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!

▲ 3eb7988a1663 6 hours ago | parent | next [-]

The appendix lists the equations transcribed from the raw answers.

  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

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kurthr 6 hours ago | parent [-]

It looks like a lot of them are missing something big. I'd think the two big ones are the evaporative cooling as you pour into the cup, and heating up the cup (by convection) itself. The convective cooling to the air is tertiary, but important (and conduction of the mug to the table probably isn't completely negligible). If there's only one exponential, they're definitely doing something wrong.

I'd like to see a sensitivity study to see how much those terms would need to be changed to match within a few %. Exponentials are really tweaky!

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andai 5 hours ago | parent [-]

Is that what that first drop is? The cold cup stealing heat from the coffee?

	▲	kadoban 5 hours ago \| parent [-]
		It's a mix of course, but I think it should be mainly that and evaporative cooling. Evap is _very_ effective but will fall off rapidly as you get away from boiling. The conduction into the mug will depend a lot on the mug material but will slow down a lot as the mug approaches the water temperature. I'd be very interested in seeing separate graphs for each major component and how they add up to the total. Even asking the LLMs to separate it out might improve some of their results, would be interesting to try that too.

▲ amelius 6 hours ago | parent | prev [-]

That model doesn't explain the relatively sharp drop in the beginning.

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spiralcoaster 2 hours ago | parent | next [-]

It absolutely does. The model that came closest simply used that model twice in the same equation. One for the cup and one for the air.

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coder68 6 hours ago | parent | prev | next [-]

It does? There is a fast drop followed by a long decay, exponential in fact. The cooling rate is proportional to the temperature difference, so the drop is sharpest at the very beginning when the object is hottest.

	▲	amelius 6 hours ago \| parent [-]
		I mean that initial drop doesn't look like it is part of the same exponential decay.

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bryan0 6 hours ago | parent | prev | next [-]

Are you sure? I believe Newtown's law of cooling says the temperature will drop sharply at the beginning:

dT/dt = -k(T_0 - T_room)

so T(t) = T_room + (T_0 - T_room) exp(-kt)

exp(-x) has a fast drop off then levels off.

	▲	amelius 5 hours ago \| parent \| next [-]
		https://www.electronics-tutorials.ws/rc/time-constant.html scroll down, these graphs just don't look similar.
	▲	cyberax 3 hours ago \| parent \| prev [-]
		Ha. My university professor used this in a lab to catch people who slack off. There is another factor here: convection. Its speed depends on the viscosity of the fluid and the temperature difference both. And viscosity itself depends on the temperature, so you get this very sharp dropoff.

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lacunary 5 hours ago | parent | prev [-]

probably dominated by the cup as the ambient temperature initially and then as air/the counter top as the ambient temperature on the longer time scale, once the cup and the liquid near equilibrium