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I have a stainless steel kettle that I let boil dry. That is, I left it on a hot coil burner with nothing in it, and noticed a few minutes later that the bottom was red hot and smoking. To make matters worse, I panicked and quenched it with cool water.

Both the inner and outer surface were left discolored with various shades of yellow and purple. This seems to be a known consequence of overheating, but I can't find an explanation of what actually happened.

  1. I understand why heat (especially combined with quenching) would warp the metal, but why would it cause discoloration (that persists after the metal has cooled)?

  2. After hunting around, I saw several suggestions that boiling white vinegar could remove the discoloration, and to my suprise, it worked pretty well. Why does this work?

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  • $\begingroup$ Welcome to chemistry.SE! If you had any questions about the policies of our community, please ‎visit the help center. $\endgroup$
    – M.A.R.
    Commented Dec 22, 2015 at 13:35
  • $\begingroup$ It is all about thin oxide layer on the surface. Looks like stainless steel is not absolutely stainless, after all. Strange enough, the Wiki page about it does not exist in English. $\endgroup$ Commented Dec 22, 2015 at 13:40
  • $\begingroup$ Steel (even stainless) will oxidize at high temperature. Vinegar, which is acidic, will etch the oxide, leaving a slightly thinner and probably warped kettle. $\endgroup$
    – Jon Custer
    Commented Dec 22, 2015 at 13:40

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Stainless steel have many different compositions. If you heat it up the surface undergoes oxidation. Diffusion at higher temperatures plays also its roll - shifts surface composition of the kettle. The various shades of yellow and purple are metal oxides. The vinegar removes the oxides, thus removes the coloration. Vinegar should remove the oxides even without boiling. It will just take more time.

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