# How much oxygen would be consumed on a 1 cm squared surface which is on fire?

I'm trying to figure out how much oxygen the Human Torch produces when he is on fire. I figure if I knew how much oxygen on average (per second?) is consumed by a 1 cm squared surface which is producing flame ( rapid combustion) I would then be able to take the average surface area of a human and figure out his oxygen burn rate.

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It really depends upon the fire. Anyone who's worked with bunsen burners will know that the type of flame(oxidizing/orange or reducing/blue) can be varied via the amount of oxygen supplied. Also, the amount and type of fuel makes a difference. For a given color of flame and a given fuel, the oxygen is proportional to the amount of fuel. More fuel-->more fire-->more oxygen. So we'd need to do an analysis on the Human Torch before we find out :/. That being said, I guess someone could give an approximate value, after checking the flame height and color of the Torch and assuming ethanol fuel. – Manishearth Mar 30 '12 at 3:08

Let's suppose that the Human Torch produces heat by burning methane (maybe he eats a lot of chilli), and suppose he produces a total heat output of 10,000W - I pulled this figure out of the air so feel free to modify it up and down.

The enthalpy of combustion of methane is 882kJ/mol, so, to generate 10,000W, he needs to burn 0.011 moles of methane per second.

The equation for the combustion of methane is:

$$CH_4 + 2O_2 \to CO_2 + 2H_2O$$

so one mole of methane requires 2 moles of oxygen to burn. That means the Human Torch consumes 0.022 moles or 0.7 grams of oxygen per second. The area of skin per human is about 2m$^2$ or 20,000cm$^2$ so the Human Torch consumes about 1.1 x 10$^{-6}$ moles or 3.6 x 10$^{-5}$ grams of oxygen per cm$^2$ per second.

Later:

Let's revisit that power output of 10kW that I guessed at the start of the calculation. maybe it would be better to ask what the Human Torch's surface temperature is, and use this to calculate the power. Assume the Human Torch is a black body. This probably isn't a good approximation at ambient temperatures, but is probably OK when he's really hot. The power output of a black body is given by the Stefan–Boltzmann law:

$$j = \sigma T^4$$

where $\sigma$ is about 5.67 x 10$^{-8}$Js$^{-1}$m$^2$K$^4$. So what temperature would my guess of 10kW correspond to? Taking the area as 2m$^2$ we get:

$$T = \left(\frac{5000}{\sigma}\right)^{-4} = 545K = 272^o C$$

so not that hot really. Good if you want to make a cup of tea, but not great for burning through steel. Suppose the Human torch is really going for it and burns as hot as the surface of the Sun - 6000K to keep it a round number. The power is just:

$$j = 2 \times \sigma \times 6000^4 = 1.5 \times 10^8W$$

Using the working above he now consumes 340 moles or 10.9kg of oxygen per second or about 0.55g per cm$^2$ per second.

So you wouldn't want to be in the same room as him. Not only would you be roasted, you'd be suffocated too!

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+1 for chili :). Seriously though, the question still boils down to knowing the parameters of the Human Torch. I guess we could speculate from the time it took for him to melt the ice in the movie amd similar stuff :/ – Manishearth Mar 30 '12 at 6:37
I extended my answer to calculate oxygen consumption based on surface temperature. It's a long time since I read any Marvel comics, so I can't comment further :-) – John Rennie Mar 30 '12 at 6:51
Yeah, it just occurred to me that he can probably control his temperature. So just giving th OP and idea of the bounds is good enough. Even I've not read Marvel comics in a while ;). Just a note:when using parentheses on fractions/large expressions, use \left and \right. Example:$(\frac12)\to\left(\frac12\right)$. Right click to see $\TeX$code. I've made the fix in your answer, hope you don't mind :) – Manishearth Mar 30 '12 at 6:57
@Manishearth or \biggl and \biggr etc. (which are preferred for actual LaTeX documents). – David Z Mar 30 '12 at 7:33
@David I don't really use TeX formally (sometimes for note-making), but I personally prefer the compiler to do the work of picking the size ;-) – Manishearth Mar 30 '12 at 8:16