Let's say there's a puddle of water on the ground. I use a magical device to give it enough thermal energy to vaporize into water vapor. The water vapor floats up into the sky. I then use the magical device to absorb the same amount of thermal energy I previously gave it. The water vapor then condenses to water and falls back on the ground and forms the same puddle of water on the ground.

My device absorbed and gave the same amount of energy, therefore the net energy in the system should be the same. However, it seems like the system's energy should increase from the water vapor floating up into the sky and producing thermal energy from friction with the air molecules. It should also produce more energy from friction with the air when it falls as rain drops towards the Earth and also when it hits the ground and disperses more thermal energy from its kinetic energy.

This breaks the law of conservation of energy, but I don't see what's wrong with my model. I thought about this when I read that the rain produces a lot of thermal energy from friction with the air.

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    $\begingroup$ Possible explanation (not a physics expert, so grains of salt, etc.): will the water be the same temperature after you absorb the vapor's energy as before you put the energy into the puddle? $\endgroup$ Commented Oct 6, 2021 at 18:41
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    $\begingroup$ The water vapor floats up into the sky. Some work here is being done; where does the energy come from? Perhaps the vapour cools? $\endgroup$
    – Qwerky
    Commented Oct 7, 2021 at 17:16
  • $\begingroup$ A different amount of energy is absorbed when the water condenses from the amount that was required to vaporize it, because pressure changes in the atmosphere. $\endgroup$ Commented Oct 8, 2021 at 2:03
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    $\begingroup$ The system you are describing is not a closed system. The rising vapor gains gravitational potential. $\endgroup$
    – Markoul11
    Commented Oct 8, 2021 at 8:32
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    $\begingroup$ The system you're describing is only about two steps away from being a steam engine, and steam engines don't violate conservation of energy. $\endgroup$
    – Hearth
    Commented Oct 8, 2021 at 15:49

7 Answers 7


The first device is not so magical - you can accomplish the same result with a fire under a pot. The second device, on the other hand, is indeed magical: you are converting heat into usable energy without any side effects. This is prohibited by the second law of thermodynamics.

You are also violating the first law, though, and the thermodynamics police are coming for you ;)

The problem lies in making an unwarranted assumption. The vapour rises in the air, and you say you want to consider effects such as "friction", or the exchange of heat between water and other air molecules. A proper description of this phenomenon will show that any energy gained by the air is lost by the water, so when you activate your second-law-violating device the vapour will yield less energy than what you put in initially - unless, of course, you are also able to retrieve the energy which was lost to the air.

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    $\begingroup$ The concept is critical: even though energy is conserved, useful energy in the form of available work is not. Otherwise, among other absurdities, you could simply transfer heat from cold to hot, and generate both absolute zero and core-of-sun temperatures simultaneously for free. $\endgroup$
    – obscurans
    Commented Oct 6, 2021 at 18:30
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    $\begingroup$ Could Maxwell's demon save it? $\endgroup$ Commented Oct 7, 2021 at 10:36
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    $\begingroup$ I think even if we neglect friction, we find that the air gains potential energy when the water is vaporised at ground level. When the water was vaporised, its volume increased. The increase in volume displaced the air, which has nowhere to go but up, therefore, the air gains potential energy. $\endgroup$
    – Clumsy cat
    Commented Oct 7, 2021 at 15:33

I use a magical device to give it enough thermal energy to vaporize into water vapor. The water vapor floats up into the sky. I then use the magical device to absorb the same amount of thermal energy I previously gave it.

There is nothing particularly magical about these devices. An ordinary heat pump will suffice. Or you can go even simpler and just have a warm thermal reservoir at the bottom and a cool thermal reservoir at the top.

What you are missing is pressure. The enthalpy of vaporization and the enthalpy of condensation are equal (opposite) but they depend on the pressure: $$\Delta H_{vap} = \Delta U_{vap}+ p \Delta V$$

So as the vapor rises the work that you can extract is equal to the change in the $p \Delta V$ term. The $\Delta U_{vap}$ term is the same at the top and the bottom, but the $p \Delta V$ term is not.

