it's fun to think about this in the context of science and the experimental method.
this of experiment is what we're taught in grade school is science. there are a limited number of variables and you manipulate them in a controlled setting to figure out cause and effect.
this works reasonably well in the physical sciences, sometimes in the biological sciences, and almost never in the social sciences (to the dismay of economists and other social scientists).
the number of omitted variables and heterogeneity in outcomes explodes as you move from the physical sciences to the social ones so you need a different toolkit to handle them (RCTs!).
economics and other social sciences still want to get at explainable mechanisms like you would get in physics but it's just not that simple and generalizable. it reminds me of trying to get to ai explainability, it's really hard because you're trying to project a high-dimensional space into a lower one that never fully explains the higher one. capeesh?
I had similar thoughts. When I got to the end of this and fit the regression, I though, "wait, don't I hate regression!?" But in this particular case, there are strong reasons to think that the temperature truly only depends on the height and pour-speed, so a regression seems totally justified. (At least if we assume room temperature, the kettle shape, the thickness of the cup, etc. are all constant.)
Your experiment reminds me of the Mpemba effect, which is "...the name given to the observation that a liquid (typically water) which is initially hot can freeze faster than the same liquid which begins cold, under otherwise similar conditions. There is disagreement about its theoretical basis and the parameters required to produce the effect." https://en.wikipedia.org/wiki/Mpemba_effect
Based on wild extrapolation, this means pouring a cup of boiling water from 5 meters above ground would cool it so much it turns into snow (regardless of time spent pouring).
I found a video online of a guy tossing a pot of boiling water into -29°C air, and it's not too far off!
I didn't measure them, if that's what you're asking, but I'd expect they were nearly constant, given that I took all the measurements in a short period of time in a temperature-controlled environment. Certainly, these would change the amount of cooling, although the range of likely indoor temperatures varies in a range that's fairly small relative to the temperature boiling water. (100 ℃ vs. 15-30 ℃?) No idea how much evaporate cooling might be a factor here, though!
In India in the 1940s and 1950s (and for all I know possibly still today..) one used to be able to buy a cup of tea on railway platforms, whilst the train was stopped at the station. Because the water came out of the tea urn boiling hot and you were going to drink it from a heat-conducting tin mug (and only had a few minutes whilst the train was stopped) the tea urn man would cool the tea for you by pouring it between two receptacles before pouring it into your tin mug.
Rather than pouring from an extreme height he would pour it several times back-and-forth, and rather than trying to "aim" the pour he would begin the pour with the two receptacles held close together, one in each hand, then (with practiced swiftness) raise one up whilst keeping his eyes on the lower one to keep the stream roughly in the centre of its opening. Then he'd bring the two receptacles back together, all the while "connected" by this stream of tea, and pour back to the first one. The whole process looked a bit like somebody stretching out some ductile, toffee-like substance between their hands.
[Also: to know how significant your results are, I'd love to know what temperature the water started off at? (ie. what was the ambient pressure in the room it was boiled in, and was boiling-point measured visually, by thermometer, or by a steam-pressure-switch like you'd find in a kettle..?)]
I can confirm that the "pour between two metal cups" method is indeed still used in India today. Even in more circumstances—if you order coffee in a restaurant, it might be served to you with two cups even though you're sitting down and have plenty of time. I don't understand why steel cups are used, though, given that steel is such a good heat conductor...
Anyway, for these experiments, the water started at very close to 100 ℃ as measured with the same digital thermometer as used for the other measurements. Each time I need to refill the kettle, I let it reach the maximum temperature the thermometer could show, and left the kettle on the heat at all times between measurements.
tl;dr: I suspect they had both situations in different runs of the experiment.
In the continuous case, the observed temperature at the base should be an exponential decay in the product of the height h₀ and the initial outlet radius r₀. That is: if air has temperature T_a and the water is initially at temperature T_b but has temperature T at the base, then
T/T_a≈1+(T_b/T_a-1)exp(-Ch₀r₀)
for some constant C (derivation below). But we don't see a corresponding shift in the initial experiment; big and medium exhibit roughly the same temperatures. Presumably this is because they shift from a laminar stream in the small and medium cases to droplets in the big?
