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:

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


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?

  • $\begingroup$ Aluminium oxide does form large transparent crystals e.g. in the Ruby laser, in which the crystal is doped with a few chromium ions, so it is possible that at high temperature the micro crystals are combining to become more transparent and less light scattering. At high temperature the radiation losses would be that due to a 'black body, as with any other material. Have you tried to heat a crucible up with an acetylene torch to see if transparency is changed? $\endgroup$
    – porphyrin
    Jul 1, 2018 at 21:05
  • $\begingroup$ What temperatures are we talking about?!? What does the manufacturer say about accuracy at high temperatures? Why don't you test this, by measuring a standard substance? $\endgroup$
    – Karl
    Jul 8, 2018 at 10:04
  • $\begingroup$ between 25 °C and 1500 °C. The manufacturer says that it should be avoided but not why. I cannot test this at the moment and I want to understand the reasons. $\endgroup$
    – user65662
    Jul 8, 2018 at 10:08

1 Answer 1


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.

  • $\begingroup$ Thanks, this answers question #1. But I need a source for this, wikipedia and websites are not really quotable. Do you (or anybody else) have a valid source? $\endgroup$
    – user65662
    Jul 2, 2018 at 6:55
  • $\begingroup$ Rayleigh light scattering is considered common knowledge and should not require a source. Here is an article for the transmission of sapphire: osapublishing.org/ao/abstract.cfm?uri=ao-27-2-239 $\endgroup$
    – A.K.
    Jul 2, 2018 at 12:45
  • $\begingroup$ Ok, I've skimmed through the paper and some other sources. They all write, that with rising temperature the absorption coefficient also rises. So shouldn't the losses be lower? Despite the fact, that your explanation is understandable, the larger absorption coefficients are confusing me again. This might be due to some misunderstandings between absorption, transmission, emission and scattering... $\endgroup$
    – user65662
    Jul 7, 2018 at 9:10
  • $\begingroup$ The absorption can rise, but the intensity of incident light from the sample increases at a faster rate. Thus the relative amount of ligt (the ratio) is less but overall the absolute amount of radiation that transmits through the alumina is greater $\endgroup$
    – A.K.
    Jul 10, 2018 at 15:04
  • $\begingroup$ Ok, but why is there almost no literature which describes this? Maybe this is only an indirect conclusion form basic heat transfer and radiation theory? $\endgroup$
    – user65662
    Jul 12, 2018 at 6:54

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