Definition question

Mike T., Swampeast MO

Different sources say burning hydrogen produces water in some form--some say "water" some say "vapor" some say "steam".

http://www.hyperdictionary.com/search.aspx?Dict=&define=vapor&search.x=8&search.y=9

"Vapor is any substance in the gaseous condition at the maximum of density consistent with that condition. This is the strict and proper meaning of the word vapor."

Does this mean (in the terms of water) that vapor could be called "steam without latent heat???"

imatellerslie

definition of \"latent heat\"

> Different sources say burning hydrogen produces

> water in some form--some say "water" some say

> "vapor" some say

> "steam".

>

> http://www.hyperdictionary.com/search.

> aspx?Dict=&define=vapor&search.x=8&search.y=9

>

> "

> Vapor is any substance in the gaseous condition

> at the maximum of density consistent with that

> condition. This is the strict and proper meaning

> of the word vapor."

>

> Does this mean (in the

> terms of water) that vapor could be called "steam

> without latent heat???"

imatellerslie

definition of \"latent heat\"

Mike,

Latent heat is actually the energy required to change the phase of the substance. So, even if the vapor is cooled to the point of condensing, which is called "saturated", it would still contain latent heat. Only when the steam condensed would it give up the latent heat.

imatellerslie

burning hydrogen

As for the process of burning hydrogen, when the Oxygen and Hydrogen molecules interact, and the oxygen oxidizes the hydrogen, yielding H2O and heat, the product molecule would certainly be in the gaseous state. The molecule may not have enough kinetic energy or vapor pressure to be considered steam (the water vapor in the room in which you sit would not be considered steam), but I would think that those molecules would certainly give up considerable energy in going to the liquid state. The process certainly would not directly produce liquid water, although the product molecules may relatively quickly condense with other product molecules to produce condensate. The rate of this condensation would depend on the temperature of the environment and density of the gas.

Mike T., Swampeast MO

Thank you.

This really is absurdely confusing to me...

If it wasn't water before (it was hydrogen and oxygen) and the oxidation of hydrogen results in a gasseous molecule of water how did that water molecule "pick up" latent heat when it never existed in the liquid state to begin with?

imatellerslie

Latent heat

Mike,

Latent heat is a statistical measure of the work required to pull a water molecule away from other water molecules in the liquid form, so that it is free from the bonds of liquidity and can roam as a gaseous water molecule.

When this work is done on the molecules, it raises their potential energy, so that they have the ability to do work in the future. This is why a pound of water in the gaseous state has more energy than a pound of water in the liquid state, even if the vapor is at the same temperature and pressure as the liquid water.

Keep in mind that thermodynamics is statistical, so if you try to apply it to one molecule, it can become very confusing.

imatellerslie

think of it this way

Heat is just the flow of energy from one place to another (usually from a region of high temperature to lower temperature). Temperature is a measure of the random kinetic energy of the molecules of a substance; it is statistical, and to talk about the temperature of one molecule is nonsense. A water molecule in vapor form transfers heat to a radiator or pipe by running into it and making one or more molecules in the radiator vibrate. Now, here's where latent heat becomes important. There is an attractive force pulling the water molecule into the water coating that radiator. It doesn't act over a very big distance, but it has the effect of speeding that gaseous water molecule up so that it's going faster when it runs into the surface, thereby causing more vibration. This extra energy is equal to the potential energy I mentioned in the last post. This is why a pound of steam has more energy than a pound of water; A LOT MORE.

scrook_2

The energy (heat)

comes from the energy released by the oxidation of the hydrogen. "Energy of vaporization" might be a better (more general) term than heat of vaporization, heat just being a specific manifestation of energy, but the latter is the phrase commonly used.

As for vapor vs steam vs water: to say vapor in this contest id to imply "[water] vapor" to say steam is to indicate the oxide of hydrogen in the vapor or gas phase, to say water doesn't give explicit information about it's phase.

jerry scharf

where's that latent energy coming from?

Mike,

When dealing with agregates of molecules, there are three types of energy to accounted for. First and largest are the bonds that hold the molecules together. Second is the bonds that attract the nolecules to each other. Finally there is the kinetic energy of the molecules moving around.

Heat is the kinetic (motion) energy of the molecules bouncing off each other even though there is no net velocity to the aggregate. Latent heat is the change in molecule to molecule attraction as the phase is changed. Combustion and gavlanic action are good examples of energy derived from rearranging the bonds within the molecules.

So when you burn pure hydrogen and oxegen is the ideal 2:1 mix, you release a whole bunch of energy because the bond patterns for H2O is lower energy than H2 and O2. This energy change is split between the intermolecular and heat energies. As you remove heat energy, the water con condense and at that point, the intermolecular bond energy changes significantly and more heat is released.

