LLMs predict my coffee (dynomight.net)

The title is very misleading. This has almost nothing to do with coffee. I was expecting that the input would be the parameters of a coffee recipe (like quantities of coffee and water, grind size, etc for a given type of preparation), and the output to have something to do with coffee too (like extraction time, rate, etc.). It actually is just about water cooling down. Also, it doesn't actually ask the LLM for actual prediction about the result of the experiment, only to generate a ±textbook formula for the situation (which is a good point since LLMs aren't made for that at all, but contributes to make the title misleading).

  Does that seem hard? I think it’s hard. The relevant physical phenomena include at least..,

In most engineering problems, the starting point is recognizing that usually one or two key things will dominate and the rest won’t matter.

To me the neat bit isn't that it got the exponential decay right - that's pretty standard, its that it realised there were two different timescales for the decay and got ball-park numbers for them pretty well.

This is the kind of model you would expect from a simple cylindrical model of the coffee cup with some inbuilt heat capacity of its own.

However, those decay coefficients are going to be very dependent of the physical parameters of your coffee cup - in particular the geometry and thermal parameters of the porcelain. There's a lot of assumptions and variability to account for that the models will have to deal with.

On a related note, I have been working on an app that helps determine the correct grinder setting when dialing in espresso. After logging two shots with the same setup (grinder, coffeee machine, basket etc), it then uses machine learning (and some other stuff that I am still improving) to predict the correct setting for your grinder based on the machine temperature, the weight of the shot etc.

https://apps.apple.com/ph/app/grind-finer-app/id6760079211

Its far from perfect when it comes to predictions right now but I expect to have massive improvements over the coming weeks. For now it works ok as an espresso log at least.

I'm hoping after a few tweaks I can save people a lot of wasted coffee!

There's a simple differential equation often taught in intro calc courses, "Newton's Law of Cooling/Heating," which basically says that the rate of heat loss is proportional to the difference in temperature between a substance and its environment. I'm curious what that'd look like here. It's a very simple model, of course, not taking into account all the variables that Dynomight points out, but if a simple model can be nearly as predictive as more complex models...

I'm also curious to see the details of the models that Dynomight's LLMs produced!

  LLM  T(t)  Cost
  Kimi K2.5 (reasoning)  20 + 52.9 exp(-t/3600)+ 27.1 exp(-t/80)  $0.01
  Gemini 3.1 Pro  20 + 53 exp(-t/2500) + 27 exp(-t/149.25)  $0.09
  GPT 5.4  20 + 54.6 exp(-t/2920) + 25.4 exp(-t/68.1)  $0.11
  Claude 4.6 Opus (reasoning)  20 + 55 exp(-t/1700) + 25 exp(-t/43)  $0.61 (eeek)
  Qwen3-235B  20 + 53.17 exp(-t/1414.43)  $0.009
  GLM-4.7 (reasoning)  20 + 53.2 exp(-t/2500)  $0.03

The fact that near boiling water cools down quicker than warm water used to be a well-known kitchen knowledge bit. Like my grandma who wasn't a physicist at all knew it. I guess in some places (particularly those where people microwave water) that part of culture is lost cause there's at least a whole generation which hasn't done cooking.

" Does that seem hard? I think it’s hard. The relevant physical phenomena include at least"

Imo no, this seems like something that would be in multiple scientific papers so a LLM would be able to generate the answer based on predictive text.

The problem is both highly complex, but fairly easy to model. Engineers have been doing this for over a century.

Of all the cooling modes identified by the author, one will dominate. And it is almost certainly going to have an exponential relationship with time.

Once this mode decays below the next fastest will this new fastest mode will dominate.

All the LLM has to do, then, is give a reasonable estimate for the Q for:

$T = To exp(-Qt)$

This is not too hard to fit if your training set has the internet within itself.

I would have been more interested to see the equations than the plots, but I would have been most interested to see the plots in log space. There, each cooling mode is a straight line.

The data collected, btw, appears to have at least two exponential modes within it.