Derivation: Assume that the stream is axisymmetric (circular cross-sections of varying radius r), falling with local velocity v sufficiently large that the angle of the water surface is nigh-vertical. As long as sufficient water falls to avoid pinch-off, conservation of mass implies that vA is constant, where A is the cross-sectional area of the stream (A=πr²). During an infinitesimal time interval [t,t+dt], water falls a distance v dt in a stream with radius r, and so generates a surface area 2πrv dt=2π√(v*vr²) dt. By Newton's law of cooling, the temperature flux generated during that time is proportional to the product of the surface area and the temperature difference across the surface. Thus our laws of motion are:
vr²=C₁
dv/dt=g
dh/dt=-v
dT/dt=C₂√v*(T_a-T)
where C₁ and C₂ are constants of geometry and material and "down" is positive for v but negative for h. Note that these equations hold even if the stream has sufficient internal structure ("laminar flow") to avoid thermal equilibrium, since then the effects of conduction are negligible during the brief fall; the equations still describe the outer surface of the stream; and then turbulence at the base redistributes the "coolness" flux evenly through the water. The net effect is that, relative to the non-laminar case, the total shift in temperature changes by a multiplicative factor corresponding to the proportion of the stream in the outer well-mixed surface layer — but we can absorb that factor into C₂.
Rearranging our fourth equation and letting C₃=C₂/g,
dT/(T_a-T)=C₂√(v) dt=C₃√(v) dv
Thus T_a-T=-C₄*exp(2C₃*v^³/₂) for some constant of integration C₄.
At time t=0, v=C₁/r₀² but T is some boiling T_b. Thus
C₄=(T_b-T_a)exp(-2C₃(√(C₁)/r₀)³)
As is well-known, the second and third equations integrate to give
v=C₁/r₀²+gt
h=h₀-C₁/r₀²t-½gt²
where the initial height is h₀ and radius of the outlet is r₀. However, the height of the fall is unlikely to be so large that the water experience substantial acceleration during the drop, and we can approximate the time of arrival at the base as t≈h₀r₀²/C. The corresponding velocity is C₁/r₀²+gh₀/(C₁/r₀²).
Since we've already assumed the gravitational velocity-shift gh₀/(C₁/r₀²) small relative to the initial velocity C₁/r₀², we can approximate the difference of square-roots-cubed with the derivative: (x+dx)^³/₂-x^³/₂=³/₂√x dx. Then
(Note that g drops out entirely! Pour water in a pressurized moon base and you should observe the exact same temperature drop — except, of course, that then the water might fall slow enough for the gravitational velocity-shift neglected above to become relevant.)
Depending on how small u=C₂/√(C₁)∙h₀r₀ is, the final result above can be well-approximated as a linear decrease (u very small), a quadratic decrease (u moderate), or neither (u large). In the initial experiment, Dyno-might observed a quadratic decrease, but in the kettle experiment they observe linear decrease. I assume this is because the kettle experiment did not have a well-mixed stream, reducing the effective C₂ as discussed above.
The "insect-bitten tea" link doesn't work for me.
Ack, thank you! I updated the link—in case it doesn't work, I was trying to link to #14 here: https://dynomight.net/thanks-3/
it's fun to think about this in the context of science and the experimental method.
this of experiment is what we're taught in grade school is science. there are a limited number of variables and you manipulate them in a controlled setting to figure out cause and effect.
this works reasonably well in the physical sciences, sometimes in the biological sciences, and almost never in the social sciences (to the dismay of economists and other social scientists).
the number of omitted variables and heterogeneity in outcomes explodes as you move from the physical sciences to the social ones so you need a different toolkit to handle them (RCTs!).
economics and other social sciences still want to get at explainable mechanisms like you would get in physics but it's just not that simple and generalizable. it reminds me of trying to get to ai explainability, it's really hard because you're trying to project a high-dimensional space into a lower one that never fully explains the higher one. capeesh?
