The two cup trick is also used in Japan when making green tea, because the fragile tea leaves can be damaged if the water is too hot.
I once heard a tea master say that every time you pour the water from one cup to another, it cools down by 10 degrees (celsius). I'm not sure I fully believe him because pouring the water between cups 10 times didn't freeze the water. ;-)
It won't be a fixed "10 degrees". More like dt = (T_cup - T_room) x cooling factor.
I'll leave it as an exercise, as to what 10 pours will do, but it won't freeze the water; as I don't remember how to solve it and I don't have a piece of paper handy.
And that's not accounting for the specific heat in the state change.
From reading the Laura Ingalls Wilder books, when I was a young'n, I remember that 'saucering' was considered rural, unrefined behavior. That kind of thing goes out quick; I bet none of the Silver Screen movie stars were seen saucering their tea.
I always stir my coffee in a random way so that you see turbulent movement in the cup as opposed to the whirlpool you get from moving the spoon in smooth circles.
My intuition tells me that this should give more efficient cooling. I basically have nothing to back this up. Thinking about it, it probably makes more sense for dissolving the sugar than cooling the coffee.
Your aim is to make sure that every bit of the coffee eventually makes it to the surface (so that the highest-energy molecules there can escape and cause cooling). If you stir regularly then (roughly) each bit of the coffee just moves in a circle, and parts that start out below the surface stay below the surface. If you stir it at random, then (this is probably a theorem, if you define "random" with a bit of care, and it probably doesn't require all that much randomness) any portion of the coffee ends up eventually at any given place.
Someone on physics.stackexchange already mentioned that we should probably ask chemists. IIRC, the most commonly used stirring method in industry is, basically, figures-of-8.
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."
Similarly I've often wondered if it results in hotter coffee to mix your milk with the coffee when it goes into the thermos, rather than when it comes out. I'm pretty sure it is better to do it this way - my reasoning is that the hotter black coffee, because of a greater temperature difference with the surroundings, loses heat more rapidly through the thermos walls than the milk + coffee mixture. But I'm unsure how significant the effect is. Perhaps I'll measure it some time.
This is related to the phenomena that tea leaves tend to group in center of cup when you stir it with spoon.
Due to centrifugal forces the level of water near the border is higher that at the center so the water start to flow from borders to center at the bottom of the cup.
This speed up heat exchange as the water gets constantly mixed in addition of usual thermal exchange flow.
I remember similar physics puzzle from high school days. Given a cup of hot tea and few spoons of sugar how does one go about sweetening the tea so that it would cool off the most in the process.
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.
Hahaha yes, while I liked the thought exercise I kept thinking: "Why would I want to cool it?" In fact, I often pre-heat my cup with hot water before I put coffee in it, so it doesn't cool down as fast.
[+] [-] rgrieselhuber|15 years ago|reply
I once heard a tea master say that every time you pour the water from one cup to another, it cools down by 10 degrees (celsius). I'm not sure I fully believe him because pouring the water between cups 10 times didn't freeze the water. ;-)
[+] [-] wisty|15 years ago|reply
I'll leave it as an exercise, as to what 10 pours will do, but it won't freeze the water; as I don't remember how to solve it and I don't have a piece of paper handy.
And that's not accounting for the specific heat in the state change.
[+] [-] edge17|15 years ago|reply
[+] [-] kingsley_20|15 years ago|reply
[+] [-] ern|15 years ago|reply
I wonder why the practice went out of fashion. Do paper doilies send a signal that the saucer is not to be used to hold liquids?
[+] [-] samatman|15 years ago|reply
[+] [-] JonnieCache|15 years ago|reply
My intuition tells me that this should give more efficient cooling. I basically have nothing to back this up. Thinking about it, it probably makes more sense for dissolving the sugar than cooling the coffee.
[+] [-] gjm11|15 years ago|reply
So I think your intuition is doing pretty well.
[+] [-] Confusion|15 years ago|reply
[+] [-] gus_massa|15 years ago|reply
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:
Taking logarithms 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."
[+] [-] spullara|15 years ago|reply
[+] [-] kschua|15 years ago|reply
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
[+] [-] phaedrus|15 years ago|reply
[+] [-] maxbRuns|15 years ago|reply
http://maxbrunsfeld.wordpress.com/2011/02/17/on-the-mixing-o...
my result agrees with your intuition.
[+] [-] unknown|15 years ago|reply
[deleted]
[+] [-] thesz|15 years ago|reply
Due to centrifugal forces the level of water near the border is higher that at the center so the water start to flow from borders to center at the bottom of the cup.
This speed up heat exchange as the water gets constantly mixed in addition of usual thermal exchange flow.
[+] [-] indrax|15 years ago|reply
[+] [-] maxbRuns|15 years ago|reply
[+] [-] thret|15 years ago|reply
[+] [-] eps|15 years ago|reply
[+] [-] neworbit|15 years ago|reply
[+] [-] gregschlom|15 years ago|reply
Reminded-me of this xkcd strip: http://xkcd.com/435/
[+] [-] juiceandjuice|15 years ago|reply
[+] [-] JoeAltmaier|15 years ago|reply
[+] [-] gus_massa|15 years ago|reply
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.
[+] [-] ramki|15 years ago|reply
[+] [-] hencq|15 years ago|reply
[+] [-] BoppreH|15 years ago|reply
[+] [-] 51Cards|15 years ago|reply
[+] [-] nivertech|15 years ago|reply
[+] [-] Pooter|15 years ago|reply
2) Use spoon to extract the ice from their glass
3) Put ice in coffee
Also, you may need to threaten the person with the spoon to get the ice.
[+] [-] harshpotatoes|15 years ago|reply
[+] [-] drstrangevibes|15 years ago|reply