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Why is it when I turn on the stove the flame
comes out blue? Also sometimes it gets orange on
the top? So why's that?
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Question Date: 1998-11-12 | | Answer 1:
Blue flames come from very hot, complete
burning of air and fuel. Orange color comes from
cooler spots in the flame where fuel isn't being
totally burned and these produce a little bit of
soot and smoke. The color of the flame is related
to the heat of the flame. Think about the colors
of different hot things: the Sun is white or
yellow, a hot stove is red, a camp-fire is orange,
and a gas flame is usually blue. These colors
come from different kinds of a light and different
places in the electromagnetic spectrum. This
spectrum goes from relatively cool red things to
very hot blue and violet things:
Red - Orange - Yellow - Green - Blue - Indigo -
Violet (think "ROYGBIV") relatively cool
very hot
a stove top or toaster a gas flame
When the fuel in your gas flame doesn't
burn completely, it doesn't get as hot and the
flame is red/orange instead of blue. So, as a
question for you, will food cook more quickly over
a roaring orange campfire or a quiet blue gas jet?
Why? Hope that helps. | | Answer 2:
A flame is a region of strong chemical
reaction. For example, a fuel, like say propane
gas, a hydrocarbon can react with diatomic oxygen
to produce carbon dioxide plus water vapor and
lots of heat (and some light also). This is called
a COMBUSTION reaction. (LOOK THIS WORD UP
IN A SCIENCE ENCYCLOPEDIA) So when you see a flame
you are looking at this reaction zone where the
fuel is being converted to the products of
COMBUSTION. Now the rate of the reaction, and
hence the temperature and COLOR of the flame, is a
function of the REACTION rate...that is, how fast
the oxygen can combine with the fuel (natural gas
or propane or gasoline or oil , etc.).
So at the edges of the reaction region where
oxygen is more abundant (since air in the room is
21% by volume of oxygen) is where the temp is
highest. | | Answer 3:
Blackbody radiation is the emission of
light from a hot body; higher the temperature
larger the frequency of light emitted (light
emitted will shift from infrared, red, green, blue
towards UV as temperature increases). This is what
the answer on science line points to: more
complete the combustion process, more the heat
released, and higher the temperature, which makes
it look blue.
Although the above reasoning of the answer and
the physics is correct, this is not typically what
happens in gas flames. There is another mechanism
by which gases/molecules or even solids emit light
(LED: light-emitting diode is a great example of
solid emitting light). This is a quantum
mechanical phenomena, where the molecule gets into
an excited state with the energy supplied by the
combustion process and subsequently relaxes to a
lower energy state by releasing the extra energy
in the form of light. The color of the light
emitted in this process is characteristic of the
molecule excited.
Both, blackbody radiation and molecular
excitation should be discussed in the context
of color of the flame. Orange flame is due
to blackbody radiation coming out from an
incomplete combustion or slightly cooler area,
where the released energy is not enough to excite
the molecules. When the combustion process is
complete and produces enough heat to excite
molecules, the molecular excitation produces a
distinct bluish-green spectrum.
A simple and beautiful experiment to see
molecular excitation in play is to hold a salt in
flame and see the color emitted. This is often
used to identify a salt, since the color is
characteristic of the molecule. Calcium salts emit
brick-red color, Copper compounds emit greenish
blue flame. Here is a youtube link: click
here to watch
| | Answer 4:
Experiment suggested by a reader:
I just thought of an experiment to suggest to your
audience. This can be carried out in an
undergraduate lab.
Get a Bunsen burner. Take some small (small
compared to the size of the burner flame) piece of
something that will not melt or burn, ceramic,
metal, a piece of rock, it doesn’t make much
difference. Fashion some sort of support out of
very thin platinum wire (thin to reduce the heat
transfer through the support). Hold the object in
the hottest part of the flame for a minute or so
until it reaches thermal equilibrium with the
flame. Note color. If the color of the flame is
due to black body radiation, then the radiation of
an actual body in thermal equilibrium (just to say
the same temperature) should radiate the same,
that is, the object should glow blue (In fact, it
might actually disappear against the blue
background; this is the idea behind a radiant
pyrometer ). It will not. It will, at best,
glow reddish. One cannot argue that the object is
cooler precisely because it is radiating since (1)
the premise of the flame color is that it is
radiating black body as well (although it could be
less efficient, i.e., less “black”) (2) the
surface area through which it is absorbing heat is
the same as the surface area radiating, same as
for a soot particle, and those do glow
yellow-orange.
If anyone does this experiment, please let me
know. I would like to know the result.
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