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If one takes a bundle of wood up high to the mountains so its potential energy increases, would there be obtained more heat by burning it?

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I just can't make any sense of this question... –  David Z Jan 20 '11 at 23:27
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I guess its just a bundle of logs for a fireplace. Answer: No, but You can "earn" back that potential energy when You carry the oxidation products down from the mountains. –  Georg Jan 20 '11 at 23:45
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Oh my. I stared at it for several minutes until realized that "vjazanka draw" means "вязанка дров" in russian. It means "bundle of logs" indeed. –  Kostya Jan 21 '11 at 11:19
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Apart from the mangled English which could have been corrected, it is an interesting question. –  Philip Gibbs Jan 21 '11 at 15:32
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I see, Kostya, haha. Even with my Western Slavic attitude, I can now understand that it is "vázanka dřev". The "draw" is a really cute spelling, especially because it is an English word that has nothing to do with the wooden logs. :-) –  Luboš Motl Jan 22 '11 at 8:50
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6 Answers 6

Yes (at least in certain circumstances).

For simplicity, let's assume that there's no ash left, and the combustion products are gasses that have the same density as the atmosphere. Now, the state of the system when you burn wood at the top of the mountain looks very much like the state of the system when you burn wood at the bottom of the mountain, so what happened to the energy you expended when you hauled the wood up the mountain? The answer is that the air pressure at the top of the mountain is lower than the air pressure at the bottom, so the fire at the top expends less work displacing air to account for the gasses released in combustion, and so produces more heat.

How about in real life? You would also have to take into account the density of the gasses produced by combustion, and compare the potential energy of these gasses (taking into account the oxygen consumed) with the work done displacing the air. I'm not going to do the research to figure this out, but I think it very likely the answer is also "Yes".

ADDED MATERIAL:

There's been an objection to this answer, based on the fact that if you burn magnesium (say), which consumes oxygen without producing any gaseous combustion products, the fire at the top of the mountain produces less heat than the fire at the bottom. But consider what happens when you burn magnesium. You turn magnesium, Mg, into magnesium oxide, MgO. MgO is heavier than Mg. This means that if you haul magnesium up a mountain, burn it, and then bring the ashes down, you've actually gained energy in this process, since the ashes going down are heavier than the magnesium you took up. Where did this extra energy come from? It comes from the fact that the fire produces more heat on the bottom of the mountain.

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-1 Too learn what is the nonsense with Your answer, just do some circular process: Enclose the wood in a chamber, reduce pressure to the same pressure as will be on mountaintop, burn an then carry all on the mountain. –  Georg Jan 22 '11 at 15:21
    
And you're not using any energy pumping the gas out of the chamber? The energy difference caused by pressure isn't very large compared to the energy released by burning the wood, but it exists. –  Peter Shor Jan 22 '11 at 15:39
    
@Peter Shor: You make a good point regarding air pressure. The reasoning is slightly fuzzy to me, but I suspect it's correct nonetheless. –  Noldorin Jan 22 '11 at 16:15
    
You're mixing up so many things in this answer I don't know where to start. How about this: If the thing you're burning produces 10 moles of gas for every 1 mole of O2 consumed, then the volume of gas increases from reactants to products, and the enthalpy change is decreased by the need to push away the atmosphere. It's decreased less at higher altitude, so your conclusion is correct. However, if the fuel produces 1 mole of gas for every 10 moles of O2 consumed, then the volume of gas decreases, and the enthalpy change is actually greater than the energy change due to the atmosphere's pushing. –  Keenan Pepper Jan 22 '11 at 18:14
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@Peter: I think this is a good example of exactly the problem you highlighted on meta months ago. –  Joe Fitzsimons Jan 22 '11 at 18:20
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Simply, no. Gravitational potential energy cannot be released by burning (combustion).

Gravitational potential energy is quite independent from chemical potential energy (which is ultimately electromagnetic in form). As of fundamental forces, they do not interact. Burning only converts chemical potential energy (related to the binding energy of chemical compounds) to heat (thermal energy and EM radiation). Hence, gravity is basically irrelevant.

Saying this, there may be a small indirect effect of gravity. For a consideration of the possible effect of air pressure (which varies with altitude), see Peter Shor's answer.