If you add the place you're removing the heat to as part of the system, the total energy of the system is constant.

does this help or make things worse?

jerry

Mike T., Swampeast MO

Thanks again, both of you...

...I guess I'm making something much more difficult than necessary but am still not understanding the state of the hydrogen oxide when the hydrogen is oxidized.

Maybe I have it now???

When the hydrogen contained in natural gas oxidizes it produces heat and extremely superheated water. (I'm saying "extremely superheated" because I presume it's at the temperature of combustion--not the temperature of ignition?).

A portion of that superheat in the water becomes latent heat as the water vaporizes [in? just outside?] the flame with the rest of the superheat going [mainly] to the non-oxygen portion of the combustion air....

Is this anywhere near correct?

imatellerslie

latent heat NOT from interatom bonds

To say that the water molecule gets its latent heat of vaporization from the reaction between the O & H atom is not correct. The water molecule has this energy by virtue of its position relative to other water molecules. If it is not close to other water molecules (in the liquid state), but there are other water molecules available with which it could gather into a water droplet, then it has energy by virtue of the attractive force, which can do work over a distance, as the molecules come together.

I agree that the heat given by the combustion process is derived from the attractive force between the hydrogen & oxygen atoms attracting one another, and thereby doing work over distance. Same concept, but different potential well.

imatellerslie

the energy to superheat is independent of the latent heat

Mike,

As long as we're talking statistically about the product of hydrogen combustion (water molecules), and not about any particular water molecule, I'd agree that the water resulting from hydrogen combustion is going to be superheated. If you're considering a particular water molecule produced from combustion, it may be at rest, or it may have a very large kinetic energy. There will be a large range of possible velocities associated with the combustion products.

As for the reaction that's taking place, it happens in the flame. When the hydrogen (which normally exists in a diatomic state, eg H2) gets close to Oxygen (also normally mostly in a diatomic state), the Oxygen, which tightly binds its electrons, "steals" an electron from the Hydrogen. This is called Oxidation, because Oxygen does it so well, even though other substances that bind electrons tightly do it well also. The Hydrogen, which now has a positive charge, is strongly attracted to the Oxygen, which now has a negative charge, and they are bonded. In order to satiate the greedy Oxygen atom, two Hydrogens must sacrifice their electrons, and so we end up with H2O. The energy that is given off in this process is mostly due to the work done by the attractive force between the Oxygen and Hydrogen. They combine rather violently, which can impart significant translational and vibrational kinetic energy to the molecule. (It can be going very fast, and vibrating very violently) The average amount of kinetic energy the water molecules have is what determines their Temperature.

The latent heat of vaporization has nothing to do with the bonds between atoms within a water molecule. It is from the attractive forces that exist between water molecules. It is not given to the molecule during the combustion process. It is not really even meaningful to talk about the water molecule having latent heat unless there are other water molecules around for it to "bond" with and form a droplet of liquid.

Mike T., Swampeast MO

That sounds exactly like...

...high school chemistry experiment where we used electrolysis to break water into hydrogen and oxygen, collected each, recombined, added spark and got a hazy blue flame and condensation in the beaker.

That's the part that I [think] I understand pretty well...

>The latent heat of vaporization has nothing to do with the >bonds between atoms within a water molecule. It is from >the attractive forces that exist between water molecules.

Understood. If I said or implied something to the contrary, it was unintentional.

>It is not really even meaningful to talk about the water >molecule having latent heat unless there are other water >molecules around for it to "bond" with and form a droplet >of liquid.

So, I think you're saying that inside the flame that the water must be forming as water--even if it is extremely superheated water...

After it leaves the flame however, part of that superheat becomes latent heat as the water in the flame is now forced to vaporize. Correct? If so, isn't there a LOT more heat in that superheated water that has to go somewhere because it can't stay in the vapor?

imatellerslie

energy to superheat is unrelated to heat of vaporization

>So, I think you're saying that inside the flame that the water must be forming as water--even if it is extremely superheated water...<

Yes, as soon as the hydrogen and oxygen react, a water molecule is formed. The water that forms inside the flame will be at a very high temperature, and will thus be considered superheated vapor. The water is vapor as soon as it is formed not bound to other water molecules. The vapor is superheated, because it is at a temperature higher than saturation temperature for its pressure. A liquid/vapor mixture is said to be saturated if the rate of condensation is equal to the rate of evaporation (in equilibrium) and this happens at a specific temperature for every pressure. If the vapor is heated to a higher temperature, for example by compression, then it becomes superheated, and the rate of condensation will increase until equilibrium is once again reached (assuming we're talking about a closed system)

>After it leaves the flame however, part of that superheat becomes latent heat<

No, the energy associated with the superheating of the vapor is unrelated to the latent heat of vaporization. The latent heat of vaporization is given off as the water condenses. Imagine if you have two hydrogen atoms, and one oxygen, and they combine. If this occurs in a vacuum, and no other water is around, would you say that this molecule has latent heat of vaporization? It would be a meaningless statement.

