What is a flame? why does sulfur burn?

  • Thread starter Thread starter g.lemaitre
  • Start date Start date
  • Tags Tags
    sulfur
AI Thread Summary
A flame is produced when sulfur on a match is struck against a hard surface, generating friction that excites electrons and releases photons, which we see as light. The combustion process involves a chemical reaction that converts wood into ash, with the heat causing atoms to break their bonds and release gases. The visible flame results from tiny unburned particles heated to glowing temperatures, and its color can vary based on the substances involved. As wood burns, most of its mass is converted into gases, leaving behind ash, which consists of non-combustible materials and structural remnants. The overall process illustrates the transformation of substances during combustion, characterized by energy release and the formation of new compounds.
g.lemaitre
Messages
267
Reaction score
2
When we strike a match which is made of sulfur against a hard surface (I don't know what the surface is made of) a flame is produced. That flame can then cause a piece of wood to burn. We can also see the flame. Why?

Let me take my best guess since this website likes people to try. Please correct my mistakes.

When sulfur strikes the hard surface friction is produced. Friction is essentially atoms bumping up against each other. The electrons of the two different substances rub against each other and since electrons are negative it requires energy to push them close. That energy causes the atoms to jitter more and the more they jitter the more the electrons ascend into a higher orbit. When the electrons ascend into a higher orbit they release a photon, which is the flame that we see. Photons have energy so when a flame from a match made of sulfur comes into contact with wood, the photons are hitting the wood which is mostly carbon, hydrogen and oxygen. As the photons hit the wood they cause the atoms in the woods to excite which causes the electrons surrounding the nucleon to ascend into a higher orbit and release a photon.

I guess I would like to know why a piece of wood gets converted into ash when burnt.

As the atoms speed up and move faster they gradually break their bonds to the other atoms in the wood. I suppose the atoms still clump together and break away as chunks which is why we can still see ash. But the space between atoms is far greater in wood than in ash.
 
Physics news on Phys.org
Combustion (burning) is a chemical process.

One of the results of this process is usually in the form of a gas and/or of tiny particles mixed up (suspended) in the gas. We see this as the flame.

By results I mean new substances procuced by the chemical reaction process that were not present originally.

That is what a chemical reaction is. The interaction of substances to form different ones.

Chemical reactions can take place if the result releases energy. Some, such as combustion, release a great deal of energy.
This is why the products of the reaction become hot.

In many cases a small input of energy, called the activation energy, is required to start the process off. This is what happens when you strike the match. The mechanical energy of striking the match is sufficient to start a small amount of substances to reacts. The reaction releases more enrgy which enables more reaction, which releases more enrgy which...

So the flame is the hot gas produced by the reaction. The colour is due to tiny unburned particles that have been heated up to glowing temperature by the heat of the reaction.
 
A flame is an ionized gas...
see here:

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

You see it because it happens to have some emissions in the visible range of lightwaves...

Also see the illustrations under 'flame color' in the above link...
 
Studiot said:
Combustion (burning) is a chemical process.
The colour is due to tiny unburned particles that have been heated up to glowing temperature by the heat of the reaction.

Only for soot. Soot particles emit a yellow light due to thermal radiation. When the flame is for instance blue, it is due to excited radicals formed in the flame that emit light somewhere in the 400-500 nm range, depending on the radicals.

Also, you can have a flame without any ionization. It's a usual flame feature, but not necessary.
 
g.lemaitre said:
I guess I would like to know why a piece of wood gets converted into ash when burnt.

As the atoms speed up and move faster they gradually break their bonds to the other atoms in the wood. I suppose the atoms still clump together and break away as chunks which is why we can still see ash. But the space between atoms is far greater in wood than in ash.
Ash consists of those combustion products that do not readily vaporise and any original constituents that were not combustible at the temperature of the combustion. Most of the mass of wood (including any water) has left as gases, so the residue is typically quite small. Robbed of the large molecules that gave it structure, it collapses into a heap. Generally, there are large airspaces in the heap. Unless pressed down, it will be much less dense than wood. Even then, the molecules are much smaller than in wood, so I would guess the average atomic spacing is greater. OTOH, there will be very little hydrogen remaining, so the atoms on average will be much heavier.
 
A flame is the spatial region where some energetic reaction is taking place. Given the high level of energy negotiated within this reaction, its parts are frequently emitting light during the time the reaction takes.
 
Thread 'Gauss' law seems to imply instantaneous electric field propagation'
Imagine a charged sphere at the origin connected through an open switch to a vertical grounded wire. We wish to find an expression for the horizontal component of the electric field at a distance ##\mathbf{r}## from the sphere as it discharges. By using the Lorenz gauge condition: $$\nabla \cdot \mathbf{A} + \frac{1}{c^2}\frac{\partial \phi}{\partial t}=0\tag{1}$$ we find the following retarded solutions to the Maxwell equations If we assume that...
Maxwell’s equations imply the following wave equation for the electric field $$\nabla^2\mathbf{E}-\frac{1}{c^2}\frac{\partial^2\mathbf{E}}{\partial t^2} = \frac{1}{\varepsilon_0}\nabla\rho+\mu_0\frac{\partial\mathbf J}{\partial t}.\tag{1}$$ I wonder if eqn.##(1)## can be split into the following transverse part $$\nabla^2\mathbf{E}_T-\frac{1}{c^2}\frac{\partial^2\mathbf{E}_T}{\partial t^2} = \mu_0\frac{\partial\mathbf{J}_T}{\partial t}\tag{2}$$ and longitudinal part...
Thread 'Recovering Hamilton's Equations from Poisson brackets'
The issue : Let me start by copying and pasting the relevant passage from the text, thanks to modern day methods of computing. The trouble is, in equation (4.79), it completely ignores the partial derivative of ##q_i## with respect to time, i.e. it puts ##\partial q_i/\partial t=0##. But ##q_i## is a dynamical variable of ##t##, or ##q_i(t)##. In the derivation of Hamilton's equations from the Hamiltonian, viz. ##H = p_i \dot q_i-L##, nowhere did we assume that ##\partial q_i/\partial...
Back
Top