Author Options:

How does 'fire cupping' work, and what's the Physics behind it? Answered

This is another unethically homework-based question. 
I have to estimate the pressure inside the cup, but I don't get how low pressure is formed. 
I am not sure that you can heat the air inside so much that the thing will suck in your skin immediately without giving you bad burns. Anyway, they put a match in it for a really short time...
Any ideas?
The topic is Gas Laws, if it can help anyone.


The air in the cup is heated and expands - The cup is then sealed on the skin. As the air cools it forms a vacuum in the cup. Nothing complicated.

If it was like that, then the force with which it sticks should be almost zero at the beginning, but then gradually rise. BUT it sticks quite firmly right off and the force doesn't change much.

Consider the heat capacity of 'air' compared to the heat capacity of the glass -- the air will cool very quickly

Is this that Chinese practice Jackie Chan did on karate kid? It is actually not the heating and contracting of gases. From a simple Bill Nye experiment I remember as a kid, it is process of converting oxygen to carbon dioxide. Invert a cup over a candle in a bowl. Fit it with water, watch the water rise as the flames go out. Too tired to do calculations, but you could probably do stoichometry to see the difference in gases before and after the reaction.

If we estimate the candle as pure carbon then C+O2=CO2 => there is no difference in molar quantities so in the first estimation the effect is only present due to the fact that CO2 dissolves in water in terribly large amounts.

It's been to long since I've done stoichiometry. However I do remember in a combustion reaction, H2O is always a by product. Therefore if we are burning ethanol, the equation should look like CH3CH2OH + 3O2 = 2CO2 +3H2O. But then again I think you are right, since atoms cannot be created or destroyed, only rearranged.

I guess the change in temperature must be the answer.

Well, when you put your hand on the cup, you extinguish the flame. When that happens, the temperature drops, thus less pressure..... simples.....

Rickharris is exactly right.

"wouldnt it rise gradually"...

The temperature change is quite rapid considering the heat capacity of the glass and the heat capacity of the air. (within a few seconds the air will drop back to ambient temperature, chilled by the glass.

if I remember correctly, PV = nRT. Pressure * Volume = mass * constant * Temperature.
http://en.wikipedia.org/wiki/Ideal_gas_law (confirmed)

Only thing changing will be pressure and Temperature, the rest are constant. Since they are on opposite sides of a formula they are directly proportional. More temperature = more pressure, less temperature = less pressure.

Temperature is measured in Kelvin (absolute temperature) so say the temperature is around 100C (for a very short time, probably much more) (393Kelvin), and drops once the flame extinguishes back to room temperature (25C ish) = 318 Kelvin.
pressure is normally 100 Kilopascals give or take.

Volume doesn't change. Amount of gas doesn't change. Gas constant ...is a constant.

Now we know all the information we need to solve.

100kpa @ 393 kelvin == x kpa @ 318 kelvin. Cross multiply
100 * 318 / 393 = ~80 kpa, or 20kpa difference between inside and outside.

Now, Pressure is a force over an area. More area = more force. 1 Pascal is 1 Newton / 1 meter squared. Say the cup has a 5cm diameter opening (20.25 cm sq or .002 square meters) We have a pressure of 20,000 Pascals applied to 0.002 square meters. 20,000pa * 0.002 m^2 = 40 Newtons of force pushing the glass against the skin. In other words roughly equivelant to the force exerted by gravity on 4 kilograms.

nearlyright... n is not the mass it is the number of molecules messured in mol...
(and for the higher temperature id choose something more like 300°C as thats about the temperature of the flame, but thats irrelevant)

They only stick the match in for a second, so the higher temperature should be around 330 kelvin.

no, you shouldn't think of it as 'they heat the air inside' as it's more like 'they "fill" the cup with the flame'... the gases inside the 'hot' cup are more the burnt stuff than heated air

What is 'burnt stuff' and a 'cup filled with flame'?

. PVT (Pressure, Volume, Temperature). V remains constant. Temperature drops. What happens to Pressure?

Remember that the Ideal gas law only applies to ideal gases, Boyle's is a varient.


Another example: In a classic school Chemistry demonstration a piece of paper is burnt inside a bottle that is connected by a tube with a container with water. They say that Oxygen gets burnt, so water flows from the container to the bottle.
BUT, if you burn paper you get C+O2=CO2, and as N(O2)=N(CO2) (N is nu), so V(O2)=V(CO2) and there shouldn't be any pressure difference...

You don't get as much CO2 as the oxygen you burn. Most of the oxygen goes into converting the paper into ash, hence the decrease in gas pressure.

So what you need to know, for this as for your original question, is how much oxygen is acquired by the solid waste products of the burning. That'll give you a first approximation to the pressure difference. You can then apply a correction for how much CO2 is actually produced.