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This is just an interesting question a friend's uncle asked me that I was somewhat annoyed I couldn't answer.

When a material dissolves in acid there is a chemical process that causes the changes in state/composition of the materials in question. There will be a change in energy due to chemical bonds forming/breaking and potentially a change in temperature.

So we take a spring and compress it in a clamp. Obviously now there is a potential energy based on the compression of the spring as well as the chemical energy within the spring itself before compression.

This is now placed into a bath of acid that will dissolve the spring but not the clamp.

There is a change of energy based on the dissolving of the spring, but what happens to the potential energy due to it being compressed?

My thoughts were that in a realistic situation one (weak) point on the spring would be attacked, break, it would then probably just ping out of the clamp and lose the energy that way.

But in the ideal scenario where is dissolves at a uniform rate, this wouldn't happen, so what happens at the instant that it is completely dissolved? This is obviously a massive assumption, so would be unlikely to be true. I thought maybe some kind of change of temperature would occur, due to the assumption that the compression would cause the atoms to be more tightly compacted causing a repulsion. As you remove layers this would gradually decrease, but there is still an inherent increase in energy for each atom, so this would most likely be expelled as something (like) heat.

What are people's thoughts on this?


I'm mainly interested in the detail of the mechanisms for the energy tranfers on an atomic level as opposed to the broad picture. Specifically the transfer from coulomb repulsion to heat in the acid. The more low-level detail the better.

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I agree, the energy goes into the bath as heat. – Mike Dunlavey Jul 30 '13 at 14:01
So for every layer of atoms removed there is a proportional increase in temperature? The fact that (I think) the atoms are packed closer together, increasing coulomb repulsion, is then transfered from the electromagnetic potential/replusion leads to an increase in temperature of the acid in solution? – Folau Jul 30 '13 at 14:04
That's my guess. – Mike Dunlavey Jul 30 '13 at 14:05
I think the answer is given in the question. – Ali Jul 30 '13 at 14:09
You should also note that as the spring dissolves, its spring constant would change. So $\frac 12 kx^2$ would change. – udiboy1209 Jul 30 '13 at 14:20
up vote 7 down vote accepted

Your first assumption, that there would be a weakness in the material of the spring and it would suddenly break if corroded enough is what realistically would happen.

In your idealized situation each atom dissolved in the liquid had a proportional part of the potential energy of the stressed spring. As it looses its bonds with the surface, bonding with the molecules that corrode the material, its potential energy becomes kinetic energy of the new molecule and, due to the large number of atoms, it will be small enough to just turn into the heat energy of the liquid. In this process the potential energy of the spring will be gradually transformed to heat energy of the liquid, as is your guess.

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So the coulomb repulsion of the atoms in the spring is transfered into a repulsion of the product molecule away from the atoms remaining in the spring which, in turn, becomes kinetic energy as the product is free to move in solution - which is approximately heat. – Folau Jul 30 '13 at 14:18
Yes, instead of one large PING if the clamps were taken out, there are proportional small pings, as each atom changes partners, from surface atoms to corrosion atoms. – anna v Jul 30 '13 at 14:25

Start by considering a gas phase reaction between two reagents $A$ and $B$. If you were watching the molecules with a microscope you'd see the reagents $A$ and $B$ collide, and the products $C$ and $D$ emerging. If you measure the kinetic energy of the product molecules you'll find it differs from the kinetic energy of the reagents by the energy of the reaction. So if the reaction is exothermic the products will have greater kinetic energy than the reagents. So when we talk about the energy of a reaction we mean the increased kinetic energy of the products.

Now reactions in solution are a whole lot messier, but the basic idea remains the same. If you dissolve a spring not in tension you have a reaction something like:

$$ M + 2H^+ \rightarrow M^{2+} + H_2 $$

(for a divalent metal dissolving in acid) and the reaction energy will be (initially) carried away by the $M^{2+}$ ion and $H$ atoms. I say initially because of course the reagents are not interacting in isolation because they're surrounded by solvent molecules, but basically the reaction energy is carried away as kinetic energy. In a liquid this of course raises the temperature.

Now when you put the spring under compression you are pushing metal atoms together. So when they react they carry away not just the energy of the reaction but also a small amount of energy from the fact the metal atoms will spring apart from each other when they can. So the total energy of the reaction products will be slightly greater than if the spring was relaxed.

So by compressing the spring you are very slightly increasing the reaction energy with whatever is dissolving the metal, and the result will be that the solution ends up very slightly hotter. Needless to say, in practice the increase would be immeasurably small.

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