Philip Gibbs also makes an interesting point relating to General Relativity. However, this relies on space-time curvature for a fixed observer, and is not as simple as considering gravitational potential.

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But the chemical energy has a mass equivalent which is lost on burning. This affects the amount of potential energy. So this conclusion is wrong. –  Philip Gibbs Jan 22 '11 at 12:34
    
@Philip: What? The two forms of energy are independent. I'm pretty sure it is correct. –  Noldorin Jan 22 '11 at 13:52
    
What is " chemical potential energy"? I know chemical potential, but that chemical potential energy is unknown in chemistry and thermodynamics. Somebody else used this expression some days ago here. –  Georg Jan 22 '11 at 14:05
    
In general relativity no form of energy is independent of gravitational energy, because all forms of energy gravitate. Chemical energy certainly adds to gravitational pull otherwise the equivalence principle would not hold. I'm puzzled by what you mean by saying that they are independent. –  Philip Gibbs Jan 22 '11 at 14:05
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@Georg: Don't be arrogant now. You don't know everything. This is a perfectly legitimate term, go read up on it. –  Noldorin Jan 22 '11 at 19:30
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It seems to me that there is an ambiguity here that is causing some trouble in giving a definitive answer. The question does not specify how the material for the fire is burned, or what it consists of, and this can potentially give rise to different answers. Clearly if we elevate the material to 100km, some material won't burn, because of the lack of an oxygen rich atmosphere. On the other hand, some materials which are already oxygen rich can still be burned.

To lift this ambiguity, and hopefully crystallize the question, lets consider a specific case: You have a sealed insulated contained in which a certain amount of fuel is burned with a fixed amount of oxygen. Further, let's assume there are no ashes left over. (Although it is relatively straight forward to expand this treatment to deal with that case, I think this gives a cleaner question). The container is then opened either at sea level, or at some elevation $H$. The question is now whether the latter process heats the atmosphere more than the first.

The answer to this question is clearly yes. In both cases exactly the same number and type of molecules are produced, and in each case these have the same kinetic energy. However the gas in the container that has been elevated also has some additional energy: the potential energy gained by each molecule due to it's relative elevation in the gravitational field. Once the containers are opened, in each case the gas will eventually come in to equilibrium with the atmosphere. Clearly in the latter case, the total energy of the atmosphere and gas as a whole is now higher than that of the first case by a factor of $m g H$, where $m$ is the mass of the gas released. As the average energy is increased, the temperature must also increase. This is because the possible energy levels for each molecule of the system are the same in both cases, the only way to achieve a Gibbs state with an increased energy expectation value is to increase the temperature.

I hope this can finally resolve this discussion.

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Your answer seems fine, but it depends on a very specific setup. –  Noldorin Jan 23 '11 at 16:35
    
@Noldorin: Yes, but there is an ambiguity in the question. As Peter mentioned, if there is no (or little) gas emitted then the situation is different. –  Joe Fitzsimons Jan 23 '11 at 17:25
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The question is about the effects of gravity on energy and can only be answered properly in the context of general relativity. The answer is that more energy is released when the wood is burnt on the top of the mountain, provided everything is measured from the point of view of a fixed observer and we make some assumptions so that we can ignore uninteresting effects.

I'll assume that the strength of the gravitational pull at rest on the mountain top is the same as at the bottom, because otherwise we would have to consider other effects such as the tiny amount of energy in the wood due to its very slight compression under the gravitational field. This would change as the gravitational pull changed. I don't think that is what the question is meant to be about so I'll ignore it. I will just look at the direct effect of the potential energy as the question seems to suggest.

Energy in relativity is a relative concept. It depends on the relative velocity of the various frames of reference. More significantly for this question, it depends on the position of an observer in a gravitational field. This effect can be measured by observing the frequency of radiation and how the measurement changes with altitude. In practice this has been done experimentally for even quite small vertical displacements.

Consider first observers who are fixed relative to the bundle of wood and ask about the difference in released energy they can measure, if either the wood is burnt at the foot of the hill or at the top. By the equivalence principle there can be no difference unless they include the extra energy released when ashes or gasses released return to ground level. To avoid this effect let's assume that the wood is actually burnt within a sealed container so that only the radiated heat can escape. The conclusion then is that the observers must see exactly the same amount of heat released if they are fixed relative to the container.

