# Temperature resulting from the formation of water

1. Oct 27, 2009

### vjk2

Basically, I need confirmation on the heat produced when four mole of H2O is produced from Hydrogen and oxygen gas.

I'm getting something ridiculous, basically saying that the temperature goes up to ~4000 degrees C.

What I'm doing is using Hess's law to find the energies of reaction. H2 and O2 gas result in 0 KJ/mol. H2O is -285.8 KJ/mol.

4 mols x 285.8 KJ/mol = 1143.2 KJ

I'm taking this and plugging it into the equation

q=mC<>T (change in Temperature)

so,

1143.2 = (18 g/mol H2O * 4 mol H2O = 72 g) * 4.18 j/(g*c) * <>T

<>T = 3797 degrees C

I know that when water forms from oxygen and hydrogen, the result is explosive, but thousands degrees C seems way too high. However...my logic seems right. Thoughts?

2. Nov 10, 2009

### windwitch

Do not underestimate the amount of energy available in the bonds of water. If you conduct an experiment, you will be able to see that water is able to absorb and produce a large amount of heat in comparison to many other substances.

From your equations alone, it seems that there are no mistakes. HOWEVER, from your equation, it seems to assume that water rises 4000 degrees in temperature without undergoing a change in liquid to gaseous state. Remember that when you are finding a standard enthalpy of formation, you assume that the product is in it's standard state. The standard state for water is liquid. Does it make sense that water, starting from (0 or 25 degrees celsius, I am forgetful of the standard temperature) standard temperature to go to 4000 degrees without having any energy lost in a state change?

3. Nov 10, 2009

### vjk2

Yes, I've deduced that the phase change does take place.

The issue then is how to calculate it in. What I'm thinking is to

1. take the total energy released by the formation of water.

2. subtract from that figure the amount of energy required to heat water to 100 degrees C by using the equation q = (18*4)(4.18)(75)

3. take the remainder of energy and instead of using the 4.18 heat capacity figure for liquid water, use the heat capacity for vapor water.

Only problem is finding the heat capacity of vapor water.

4. Nov 10, 2009

Dont forget the specific latent heat of vapourisation that is the energy needed to change unit mass of water to unit mass of steam at its boiling point.

5. Nov 11, 2009

### Staff: Mentor

Also to be precise in this type of calculations you should take into account fact that specific heat is not temperature independent. As long as delta T is in the range of several or even small tens of degress that's usually not a large problem, but when we are talking about temperature changes in the range hundreds or thousands degress, that has to be taken into consideration.

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6. Nov 22, 2009

### vjk2

This shouldn't be all that complicated. Can anyone here do it? I'm still stumped.

7. Nov 23, 2009

### Staff: Mentor

You were already told that your calculation of heat looks OK, so don't worry about temperature. Unless what you wrote in your first post

was wrong.

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methods

8. Nov 23, 2009

### vjk2

it's not okay. The answer is 6500. How do you factor in a phase change?

9. Nov 23, 2009

### Staff: Mentor

6500 of what? Stones per yard squared?

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10. Nov 23, 2009

### vjk2

kelvin, obviously.

11. Nov 23, 2009

### Staff: Mentor

So you are trying to calculate heat produced, and you want answer in Kelvins?

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methods

12. Dec 3, 2009

### BobCivics

Umm... just by looking at the initial post, I'd have to say, check your units. Especially the units of energy. I don't think that the problem requires knowledge of heat of vaporization at all.

13. Dec 3, 2009

### BobCivics

Never mind, the units look good. Guess I should open the calculator BEFORE posting so the inside of my mouth doesn't taste like boot? Hmmm.... Everything that needs to be done has been said, but two questions pop up. Do you have an initial temperature, and how are you supposed to get the water vapor's c if it changes as the temperature increases?