Does alumina become transparent at high temperatures? Consider a normal opaque alumina crucible. I've heard that they become transparent at higher temperatures and therefore, are not appropriate for high precision DSC (caloric) measurements. The reason is, that there might be a lack of energy due to radiation losses (in addition to the bad thermal conductivity). 

I was not able to find at least one quotable source for this claim. So my two questions are:


Is this actually right?
Does someone know of a quotable source for this (scientific papers, textbooks, etc...)?


Thanks!

Edit: Based on my research, the absorption rises with temperature, so the transparency should degrease, what is a somewhat contradictory to A.K. answer, isn't it?
Well, alumina is sapphire so stand to reason that it is transparent at a low temperature too.  I think what is meant by what you heard is that the higher intensity of infrared light is transmitted by the alumina causing thermal loss. 

Alumina is transparent to UV, visible and infrared light. To our visible light vision, the scattering and reflection of polycrystalline alumina make it appear as opaque white. In the IR spectrum though it may appear transparent. This is because scattering is proportional to the inverse of the wavelength to the fourth power ($\alpha _{\text{scat}} \propto \lambda ^{-4}$).  Thus a wavelength just twice as long a red light (the longest wavelength of visible light; $\lambda = \pu{750nm}$ ), in this case, $\pu{1.5 \mu m}$, which is near infrared, would scatter one-sixteenth ($6.25 \% $) as much as the red light would. A wavelength twice as long as that wavelength ($\pu{3\mu m}$) would be in the thermal infrared spectrum and be scattered one-two-hundred-fifty-sixth as much as the red light ($0.39\%$) This reduced scattering allows more transmission. At high temperature when infrared radiation is more intense, the energy loss due to transmission becomes more significant, but the alumina was always transparent.