[The author did not list the temperature dependance of heat capacity, which for pure water is fairly constant]

Is this for real?

This is like someone with no background in physics or engineering wondering "can a LLM predict the trajectory of my golf ball". They then pontificate about how absolutely complex all of the interacting phenomenon must be! What if there was wind? I didn't tell it what elevation I was at! How could it know the air density!? What if the golf ball wasn't a perfect sphere!!? O M G

And then being amazed when it gets the generic shape of a ballistic curve subject to air resistance.

This speaks far more to the ignorance of the author than something mind boggling about the LLM.

The water temperature drops quickly because the room temperature ceramic mug is getting heated to near equilibrium with the water. If you used a vacuum sealed mug(thermos) then the water temp would drop a bit but not much at all initially.

I transcribed the data and fitted dual exponentials to it. When time t is in minutes, the data seem to follow

     T(t) = 20 + 25e^(-2.3*t) + 54e^(-0.034*t)

This is very close to what the LLMs suggested. If I wanted to make an initial guess at this as accurate as the LLMs, what would I need to know? My interpretation of the coefficients is:

(a) 20 ℃ represents the room temperature this will eventually reach.

(b) 25 ℃ is how much of the temperature the mug will absorb as it is heating up.

(c) The decay -2.3 represents how fast heat is transferred to the mug. (It will be halfway after 20 seconds.)

(d) 54 ℃ is the differential between room temperature and starting temperature once we've accounted for the loss of 25 ℃ to heat the mug.

(e) The decay -0.034 is how fast heat is transferred out of the mug to the room. (It will be halfway to room temperature after 20 minutes.)

I'm okay with (a), and I could probably have guessed (d) once I know the other parameters.

I can also sort of see myself figuring out (b): I would guess the heat capacity of the mug would be maybe 500 × 0.6 = 300 J/K, do the same for the water (4000×0.2 = 800 J/K). Some work later this comes out to a temperature loss of 20 degrees. Close enough.

But even if I tried to use my intuition for how hot the mug feels as these processes go on, I would have ended up nowhere near -2.3 and -0.034 for the decay coefficients. What would I need to know about convection, mug materials, and air properties to guess that more accurately?

Is it a neat coincidence or a good, very approximate rule of thumb that heat transfer to air is about 60× slower than that to ceramic-like solids?

It isn't that surprising that it works well, this problem is fairly well known and some simple heat equations would lead to the result, about which there is a lot of training data online.

That initial drop reminds me of one of the things that stuck to me from my thermodynamic lectures / tests: If you want to drink coffee at a drinkable temperature in t=15min, will it be colder if you add the milk first or wait 15min and then add milk? (=waiting 15 min because the temperature differential is greater and causes a larger drop). Almost useless fact, but it always comes up when making coffee.

Slightly related, I was using an LLM to help me understand whether I should add milk to my coffee before walking to my table or when I get to my table (objective to maximise coffee temperature at the point drinking). Turns out it's best to add the milk immediately when the coffee is made because the rate of cooling is higher at higher temperatures.

Irrelevant to your specific cup of coffee its giving you a generic answer.

The interesting bit about this physical experiment is that the water in the cup never starts at 100 celsius. That the act of pouring significantly reduces temperature is well-documented, so in some sense the LLM output is surprising.

This article is somewhat baffling in that it presents the graphs but not the equations the LLMs provided. Kind of implying they provided some unique models (maybe they did but I seriously doubt it).

If equations were included you'd probably see a standard equation (Integral form of Newtons law of cooling). With the time parameters known from the input and the heat transfer parameters having reasonable guesses (cup opening area, mass of water).

No estimate of uncertainty in his measurements so he can't really tell who's most right.

It looks like the author forgot to insert the joke in the third last paragraph — the author left the placeholder right there in the text! But wait... is the joke forgetting to insert the joke?

Tangential. I want a coffee maker that prepares coffee and cleans up after itself automatically. The ones I found are very expensive. Any DIY hack?

nice benchmark! coffee-based Turing test.

[dead]

... and so another benchmark is born.