I had similar thoughts. When I got to the end of this and fit the regression, I though, "wait, don't I hate regression!?" But in this particular case, there are strong reasons to think that the temperature truly only depends on the height and pour-speed, so a regression seems totally justified. (At least if we assume room temperature, the kettle shape, the thickness of the cup, etc. are all constant.)
Science!!
I don't know who you are but you are very funny
Your experiment reminds me of the Mpemba effect, which is "...the name given to the observation that a liquid (typically water) which is initially hot can freeze faster than the same liquid which begins cold, under otherwise similar conditions. There is disagreement about its theoretical basis and the parameters required to produce the effect." https://en.wikipedia.org/wiki/Mpemba_effect
Now that is an A+ awesome experiment. Well-designed, well-executed, and well-analyzed. A fun read.
Based on wild extrapolation, this means pouring a cup of boiling water from 5 meters above ground would cool it so much it turns into snow (regardless of time spent pouring).
I found a video online of a guy tossing a pot of boiling water into -29°C air, and it's not too far off!
Did you control for air temperature and relative humidity?
I didn't measure them, if that's what you're asking, but I'd expect they were nearly constant, given that I took all the measurements in a short period of time in a temperature-controlled environment. Certainly, these would change the amount of cooling, although the range of likely indoor temperatures varies in a range that's fairly small relative to the temperature boiling water. (100 ℃ vs. 15-30 ℃?) No idea how much evaporate cooling might be a factor here, though!
How much of the temperature drop from a 30 second pour would also occur if it was sitting in the cup for those 30 seconds?
I guess the best approximation for that is to look at the 8cm pour, and maybe extrapolate a little?
Fast becoming one of my favourite substacks
In India in the 1940s and 1950s (and for all I know possibly still today..) one used to be able to buy a cup of tea on railway platforms, whilst the train was stopped at the station. Because the water came out of the tea urn boiling hot and you were going to drink it from a heat-conducting tin mug (and only had a few minutes whilst the train was stopped) the tea urn man would cool the tea for you by pouring it between two receptacles before pouring it into your tin mug.
Rather than pouring from an extreme height he would pour it several times back-and-forth, and rather than trying to "aim" the pour he would begin the pour with the two receptacles held close together, one in each hand, then (with practiced swiftness) raise one up whilst keeping his eyes on the lower one to keep the stream roughly in the centre of its opening. Then he'd bring the two receptacles back together, all the while "connected" by this stream of tea, and pour back to the first one. The whole process looked a bit like somebody stretching out some ductile, toffee-like substance between their hands.
[Also: to know how significant your results are, I'd love to know what temperature the water started off at? (ie. what was the ambient pressure in the room it was boiled in, and was boiling-point measured visually, by thermometer, or by a steam-pressure-switch like you'd find in a kettle..?)]
I can confirm that the "pour between two metal cups" method is indeed still used in India today. Even in more circumstances—if you order coffee in a restaurant, it might be served to you with two cups even though you're sitting down and have plenty of time. I don't understand why steel cups are used, though, given that steel is such a good heat conductor...
Anyway, for these experiments, the water started at very close to 100 ℃ as measured with the same digital thermometer as used for the other measurements. Each time I need to refill the kettle, I let it reach the maximum temperature the thermometer could show, and left the kettle on the heat at all times between measurements.
The method is also called 'meter coffee' in southern India.
Incropera and DeWitt’s Fundamentals of Heat and Mass Transfer almost definitely has the solution to this.
Falling water droplets are covered in chapter 7, and I’m sure something in there would allow you to derive the equation for a continuous laminar flow.
Is it droplets or a continuous stream, though?
tl;dr: I suspect they had both situations in different runs of the experiment.
