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

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