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ANSWERED on Sun 28 Feb 2010 - 2:17 pm UTC by leader

Question: Can you have black flames?

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Sun 28 Feb 2010 - 3:26 am UTC

Question

kutibah
Customer

Hypothetically speaking, if the flames of a fire could be black, would it be hotter than a blue or red flamed fire? I know this probably does not exist in physical life, but just by going off the principles of chemicals which cause fire, what would cause a fire flame to be black, and if so, would it be hotter than a blue or red flame?

 
 

Sun 28 Feb 2010 - 2:17 pm UTC

Answer

leader
Former Researcher

Practically, it is difficult to produce a black flame but theoretically such a flame will exist if flame temperature is so low that it does not emit any visible color (see 1). Often, we observe black color on the outer edges of a candle flame or near the base. The blackish color on the edge is actually soot particles and residue produced from a yellow colored flame due to lack of oxygen around that particular area (see 2). You may also see black color at the inner most zone (base) of the candle flame. The black color is actually wax vapor which are not burnt due to lack of oxygen (see 3). Disregarding invisible radiation, flame temperature may be categorized as red, orange, yellow and white/blue (see 4, 5). Therefore, visible black wax vapors at the bottom of a flame or black soot particles around the edge of a yellow flame are not as hot as other flame colors.

(1) http://www.newton.dep.anl.gov/askasci/chem03/chem03656.htm
(2) http://www.straightdope.com/columns/read/2159/what-is-the-hottest-part-of-a-flame
(3) http://www.educationalelectronicsusa.com/c/fuels-VIII.htm
(4) http://ask.metafilter.com/43255/Flame-temperature
(5) http://www.coolquiz.com/trivia/explain/docs/flame.asp

For a simple illustration, see link 3, http://www.educationalelectronicsusa.com/c/fuels-VIII.htm

Thanks.

 
 

Sun 28 Feb 2010 - 4:16 pm UTC

Comment

myoarin
User

I hope Kutibah likes that answer, because I like to see questions answereed.

BUT by definition
"A flame (from Latin flamma) is the visible (light-emitting) gaseous part of a fire."  (wikipedia article "flame")

Or this::
Definitions of flame on the Web:
fire: the process of combustion of inflammable materials producing heat and light and (often) smoke.

And then the definition of black:
Definitions of black on the Web:
being of the achromatic color of maximum darkness; having little or no hue owing to absorption of almost all incident light;

A flame is by definition visible light, whatever its color, but black is the is the absorbtion of light, quite the opposite.

Perhaps  - I don't know -  a source of heat could emit "black light,"  UV-light, but "black light" is a misnomer.  Because it can't be seen, it is not black but rather invisible to the human eye.

But if you like that information, you can give credit to Leader for inspiring a rebuttal.

Regards, Myoarin

 

Sun 28 Feb 2010 - 6:46 pm UTC

Comment

leader
Former Researcher

Thanks Myoarin for constructive feedback. This is precisely why I have mentioned that the 'black' that we see in candle is not the color of the emitting light but a burning substance (residue OR wax).

 

Sun 28 Feb 2010 - 8:42 pm UTC

Uclue Researcher Comment

Roger Browne
Researcher

It all depends what you mean by "black", of course. Usually when we say something is black, we mean that it reflects no light.

But flames emit light rather than reflecting it. I think it's reasonable to say (colloquially) that if you have a flame in a dark room, and you can't see the flame, then that flame is black.

In leader's fourth reference the figure of 425 degrees Celsius is given as the cutoff temperature below which a flame won't emit visible light. There are substances which can burn below 425 C. If the heat is conducted away from such a combustion fast enough that the temperature won't rise above 425 C, I think we could say that the flame was black (although I'd prefer to describe it as "invisible").

Kutibah's question was hypothetical: IF there is such a thing as a black flame, is it hotter than a blue or red flame? The answer to that is clearly "no". If your definition of "black" allows for black flames, then those flames must be cooler than 425 degrees Celsius.

 

Mon 1 Mar 2010 - 1:58 am UTC

Comment

myoarin
User

I'll forget about the definition of flame = light,

but I don't want to be near when Roger tries to conduct away the heat from burning one of the following substances:

"Autoignition point of selected substances

Temperatures vary widely in the literature and should only be used as estimates. Factors which may cause variation include partial pressure of oxygen, altitude, humidity, and amount of time required for ignition.

    * Triethylborane: −20 °C (−4.0 °F)
    * Silane: <21 °C (70 °F)
    * White phosphorus: 34 °C (93 °F)
    * Carbon disulfide: 90 °C (194 °F)
    * Diethyl ether: 160 °C (320 °F)[3]
    * Diesel or Jet A-1: 210 °C (410 °F)
    * Gasoline (Petrol): 246–280 °C (475–536 °F)[4]
    * Butane: 405 °C (761 °F)[5]
    * Paper: 450 °C (842 °F)[6]
    * Magnesium: 473 °C (883 °F)
    * Hydrogen: 536 °C (997 °F)[7]"

http://en.wikipedia.org/wiki/Autoignition_temperature

 

Mon 1 Mar 2010 - 10:41 am UTC

Uclue Researcher Comment

Roger Browne
Researcher

Obviously it's impractical to conduct away enough heat to keep burning white phosphorus cool enough so that it does not emit light.

But it can surely be done with the hydrocarbons, because the flow rate of those fuels is easy to control. Introduce through a nozzle a steady but very small flow of the substance into an oven at the autoignition temperature.

After you have ignition, reduce the air temperature in the oven to the point where it is just enough to sustain combustion, and the "flames" (being just above the autoignition temperature) won't be hot enough to glow.

 

Mon 1 Mar 2010 - 1:18 pm UTC

Comment

myoarin
User

You could also try to spool wires of copper (better silver or gold) through the flame to carry off heat. 

And if the "flame" happens to go out, you couldn't see it anyway.

I'll still stay behind the glass shield, but credit to Roger for a possible solution.

 

Tue 2 Mar 2010 - 5:33 am UTC

Accepted and rated

kutibah
Customer

Thank you so much for all your help!

 

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