[The Spoon]

Well in theory yes

Then that is what I'm trying to do. If "u" is equal to the net electric charge in a hydrogen molecule, then "u" has to be zero.

Well yes if we're being honest , can we call this , the idol moment ?

What Billy Idol? 😆

[guest39538]

Well yes if we're being honest , can we call this , the idol moment ?

Okay then, so what would "E" represent in the equation in normal science language?

Difficult to answer but I think in your terms we can use hf³ with the same net charge .

[Kryptid]

Difficult to answer but I think in your terms we can use hf³

I assume "h" means Planck's constant. Does the "f" mean frequency? What would frequency represent for a hydrogen molecule? The frequency of the chemical bond?

[guest39538]

Difficult to answer but I think in your terms we can use hf³

I assume "h" means Planck's constant. Does the "f" mean frequency? What would frequency represent for a hydrogen molecule? The frequency of the chemical bond?

The frequency of incoming hf I'm referring to , I'm not sure about your chemical bond but I imagine the frequency speeds up some what when ''heated''  to combustion ?

And yes a constant ,

[Kryptid]

The frequency of incoming hf I'm referring to , I'm not sure about your chemical bond but I imagine the frequency speeds up some what when ''heated''  to combustion ?

It sounds like your equation assumes ignition by photons. Which means we can use any kind of photon frequency we want here, I guess?

[guest39538]

Well in theory yes

Then that is what I'm trying to do. If "u" is equal to the net electric charge in a hydrogen molecule, then "u" has to be zero.

Well yes if we're being honest , can we call this , the idol moment ?

What Billy Idol? 😆

White wedding , a mixture of frequencies .

[guest39538]

The frequency of incoming hf I'm referring to , I'm not sure about your chemical bond but I imagine the frequency speeds up some what when ''heated''  to combustion ?

It sounds like your equation assumes ignition by photons. Which means we can use any kind of photon frequency we want here, I guess?

Yes I think so , obvious the shorter the wave length , the quicker combustion is achieved .

[Kryptid]

Yes I think so , obvious the shorter the wave length , the quicker combustion is achieved .

Now we're making some progress. I assume that "V" is simply the volume of the hydrogen gas. So now on to "t", which is, I'm guessing, the time taken to burn the hydrogen. Depending on how the system is set up, you could burn hydrogen as fast or as slow as you want to. So can any number be put in for "t"? How about ten seconds?

[guest39538]

Yes I think so , obvious the shorter the wave length , the quicker combustion is achieved .

Now we're making some progress. I assume that "V" is simply the volume of the hydrogen gas. So now on to "t", which is, I'm guessing, the time taken to burn the hydrogen. Depending on how the system is set up, you could burn hydrogen as fast or as slow as you want to. So can any number be put in for "t"? How about ten seconds?

Ten seconds will do , consider also there is the volume of the hydrogen and volume external of the hydrogen unless you're talking about a compression chamber  and a hydrogen fuel engine .

[Kryptid]

Ten seconds will do , consider also there is the volume of the hydrogen and volume external of the hydrogen unless your talking about a compression chamber  and a hydrogen fuel engine .

As a gas, hydrogen will fill whatever container it's in. So those volumes can be taken as equal if you'd like.

[guest39538]

Ten seconds will do , consider also there is the volume of the hydrogen and volume external of the hydrogen unless your talking about a compression chamber  and a hydrogen fuel engine .

As a gas, hydrogen will fill whatever container it's in. So those volumes can be taken as equal if you'd like.

Arr , ok, I see what you're on about , pressure equalling density I presume ?  Obviously density also plays a part , there will be more energy per mm³ of hydrogen if pressured compared to less pressured .

[Kryptid]

Arr , ok, I see what you're on about , pressure equalling density I presume ?  Obviously density also plays a part , there will be more energy per mm³ of hydrogen if pressured compared to less pressured .

If we are assuming normal atmospheric pressure, then knowing that we have one gram of hydrogen will already give us the volume (and therefore the density).

Unless you have anything else to add, I think I have all I need to do the math. Should I start?

EDIT: I've got to go do something, but I'll be back in a little bit.

[guest39538]

Arr , ok, I see what you're on about , pressure equalling density I presume ?  Obviously density also plays a part , there will be more energy per mm³ of hydrogen if pressured compared to less pressured .

If we are assuming normal atmospheric pressure, then knowing that we have one gram of hydrogen will already give us the volume.

Unless you have anything else to add, I think I have all I need to do the math. Should I start?

Yea you can try , I'll be interested to see what you come up with .

[Kryptid]

Yea you can try , I'll be interested to see what you come up with .

I can't directly copy the equation text as you've written it (is it an image file?), but this should be an equivalent expression: (uE³)/(V/t)

"E" equals "hf³", which is the Planck constant multiplied by photon frequency. I'll use a very powerful photon in the form of a gamma ray (10¹⁹ hertz). 10¹⁹ cubed is 10⁵⁷. So "E" equals (6.62607015×10⁻³⁴ joule-seconds) x (10⁵⁷ hertz). Multiply that out and you get 6.62607015 x 10²³.