In the continuous case, the observed temperature at the base should be an exponential decay in the product of the height h₀ and the initial outlet radius r₀. That is: if air has temperature T_a and the water is initially at temperature T_b but has temperature T at the base, then
T/T_a≈1+(T_b/T_a-1)exp(-Ch₀r₀)
for some constant C (derivation below). But we don't see a corresponding shift in the initial experiment; big and medium exhibit roughly the same temperatures. Presumably this is because they shift from a laminar stream in the small and medium cases to droplets in the big?
Derivation: Assume that the stream is axisymmetric (circular cross-sections of varying radius r), falling with local velocity v sufficiently large that the angle of the water surface is nigh-vertical. As long as sufficient water falls to avoid pinch-off, conservation of mass implies that vA is constant, where A is the cross-sectional area of the stream (A=πr²). During an infinitesimal time interval [t,t+dt], water falls a distance v dt in a stream with radius r, and so generates a surface area 2πrv dt=2π√(v*vr²) dt. By Newton's law of cooling, the temperature flux generated during that time is proportional to the product of the surface area and the temperature difference across the surface. Thus our laws of motion are:
vr²=C₁
dv/dt=g
dh/dt=-v
dT/dt=C₂√v*(T_a-T)
where C₁ and C₂ are constants of geometry and material and "down" is positive for v but negative for h. Note that these equations hold even if the stream has sufficient internal structure ("laminar flow") to avoid thermal equilibrium, since then the effects of conduction are negligible during the brief fall; the equations still describe the outer surface of the stream; and then turbulence at the base redistributes the "coolness" flux evenly through the water. The net effect is that, relative to the non-laminar case, the total shift in temperature changes by a multiplicative factor corresponding to the proportion of the stream in the outer well-mixed surface layer — but we can absorb that factor into C₂.
Rearranging our fourth equation and letting C₃=C₂/g,
dT/(T_a-T)=C₂√(v) dt=C₃√(v) dv
Thus T_a-T=-C₄*exp(2C₃*v^³/₂) for some constant of integration C₄.
At time t=0, v=C₁/r₀² but T is some boiling T_b. Thus
C₄=(T_b-T_a)exp(-2C₃(√(C₁)/r₀)³)
As is well-known, the second and third equations integrate to give
v=C₁/r₀²+gt
h=h₀-C₁/r₀²t-½gt²
where the initial height is h₀ and radius of the outlet is r₀. However, the height of the fall is unlikely to be so large that the water experience substantial acceleration during the drop, and we can approximate the time of arrival at the base as t≈h₀r₀²/C. The corresponding velocity is C₁/r₀²+gh₀/(C₁/r₀²).
Putting it all together,
T=T_a(1+(T_b/T_a-1)exp(-2C₃((C₁/r₀²+gh₀/(C₁/r₀²))^³/₂-(√(C₁)/r₀)³)))
Since we've already assumed the gravitational velocity-shift gh₀/(C₁/r₀²) small relative to the initial velocity C₁/r₀², we can approximate the difference of square-roots-cubed with the derivative: (x+dx)^³/₂-x^³/₂=³/₂√x dx. Then
T/T_a≈1+(T_b/T_a-1)exp(-3C₃√(C₁)/r₀∙gh₀/(C₁/r₀²))=1+(T_b/T_a-1)exp(-3C₂/√(C₁)∙h₀r₀)
(Note that g drops out entirely! Pour water in a pressurized moon base and you should observe the exact same temperature drop — except, of course, that then the water might fall slow enough for the gravitational velocity-shift neglected above to become relevant.)
Depending on how small u=C₂/√(C₁)∙h₀r₀ is, the final result above can be well-approximated as a linear decrease (u very small), a quadratic decrease (u moderate), or neither (u large). In the initial experiment, Dyno-might observed a quadratic decrease, but in the kettle experiment they observe linear decrease. I assume this is because the kettle experiment did not have a well-mixed stream, reducing the effective C₂ as discussed above.