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I think the heat equation says that the first derivative of temperature with respect to time in a stationary solid varies as the negative of the second derivative of temperature with respect to position. I think that at very large size scale, transparent solids with low absorptivity like bubble free ice actually diffuse heat faster than the heat equation predicts by blackbody radiation. For any solid transparent to radiation with a blackbody temperature of 300 K, if it has an initial state where temperature varies sinusodially with position and the wavelength of that wave is bigger than the expected distance for an emitted photon to travel, does the rate of its amplitude dividing exponentially vary approximately as the inverse square of the wavelength with a higher thermal diffusivity used to predict that rate than the thermal diffusivity used to predict heat diffusion at small scale? Also when the wavelength of the wave in the initial state is longer than the expected distance for an emitted photon to travel, does the thermal diffusivity used to predict the rate of the amplitude dividing exponentially vary approximately as the reciprocal of the absorptivity?

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    $\begingroup$ The heat equation is just a very primitive ad-hoc approximation that works reasonably well for many system of technical importance. There is no claim to it being a fundamental equation of any sort. $\endgroup$ – CuriousOne Jan 25 '16 at 5:22
  • $\begingroup$ The heat equation is not a primitive ad-hoc equation. It is an exact consequence of energy conservation in the limit of small temperature gradients. $\endgroup$ – Thomas Jan 27 '16 at 2:34

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