seg (rough estimation). So the molecules have time to bounce and bounce. Moreover, while the drop is traveling though the surface the molecules have time to bounce and reach thermal equilibrium after some of then evaporate. So I think that a macroscopic model that only consider the temperature should work.

[experiment]

A few yeas ago, I worked at a school and we need an experiment for a physic olympiad. We measure the cooling of a plastic cup of water in different conditions. The idea was that the temperature follow a law like:

  Temp = Constant * exp(-time/tau) + Temp_external
Taking logarithms

  log(Temp-Temp_external) = - time/tau + Konstant 
So it is possible to make a graphic of the log(Temp-Temp_external) and from the slope get the value of tau. In an ideal case, tau is a constant (or almost constant) that depends on the materials and the sizes of the experiment.

Really tau is not constant. When the temperature if below 195 F/90 C the value of tau is almost constant and the graphic of the log is a nice line.

But when the water is hotter tau is not constant, the bigger the temperature, the bigger is tau (IIRC perhaps like twice the cold value). So the graphic of the log is curved and the water cooled faster than expected.

When the temperature if above 195 F/90 C something strange was happening. Looking at the cups we see that above that temperature there was a lot of steam over the cups and bellow that temperature the seam almost disappear. Looking at the graphics of the vapor pressure of water we saw that above 195 F/90 C the vapor pressure is bigger than the value bellow 195 F/90 C. So we expect to have a lot of evaporation when the water is hot and few evaporation when it is cold.

So the conclusion was that when the water is cold most of the heat is exchanged by conduction and convection, but when the water is hotter the evaporation is and additional important part of the cooling effects.

Note 1: We did some additional experiments that are consistent with this explanation, but perhaps there is something we forgot to check.

Note 2: The change between the "hot" and "cold" water above and bellow 195 F/90 C was not sharp. While the water was cooling the steam get thiner softly, tau get lower softly and the vapor pressure gets null softly. So the number 195 F/90 C is only an approximation. (And it was a few years ago.)

Note 3: Heat flow is difficult to modelate in paper, so better measure it experimentally. "In theory, practice is the same as theory. In practice, it differs."

Re: Cooling a cup of coffee with help of a spoon

#44
The indian drink stall sellers in Malaysia and Singapore have a good way to cool it down. It is called Teh Tarik

Picture here http://3.bp.blogspot.com/_0o2-S1p2MUo/Sm-YjNyRfpI/AAAAAAAAAd...

Description of what it is here http://en.wikipedia.org/wiki/Teh_tarik

Edit : Found a video of it here http://www.youtube.com/watch?v=WPuIybnQemc

Re: Cooling a cup of coffee with help of a spoon

#45

Earlier quoted context omitted.

I think he was saying that although physicists generally understand the principles involved, a chemist might have a better sense of the relative magnitudes of the different effects that come into play in cooling a solution, since chemists have to perform this task routinely.

I suppose, but any physicist worth anything should be able to remember enough high school chemistry to deduce the answer.

Fluid dynamics aren't exactly high school chemistry...
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