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The equation tells you that Δu = ΔE³ = ΔkE = Δ T / tu / 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=m³m/V= <mI won't add a big bang equation .

Quote from: Thebox on 03/04/2019 14:09:04The equation tells you that Δu = ΔE³ = ΔkE = Δ T / tu / 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=m³m/V= <mI won't add a big bang equation . What is an idol measure? Is it related to some reality TV that has obviously rotted your mind?Why not add a big bang equation. It is bound to have comedy value.

Quote from: The Spoon on 03/04/2019 17:08:52Quote from: Thebox on 03/04/2019 14:09:04The equation tells you that Δu = ΔE³ = ΔkE = Δ T / tu / 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=m³m/V= <mI won't add a big bang equation . What is an idol measure? Is it related to some reality TV that has obviously rotted your mind?Why not add a big bang equation. It is bound to have comedy value. Which big bang ? The beginning of time or the end of time ? Additionally here is the model for the cool maths . super model.jpg (49.75 kB . 744x478 - viewed 1064 times)

Lines and random letters. I've seen better Airfix models.

Quote from: Kryptid on 02/04/2019 22:32:15Quote from: Thebox on 02/04/2019 22:28:28If you already have the value and equation you suppose to use that equation, my equation just generalises the process's and is meant as an education aid in simplicity . In other words you don't have to work out an answer , it tells the ''story'' of the process's . So can it calculate the energy released when hydrogen combusts or not?Well in theory yes if you're good at maths and can work out the values . You have to work out u then you have to work out ignition energy added to the system and the kinetic response energy . Alternatively we could work backwards , 1ltr of hydrogen volumes energy combusted, divided by a volume of water , temp measure etc and reverse engineer the results to get an exact measure .

Quote from: Thebox on 02/04/2019 22:28:28If you already have the value and equation you suppose to use that equation, my equation just generalises the process's and is meant as an education aid in simplicity . In other words you don't have to work out an answer , it tells the ''story'' of the process's . So can it calculate the energy released when hydrogen combusts or not?

If you already have the value and equation you suppose to use that equation, my equation just generalises the process's and is meant as an education aid in simplicity . In other words you don't have to work out an answer , it tells the ''story'' of the process's .

Quote from: The Spoon on 03/04/2019 18:13:49Lines and random letters. I've seen better Airfix models. Of course you have ! You couldn't do no better put it that way .

Quote from: Thebox on 03/04/2019 18:18:03Quote from: The Spoon on 03/04/2019 18:13:49Lines and random letters. I've seen better Airfix models. Of course you have ! You couldn't do no better put it that way . Depends what it was trying to represent. As yours does not represent anything it is hard to judge.

Quote from: Thebox on 02/04/2019 22:36:35Quote from: Kryptid on 02/04/2019 22:32:15Quote from: Thebox on 02/04/2019 22:28:28If you already have the value and equation you suppose to use that equation, my equation just generalises the process's and is meant as an education aid in simplicity . In other words you don't have to work out an answer , it tells the ''story'' of the process's . So can it calculate the energy released when hydrogen combusts or not?Well in theory yes if you're good at maths and can work out the values . You have to work out u then you have to work out ignition energy added to the system and the kinetic response energy . Alternatively we could work backwards , 1ltr of hydrogen volumes energy combusted, divided by a volume of water , temp measure etc and reverse engineer the results to get an exact measure . I have a simpler suggestion for an equation that is just as much use as yours.anything = 1-xObviously, to calculate it you need to know xBut, as you have suggested, that's easy- you just work backwards from the answer you want.For example, if you want to know the heat of combustion of hydrogen you start off by finding out what the heat of combustion of hydrogen is- you can measure it or google it, or whatever.It's 286 JK/molNow, from that we can calculate xx = 1- 286 So x (for this particular question) is -285And then, using my formula we can calculate the heat of combustion for hydrogen.we know that x = -285 so we calculate 1- (-285)And that gives the answer 286.Do you see any problems with this method which do not apply to yours?

Quote from: The Spoon on 03/04/2019 18:35:46Quote from: Thebox on 03/04/2019 18:18:03Quote from: The Spoon on 03/04/2019 18:13:49Lines and random letters. I've seen better Airfix models. Of course you have ! You couldn't do no better put it that way . Depends what it was trying to represent. As yours does not represent anything it is hard to judge. That's strange it represents almost everything , thermodynamic process for a start , the process of time and aging, how the universe works , big bang singularity etc. Demonstrate it doesn't ?

That's strange it represents almost everything , thermodynamic process for a start , the process of time and aging, how the universe works , big bang singularity etc. Demonstrate it doesn't ?

Quote from: Thebox on 03/04/2019 19:24:48That's strange it represents almost everything , thermodynamic process for a start , the process of time and aging, how the universe works , big bang singularity etc. Demonstrate it doesn't ? It gave an answer that was wrong by about 50 orders of magnitude.That's a pretty convincing demonstration of failure

I ask again demonstrate it's failure to explain ?

It is your job to prove that it does explain things.

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 combustionHUh what's that suppose to mean ?

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 .

The stated purpose of the calculation was to determine the amount of energy released when hydrogen burns. Regardless of where the energy actually comes from, energy is released in the process. So either your equation is wrong or you gave me the wrong information to use in the calculation.

The problem is no matter what value to derive , it is relative . I could express the combustion of hydrogen releases 1 in energy , 1 would be equal to a more derived value .

Quote from: Thebox on 03/04/2019 22:40:06The problem is no matter what value to derive , it is relative . I could express the combustion of hydrogen releases 1 in energy , 1 would be equal to a more derived value .So then the equation is useless for answering questions that we don't already have the answer to.

What don't you have an answer too ?