>isn't there a LOT more heat in that superheated water that has to go somewhere because it can't stay in the vapor?<

Yes, the temperature of superheated vapor is higher than the temperature of vapor at saturation temperature, so the molecules have more kinetic energy, and thus more energy to impart to their surroundings. However, this energy can stay in the vapor, as long as the vapor stays superheated. It will stay superheated unless pressure is raised (thereby raising saturation temperature) or the vapor gives up energy to a cooler object with which it has contact.

jerry scharf

one more shot at explaining the latent heat

Mike,

I'm sure we've bored just+ about everyone leso to tears at this point.

When the water atoms are formed, they are at elevated temperature. They are boucing around way too fast to care very much about the other water atoms around them. This is the loose definition of any gas.

As you start taking the heat out of the gas (it's going somewhere) you get to a point where the kinetic energy is overcome by the molecular attractive force. The atoms may be close but not organized, a liquid, or close and well organized, a solid.

If you think about the pull between things as some sort of funnel with curved side walls, the closer things come, the farther down the funnel they fall. As they fall down, they give up energy, and that's the latent energy of condensation phase change. Boiling is the opposite procecc of putting enough kinetic energy into the water to force it up the funnel.

So in this example, the water molecule is formed inside the flame. It is not in equilibrium, so talking about saturation is not appropriate. The gas is too hot for it's surroundings, and the energy flows out in an effort to reach equilibrium. The boiler's job is to catch as much of that excess energy as possible and put it into a transport medium for use elsewhere (steam, water, turbine shaft...)

jerry

Eric Taylor_38

You folks listen to Jerry...

He knows what he is talking about.

For some heavy reading:
http://telstar.ote.cmu.edu/environ/m3/s3/08formations.shtml

The energy in a reaction comes from making and breaking bonds between atoms. If you break more than you make then the reaction is endothermic (absorbs energy). If you make more than you break then it is exothermic (releases energy).

Notice I haven't said heat yet. Most of the energy of combustion will manifest itself as heat (duh). Those newly formed water molecules will absorb some of it (double duh). Since we formed the water molecules at elevated temperature and they were never in a liquid state to start with I'm going to say that we have superheated vapour after combustion and it absorbs even more heat from the nearby reactions of continuing combustion and gets even more superheated. If you then condense that vapour to liquid you remove all the heat energy down to whatever temperature you get the condensate to. Keep in mind though that we are not talking about very much water at all. Perfect combustion of hydrogen is highly energetic for the input mass. The flue gas would be extreamly hot, but not to dense and you wouldn't be condensing hugh amounts of water.

Eric

Mike T., Swampeast MO

>When the water atoms are formed, they are at elevated >temperature. They are boucing around way too fast to care >very much about the other water atoms around them. This is >the loose definition of any gas.

But doesn't that describe dry or "superheated" steam?

{Superheated steam}, steam heated to a temperature higher
than the boiling point corresponding to its pressure. It
can not exist in contact with water, nor contain water,
and resembles a perfect gas; -- called also {surcharged
steam}, {anhydrous steam}, and {steam gas}.

--------------------------------

That's where I keep getting confused.

I don't understand the idea of vaporizing a vapor.

jerry scharf

I lied about quitting

Mike,

Wet steam is 100% water vapor at approximately atmospheric temperature. Running it through a second heating process separate from the boiling raises the temperature above boiling, and it's still 100% water vapor. Thet's why it's called superheated, you can't ever boil and superheat in the same pass. The pressure is elevated with the temperature and the increase in transfer capacity is quite large. This also explains the terms wet steam and dry steam. Both wet steam and dry steam are roughly in equilibrium when they're in an insulated pipe.

So the water vapor in the combustion product is very hot, but it's not superheated in the above sense. It's not at equilibrium and it's pressure is not elevated.

imatellerslie

superheat

The only reason I mentioned saturation is that an understanding of the concept of saturation is necessary to understand superheat, which is the state of the gas in the flame. I have tried to avoid gravitational potential energy analogies and anthropomorphic talk of atoms "wanting" or "caring" about anything because it clouds the issue. My explanation is in terms of the basic underlying physics.