When heating a metal tube with polypropelene in it, i found that i still had brittle black foam and flakes in it even though it had been in a blue flame at around 500'C

  1. What happens to complex hydrocarbons/plastics when they turn into black foam at high temperatures?

  2. At what temperature do tars and heavy hydrocarbons completely lose all their Carbon and only ashes are left?

  • 1
    $\begingroup$ In short, there is no such temperature. You can't remove carbon by heating alone. Or rather, you can, but first you will vaporize your metal tube and everything else, so that in the end there would be no ashes left. No, the presence of oxygen is the key. $\endgroup$ Nov 16, 2015 at 12:19

1 Answer 1


What happens depends whether it is being heated in the presence or absence of oxygen. For example, in the manufacture of coke (mostly carbon), coal is heated without oxygen and from Coke Production For Blast Furnace Ironmaking:

The coal-to-coke transformation takes place as follows: The heat is transferred from the heated brick walls into the coal charge. From about 375°C to 475°C, the coal decomposes to form plastic layers near each wall. At about 475°C to 600°C, there is a marked evolution of tar, and aromatic hydrocarbon compounds, followed by resolidification of the plastic mass into semi-coke. At 600°C to 1100°C, the coke stabilization phase begins.

Without oxygen, the carbon will remain upto its boiling point of 3825 °C.

With oxygen, carbon as coke with start combustion at its autoignition temperature of about 700°C

  • $\begingroup$ I wanted to add on to DavePhD's fine answer with a simple point. "Ashes" in general refers to the solids left over after oxidative combustion of hydrocarbons or carbohydrates. The primary components of "ashes" are inorganic salts. Sodium and potassium oxides, phosphates, sulfates, and carbonates, along with lower amounts of calcium and magnesium salts of these same anions, would be the primary constituent. $\endgroup$
    – Curt F.
    Nov 16, 2015 at 22:44

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.