The more interesting and correct way to look at the situation is from the point of view of observers fixed relative to the mountian. They could be high up in space collecting all the emitted radiation to measure its total energy or they could be doing the same thing from a lower altitude. The height will certainly make a difference to the total energy gathered because radiation is redshifted as it escapes from a gravitational mass. Someone higher up will measure less energy than someone lower down, but that was not the question.

The question is about how a fixed observer will measure the amount of radiated energy released from the combustion. (We can take this to be a fixed observer at large distance so that even people who do not agree that energy conservation makes sense in general relativity will agree that the ADM energy is well defined for this situation)

The answer in this case has to be that when the wood is burnt on the mountain top, more energy will be released. This is because the radiation released from lower down will be redshifted more when it escapes to infinity so less energy will be measured.

Another way to look at this is to consider what happens when someone uses energy to carry the wood to the top of the mountain where it is burnt, and then he returns the ashes and released gasses to ground level harvesting the energy released in doing so. Because the wood was burnt and it released energy, the ashes and gasses will actually weight slightly less afterwards according to mass energy equivalence, even when all the released ashes and gasses are collected. This means that less energy is released when the remains are returned to ground than were expended in lifting the original wood to the top. This would not have happened if the wood was burnt at ground level. By conservation of energy we have to conclude that more energy was released when the wood was burnt on the mountain top. Notice that I am not saying that more energy is released because the ashes return to ground level. The argument tells us that more energy must have been released during the combustion at altitude even before the ashes are returned.

An exercise for the reader is to calculate the extra energy released using both the explanations I have given and show that the answer is the same.

Edit: To show this is not just hot air I'll give a formula.

If $M$ is the mass of the before burning and energy $E$ is released in heat when burnt on the ground then the mass after burning is $M' = M - \frac{E}{c^2}$.

The energy required to lift the wood up the hill a height $h$ is $V = Mgh$ and the enrgy released on taking the wood back down is $V' = M'gh = V - \frac{Egh}{c^2}$

The extra heat generated from burning at altitude is therefore $V-V' = \frac{Egh}{c^2}$

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This is just wrong. There is no need to talk about relativity here. –  Noldorin Jan 22 '11 at 15:00
    
@Noldorin, General relativity is exactly what is required to fully understand this problem. Your thinking is too Newtonian which is why you think there is no energy difference. In GR all forms of energy interact with gravity. Taking this into account you find the small difference of energy that the altitude makes from the perspective of a fixed observer. –  Philip Gibbs Jan 22 '11 at 16:02
    
How do you know that? General relativity has not been quantised/described under QFT. This is a highly contentious area, about which no-one can speak properly. Perhaps you could explain in more detail how gravity supposedly affects chemical potential energy? I see now way this is possible. –  Noldorin Jan 22 '11 at 16:11
    
How does your answer accord with the equivalence principle? If the atmospheric pressure and gravitational force have the same values at the top of the mountain as at the bottom, shouldn't the physical processes be exactly the same, and the exact same amount of heat released? –  Peter Shor Jan 22 '11 at 16:43
    
@Peter, as I explained in the answer, the heat released as observed by different observers in a fixed position relative to the fire would be the same. This is a clear consequence of the equivalence principle. However, the question only really makes sense if answered from the perspective of a single observer who sees the fires in different relative positions. In that case he sees more energy coming from fires at higher altitude. –  Philip Gibbs Jan 22 '11 at 16:55
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The energy released in the fire is the difference between the energy of the fuel and the energy of the ash.

If you carry wood up a mountain, it gains energy.

But if you burn wood on a mountain, the ashes also have more energy than the same ashes would have at sea level.

The wood has more energy on the mountain, but the ashes also have more energy. If you burn the wood and end up with the ashes, it doesn't matter whether you burn it up high or burn it down low, because the gravitational energy cancels out.

The energy you put into the wood by carrying it up the mountain didn't go away, but the only way you can get it back is by letting the ashes fall down the mountain. (For example you could make the ashes pull a rope attached to a generator.)

(Note that in real life, the "ash" also includes the CO2 and other gases given off by the wood, so to do the experiment properly you'd have to burn it in some kind of large airtight container.)

APPENDIX: It seems like there are two other issues I need to address to satisfy some people.