So not as much energy gets transferred into the top reservoir. The difference is equal to whatever work you could have extracted from the cycle.

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    $\begingroup$ Thank you, this is the correct answer. Since it is the buoyant force that makes the vapor rise, we need to look at what causes the buoyant force to change, namely the change in volume. Things like the entropy change or heat transfer with the atmosphere as discussed in other answers are red herrings. $\endgroup$
    – Carmeister
    Commented Oct 7, 2021 at 0:42
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    $\begingroup$ Thank you. This answer is perfect. I didn't know that the enthalpy of vaporization and the enthalpy of condensation are dependent on pressure. $\endgroup$
    – John
    Commented Oct 7, 2021 at 4:31
  • $\begingroup$ @John you are welcome, I am glad this was helpful $\endgroup$
    – Dale
    Commented Oct 7, 2021 at 10:44

The reason why energy appears not to be conserved is that your scenario is utterly unrealistic.

When you say there is a puddle, what you mean is that there is a localised concentration of something like a billion billion billion water molecules each with such a low kinetic energy (on average) that they remain held together by intermolecular forces.

When you vaporise the puddle, you add kinetic energy to all of the molecules so that they are no longer tied to each other. As a result, they individually move off in random directions and at random speeds, and then undergo random interactions with the molecules that form the atmosphere. Within no time at all the molecules will be completely dispersed and none of them will have the same energy that had when they first left the puddle, having either gained or lost energy through collisions.

If, somehow, at some later instant you were able to identify all the billions of molecules that had formerly comprised the puddle, collectively they would no longer have the same energy they had collectively posessed when you first vaporised them. You say that you would now extract from these molecules the same amount of energy you initially added to them. That means you would now slow them all down by some amount. That does not mean that they would all miraculously retrace their complicated random steps and return to being a puddle, but simply that out of countless trillions of molecules in the local atmosphere, a very small proportion would reduce their average speed momentarily.

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    $\begingroup$ Good answer, though one minor quibble: it's way more than billions of molecules of water. If the puddle had, say, 1 kg of water, that would be 3.34 x 10^25 molecules of water. :) Which, of course, just makes your point even further. $\endgroup$
    – reirab
    Commented Oct 6, 2021 at 18:40
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    $\begingroup$ Reading your answer makes it seem as though you did not read OP's question at all. $\endgroup$
    – Muuski
    Commented Oct 6, 2021 at 20:20
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    $\begingroup$ Moving all the vapor molecules into one spot may be a violation of the "entropy always increases" law, but the law of conservation of energy is independently true even if you run systems backwards (and so apparently are always decreasing entropy), isn't it? It's not clear to me why the crazy improbability of the "water vapor re-coalesces a mile up" scenario should mean that, in the unlikely event of a water landing, energy should not be conserved. Let's say I'm willing to (try to) run this experiment 10^whatever times. In one of those times it works; is energy not conserved that time? $\endgroup$ Commented Oct 6, 2021 at 20:50
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    $\begingroup$ @MarcoOcram Right. And this answer does nothing to explain how/why energy is conserved despite the argument given in the question, you see what I'm saying? $\endgroup$ Commented Oct 6, 2021 at 21:55
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    $\begingroup$ @MarcoOcram But I do not magically reform the puddle with the exact same energy it had originally. I magically take as much energy out of the puddle as I first magically put into it. If energy is conserved between the two magical events, energy will be conserved across the whole combination of magical+nonmagical events. And yet it appears not to be (to the question asker, anyway); why? $\endgroup$ Commented Oct 6, 2021 at 22:58

You've described a scenario that adds kinetic energy to particles in the air, so there is a net increase in energy. Of course, we can regard this as "non-magical" if we admit you initially warm water (e.g. by burning oil to get the heat needed), then cool it with "the same" system (even if it's something different, we can define a system as two devices combined). Since the air warms, there is a net consumption of the machine's original energy supply. In particular, if the absorbed heat were used to reverse a process that had released some, we wouldn't quite be able to reverse it all.


