LLMs Predict My Coffee

5 comments

• amha 1 hour ago
  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!

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  • 3eb7988a1663 1 hour ago
    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 1 hour ago
      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 7 minutes ago
        Is that what that first drop is? The cold cup stealing heat from the coffee?
  • amelius 59 minutes ago
    That model doesn't explain the relatively sharp drop in the beginning.
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    • bryan0 41 minutes ago
      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.

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      • amelius 0 minutes ago
        https://www.electronics-tutorials.ws/rc/time-constant.html

        scroll down, these graphs just don't look similar.
    • coder68 48 minutes ago
      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.
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      • amelius 43 minutes ago
        I mean that initial drop doesn't look like it is part of the same exponential decay.
• andy99 1 hour ago

    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.
• kaelandt 1 hour ago
  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.
• IncreasePosts 13 minutes ago
  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.
• leecommamichael 36 minutes ago
  ... and so another benchmark is born.