The first issue could actually result in a measurable difference in the heat released from burning, and that's the difference in atmospheric pressure. What's going on is that when you completely burn the wood in O$_2$, the same amount of internal energy is always released, but what you actually measure is the enthalpy change $\Delta H = \Delta E + p \Delta V$. Let's look at the cases $\Delta V > 0$ and $\Delta V < 0$ separately.

If the burning of the fuel releases more moles of gas than the moles of O$_2$ it consumes (which happens to be the case for burning real wood), then $\Delta V > 0$ so $\Delta H$ is a less negative number (smaller in magnitude) than $\Delta E$. What this means is that some of the energy from the fire went into pushing away the atmosphere rather than producing heat. The pressure $p$ is less on top of the mountain, so from this you would conclude that you get MORE heat after you carry it up the mountain.

You might GUESS that this extra heat energy is somehow related to the gravitational potential energy, but that would be WRONG as I'm about to show.

Let's consider a different fuel that consumes more moles of gas than it releases when burned. (For example, consider burning a strip of magnesium, which undergoes the simple reaction 2 Mg + O$_2$ $\rightarrow$ 2MgO (s).) In this case $\Delta V < 0$, so $\Delta H$ is more negative (greater in magnitude) than $\Delta E$, which means that you actually GAIN energy from the pushing of the atmosphere. Since the pressure is less on top of the mountain, this means that burning a strip of magnesium would provide LESS heat up there than it would at sea level.

What's going on here? The wood and the strip of magnesium both gain potential energy from being carried up the mountain, but one provides more heat up there, and the other provides less! This is because in both cases, the difference between the amounts of heat released at different altitudes is due to the difference in the energy of the atmosphere, NOT the gravitational energy of the solid fuel.

APPENDIX 2:

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This is wrong because the mass of the ashes plus gasses is slightly less after burning than the mass of the wood before. This is due to the release of energy which has a small mass equivalence. See my answer for the correct analysis. –  Philip Gibbs Jan 22 '11 at 12:29
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-1 This answer is hugely misinformed and misleading in my view. –  Noldorin Jan 22 '11 at 13:50
    
Keenan, your additional analysis totally neglects the effects of gravity. Your model simply doesn't include it at all. If you really want to do this rigorously, write down the partition function for the gas in a gravitational field and do the full calculation. –  Joe Fitzsimons Jan 22 '11 at 19:04
    
@Noldorin, what's your complaint? –  Keenan Pepper Jan 22 '11 at 19:04
    
Well you didn't really differentiate between forms of energy. Your appendices help however, so I'm tempted to remove the down-vote now. Still, if you could clarify this, it would help I think. –  Noldorin Jan 22 '11 at 19:32
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I think the answer is very simple.

Suppose the combustion happens without smoke or gas rising for simplicity. All the mass stays at the same height, but it changes chemical form. There will be some slight variation due to binding energy, etc, but that behaviour is consistent at all heights and independent from the gravitational potential.

So, it is clear that whether you burn your fire at sea level or at 1000m, nothing changes from a gravitational potential standpoint.

This said, there are differences, due to enthalpy, due to the concentration of oxygen in the atmosphere, and so on. Some of these have been considered in other answers and are correct - i think they are unrelated to the basic question though.

Will the increased potential energy of the wood make a difference? No, it can't because the same mass ends up in the same relative place after the combustion, no matter at what height you perform the experiment.

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The mass does not remain in the same place if there is gas produced. In that case some of the mass becomes gas, and the gas then falls into equilibrium. –  Joe Fitzsimons Jan 22 '11 at 21:02
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@Sklivvz: No, it doesn't. The molecules literally fall (on average). –  Joe Fitzsimons Jan 22 '11 at 21:11
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@Sklivvz: No, its not. It's due to the fact that the Gibbs states for a gas in a gravitational potential are different to those for a gas free from gravity. –  Joe Fitzsimons Jan 22 '11 at 21:22
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@Sklivvz: You do have them at a lower height, that's the whole point. The charred remains may stay at the top, but anything that has literally gone up in smoke will reach equilibrium with the atmosphere, which will alter its average elevation. –  Joe Fitzsimons Jan 22 '11 at 21:38
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@Sklivvz: of course you don't see an effect if you model the whole atmosphere without gravity (which is what you are doing here), but that model does not reflect reality. –  Joe Fitzsimons Jan 23 '11 at 5:42
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