Let's say the water starts at temperature $T_{water}$, and you give it $E_{magic}$ joules of thermal energy, vaporising it and leaving it at $T_{vapor}$.

Then you wait a while, letting the vapor rise and interact with the air. This results in the air gaining some amount of thermal energy $E_{rise}$.

Finally you remove $E_{magic}$ energy from the vapor, which condenses back and falls back into the puddle (also producing $E_{fall}$ thermal energy in further interactions with the air and ground).

Now, your claim is that you added $E_{magic}$ to the system and then removed the same amount, so there should be no net increase in energy in the system. But also the system gained $E_{rise}$ and $E_{fall}$ thermal energy, which contradicts the reasoning that the system did not undergo any net increase in energy.

The obvious problem1 with your scenario is neglecting to keep track of the temperature of the water/vapor.

I'm assuming the water started in thermal equilibrium with its environment, so as soon as you increase its temperature above the initial $T_{water}$ it will start transferring heat to the ground and the air. This means that after waiting a while to let the vapor rise and interact with the air, it is no longer at $T_{vapor}$. Removing $E_{magic}$ energy from the vapor therefore does not return it to $T_{water}$, it must end up temperature $T_{final}$ where $T_{final} < T_{water}$.

If it is true that this whole process conserves energy, then the extra thermal energy that ends up in the air and ground must be balanced by this "energy deficit" comparing the final temperature of the water to the initial temperature.

On the other hand, if you don't remove exactly $E_{magic}$ energy from the water, but only enough to return it back to temperature $T_{water}$, you have removed some amount of energy less than the $E_{magic}$ you added, so you increased the energy in the system.

1 There are also thermodynamic problems with assuming that it's possible to add and remove thermal energy from one targeted part of a system with equal ease, and without needing to account for external energy.


Suppose we have water at boiling point for some pressure and temperature inside a vessel, from which mass and energy can not escape. Water in liquid and vapour state would be present, in a continuous loop of liquid -> vapour and vapour -> liquid. The energy liberated from the second phase transformation is absorbed by the first one.

All the friction and stuff like that doesn't matter because energy doesn't escape from the vessel, and will be always available as heat.

The only meaningful difference from OP scenario is the vanishing small probability that first all molecules changes from liquid to vapour state, and after that all return to liquid, keeping that process as a kind of oscillator.

The real situation is the presence of vapour and water at the same time, without macro oscillations, and just because the probability of this state is much greater.


Your system is larger than you have considered.

When you added energy to the water, it will immediately begin dissipating that energy through any and all means possible including convection, conduction, and radiation.

Let us consider only the radiation, and assume some of it has not been absorbed. Those photons are leaving you at the speed of light. Assuming you waited 1 minute between spells your system is 1 light minute in radius... far beyond the confines of the planet.

After retrieving all your stray photons, undoing all the interactions that have occurred in that minute, and otherwise retrieving all the energy originally input, you have managed to do it with 100% efficiency. Magic indeed.

  • $\begingroup$ I'm confused. The question appears to explicitly request that we do not retrieve all of our stray photons. $\endgroup$ Commented Oct 7, 2021 at 0:55
  • $\begingroup$ Sort of... he says he pulls out the same net energy, but that it is the same puddle. There is no location provided for where the magic energy came from or was collected to. So I think my answer stands... the system to be considered must be expanded to include all energy and and all interactions including magic, not just the select few in the question. To retrive that energy with 100% efficiency (including the magic realm) would require running time backward, so magic indeed. I'm not strong on this part, but I believe the system would need to be the light cone of the initial blast? $\endgroup$
    – Doug
    Commented Oct 8, 2021 at 18:22

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