A More Detailed Response To @trumpman

kryzsec (59)in #science • 7 years ago (edited)

This is a proper response to the post made by @trumpman here as I will go through and reference thing and keep data. This will be a mathematical post, not educational. If I get some of the chemistry wrong, please inform me (I am not a chem major) but I will try the best. My math in steemit chat was incorrect as I only looked at 1 side of the equation. So here we go. For reference I will link the video here:

Now the first question proposed by trumpman is what is the probability of this actually happening. But before I answer that, I want to look to see if it is even possible... using chemistry and enthalpy!

Equation

CH₄(g) + 2O₂(g) -> 2H₂O(g) + CO₂(g)

2H₂(g) + O₂(g) -> 2H₂O(g)

Now lets list the standard enthalpy of formation of each of these (kJ/mol):
CH₄(g) = -74.8
CO₂(g) = -393.5
H₂O(g) = 241.8
O₂(g) = 0
H₂(g) = 0
Now we need to cover our limiting reagent (methane) as we will assume an open room with ample oxygen. Now using this source here we can determine that a fart is roughly 7% methane and 21% hydrogen (another thing that can burn) so we will basically add the energies together at the end. Now we need to find the volume of a fart which according to wikipedia is 375ml (high end average) which can be found here meaning that roughly 7% of 375 ml is methane and roughly 21% is hydrogen (26.25ml of methane and 78.75 ml of hydrogen).

Source: wiki

Methane

Calculating the energy released from methane is done so with the following equation:
[heat of formation of CO2 (g) + 2 x heat of formation of H2O (g)] - [heat of formation of CH4 (g) + 2 x heat of formation of O2 (g)]
or
[ -393.5 kJ + 2x(-241.8 kJ)] - [-74.8 kJ + 2x(0 kJ)] = - 802.3 kJ
What this means is that there is 802.3 kJ of energy released per mol of methane, so now lets calculate how many mols of methane we have. Now we know we have 26.25 ml of methane and can convert it to to save time we will use a converter online here which gives us an answer of 0.6313224 kg/m³ (26.25ml = 2.625*10^-5 m³) meaning that there are 0.00001657221kg of methane. Since methane has has a molar mass of 16.04 (4*1.01 +12.00) kg/mol we can determine that there are 0.00000103318 mols of methane and that means 0.0008289206 kJ would be released from the oxidation of the methane (assuming 100% burn)

Source: wiki

Hydrogen

Hydrogen is a little easier to calculate the energy per mol as it is just 2x241.8 or 483.6 kJ/mol. But again, lets calculate the number of mols of hydrogen: 78.75 ml which is roughly 0.657 kg (again using an online converter) and since hydrogen has a molar mass of ~2.02 we can determine that there are 0.325 mols of hydrogen which would produce 157.17kJ of energy.

was notified (roughly 4 days ago form this edit now... sorry) that the conversion for hydrogen was incorrect and would be seeing closer to 0.00325 mols (0.325 mols) and as such the energy released by the hydrogen would be reduced to roughly 1.5717 kJ which definitely makes more sense now that I am reading it. Thank you for that correction @replichara

Total Energy

The total energy released (estimated at 100% burn efficiency) would be 157.170828921 kJ which would heat 1 kg of air up by around 157 K (157^oC) but since they are in a large room that heat would disperse quite quickly. The better question is will it burn the cat hair? Well using this source we can determine that (157 kJ = ~37 cal) that the hair would only be heated p by around 37 degrees Celsius (at most) and searching over the internet I could find hair burning at temperatures greater than 200 degrees Celsius. So the second question of is it possible? I am going to say no, not very likely at all.

so @trumpman, since I still do not think it is possible for this to work under perfect conditions (let alone realistic conditions) I will have to say there is a probability of this happening of pretty much 0 without the use of extra propellants, it could also be some strangely hydrogen potent fart. Now I am not certain about the math being completely correct (it is 2 AM here as of writing this) but worst case scenario there would be twice as much energy as calculated (100% error) meaning that the hair would heat up by 60 degrees (which sounds unrealistic from a single fart without sustained burn) so I will stand by my answer of this being faked.

References: (not including the trumpman post)

Link 1: nshs-science.net/chemistry/common/pdf/R-standard_enthalpy_of_formation.pdf
Link 2: www.muscleandfitness.com/features/edge/anatomy-fart
Link 3: en.wikipedia.org/wiki/Flatulence
Link 4: www.peacesoftware.de/einigewerte/methan_e.html
Link 5: www.researchgate.net/publication/266454223_Specific_heat_capacity_of_cosmetically_treated_human_hair
Pictures are linked directly

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#farts #steemstem #chemistry #enthalpy

7 years ago in #science by kryzsec (59)

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replichara (55) 7 years ago (edited)

It's been a while now since I made calculations like that for far less interesting chemical processes, but I'm pretty sure that

78.75 ml which is roughly 0.657 kg (again using an online converter)

is not the case, unless this human combustion device has a built-in high-end air compressor in it.

On the other hand, my guess is that a sudden release of 157 kJ would be enough to burn the cat's hair. Let's say the chemical reaction takes 0.2 seconds to complete. That would be an equivalent of 785 kiloWatt released (157 kJ/0.2s). And since the energy should be very concentrated on a small volume of air before it dissipates through the room, assuming 157 kJ is correct, my guess is, that it would burn through the cat's hair.

Poor cat, though.

Also you converted kJ to calories and then used the resulting value of 37 to apply it to air. Calories are defined by the specific heat capacity of water. Specific heat capacity of air is approx. four times less than of water which would result in a temperature increase of about 150 K.

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kryzsec (59) 7 years ago

The specific heat that I found for hair though was in calories, hence the conversion. Another problem with your assumption here is that you assume a rather static environment where the cat would feel all of the energy but as seen in the video the cat wouldn't be in the entire burn.

You may be correct that 78.75 ml of hydrogen isn't 0.657 kg (as I used an online converter and put in atmospheric pressure for the pressure and 300 kelvin for the temp) but I may have done something wrong (again 2 AM calculations)

While I definitely agree my calculations are certainly flawed somewhere (chemistry is honestly one of my least liked subjects) I will definitely say that if you can find flaws in this than write a post debunking my post and correct the answer and you will get some pretty hefty votes from everyone! I apologize if my correction at the top is condescending, I have difficulty with tone. But we also have to keep in mind that this, by no means, is taking account the amount of burn or the fact that most farts are released over a period of time and usually aren't instant.

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replichara (55) 7 years ago

Ah, now I understand. (It didn't sound condescending, btw. I hope I didn't either). I still think you would have to give more weight to the spec. heat capacity of air because it is the transmitter of energy until it reaches the cat skin.

The conversion of hydrogen volume to the amount of hydrogen goes through the molar volume of an ideal gas which is under normal conditions 22.4 liter/mol. (0.0785 liter / 22.4 liter/mol) would give us 0.00325 mols of hydrogen. Maybe the converter you used was working with cube meters or something?

Nah, I am not going to make a post :-) Chemistry and physics are way too far back for me.

Now, I am thinking that this guy might have sucked in some lighter gas into his rectum. Some people can fart at will.