One gram of hydrogen is equal to 0.49611 moles of hydrogen, which occupies 11.1129 liters of volume (0.0111129 cubic meters). So now I have the needed values:

Combustion energy =  (uE³)/(V/t)
= ((0)*((6.62607015 x 10²³)³))/(0.0111129/10)
= ((0)*((2.909163 x 10⁷¹)))/(0.00111129)
= (0)/(0.00111129)
= 0 joules

Okay, so we've run into a problem. By making "u" equal to zero, the equation predicts that no energy is released at all by hydrogen burning. However, there is a way around this. Maybe instead of net charge you meant gross charge instead? That way, instead of subtracting the charge on the electrons from the charge on the protons and getting zero, you add them together and get a finite number (6.4087064 x 10[/sup]-19[/sup] coulombs of charge, to be precise). Here's what happens if I do that instead:

Combustion energy =  (uE³)/(V/t)
= ((6.4087064 x 10^-19)*((6.62607015 x 10²³)³))/(0.0111129/10)
= ((6.4087064 x 10^-19)*((2.909163 x 10⁷¹)))/(0.00111129)
= (1.864397 x 10⁵³)/(0.00111129)
= 1.677687 x 10⁵⁶ joules

How about that? Does that look good?

[guest39538]

Yea you can try , I'll be interested to see what you come up with .

I can't directly copy the equation text as you've written it (is it an image file?), but this should be an equivalent expression: (uE³)/(V/t)

"E" equals "hf³", which is the Planck constant multiplied by photon frequency. I'll use a very powerful photon in the form of a gamma ray (10¹⁹ hertz). 10¹⁹ cubed is 10⁵⁷. So "E" equals (6.62607015×10⁻³⁴ joule-seconds) x (10⁵⁷ hertz). Multiply that out and you get 6.62607015 x 10²³.

One gram of hydrogen is equal to 0.49611 moles of hydrogen, which occupies 11.1129 liters of volume (0.0111129 cubic meters). So now I have the needed values:

Combustion energy =  (uE³)/(V/t)
= ((0)*((6.62607015 x 10²³)³))/(0.0111129/10)
= ((0)*((2.909163 x 10⁷¹)))/(0.00111129)
= (0)/(0.00111129)
= 0 joules

Okay, so we've run into a problem. By making "u" equal to zero, the equation predicts that no energy is released at all by hydrogen burning. However, there is a way around this. Maybe instead of net charge you meant gross charge instead? That way, instead of subtracting the charge on the electrons from the charge on the protons and getting zero, you add them together and get a finite number (6.4087064 x 10[/sup]-19[/sup] coulombs of charge, to be precise). Here's what happens if I do that instead:

Combustion energy =  (uE³)/(V/t)
= ((6.4087064 x 10^-19)*((6.62607015 x 10²³)³))/(0.0111129/10)
= ((6.4087064 x 10^-19)*((2.909163 x 10⁷¹)))/(0.00111129)
= (1.864397 x 10⁵³)/(0.00111129)
= 1.677687 x 10⁵⁶ joules

How about that? Does that look good?

Wow , yes it looks good , I have no idea what it means lol , that's why I have a simple equation . Does it work ?

[Kryptid]

Wow , yes it looks good , I have no idea what it means lol , that's why I have a simple equation . Does it work ?

The enthalpy of combustion of hydrogen is 141,580 joules per gram according to https://www.engineeringtoolbox.com/standard-heat-of-combustion-energy-content-d_1987.html

141,580 joules per gram is nowhere remotely the same as 1.677687 x 10⁵⁶ joules per gram.

[Bored chemist]

I have a simple equation . Does it work ?

No, of course not.
That's what we have said all along.
Your calculation is nonsense.

[guest39538]

Wow , yes it looks good , I have no idea what it means lol , that's why I have a simple equation . Does it work ?

The enthalpy of combustion of hydrogen is 141,580 joules per gram according to https://www.engineeringtoolbox.com/standard-heat-of-combustion-energy-content-d_1987.html

141,580 joules per gram is nowhere remotely the same as 1.677687 x 10⁵⁶ joules per gram.

Perhaps the answer shouldn't be in joules , how do they derive their equation ?  Are they doing it correctly  ?

nX + mO2  →  xCO2 (g) + yH2O (l) + zZ + heat of combustion

HUh what's that suppose to mean ?

[guest39538]

I have a simple equation . Does it work ?

No, of course not.
That's what we have said all along.
Your calculation is nonsense.

Nope , my equation works , values are just final touches .

Look at this way

A glass full of water

u=1/2 ltr
E³ = +1/2ltr

Output overflow 1/2 ltr

=1/2 ltr

Combustion energy =  (uE3)/(V/t)
= ((0)*((6.62607015 x 1023)3))/(0.0111129/10)
= ((0)*((2.909163 x 1071)))/(0.00111129)
= (0)/(0.00111129)
= 0 joules

Would be correct , the energy released is kinetics .

Antoine - nothing is lost and nothing is gained .
The hydrogen atom loses no energy in the process of combustion .

[guest39538]

The equation tells you that

Δu = ΔE³ = ΔkE = Δ T / t


u / t is a constant and the idol measure , u always returns to u , nothing lost and nothing gained .

Natural temperature 0 .

u = m

m+m = m2

m*m=m³

m/V= <m

I won't add a big bang equation .