Where does potential energy go when exposed to acid?

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SUMMARY

The discussion revolves around the fate of potential energy in various scenarios involving acids and mechanical systems. In Case 1, the potential energy stored in a compressed iron spring is released through stress corrosion cracking before the spring fully dissolves in acid. Case 2 highlights the repulsive potential energy between two magnets, where the energy is released when the magnet inside the acid vessel dissolves. Case 3 addresses the energy stored in a charged capacitor, emphasizing that the energy is conserved and transformed into heat when the capacitor plates are destroyed. The discussion emphasizes the importance of understanding energy transformation and conservation in chemical and mechanical contexts.

PREREQUISITES
  • Understanding of potential energy and its mathematical representation (E= ½ * K * x²).
  • Knowledge of stress corrosion cracking and its effects on mechanical systems.
  • Familiarity with capacitor operation and energy storage in dielectric materials.
  • Basic principles of magnetism and magnetic repulsion.
NEXT STEPS
  • Explore the principles of stress corrosion cracking in metals.
  • Research energy transformation in chemical reactions, particularly in acid solutions.
  • Learn about the thermodynamics of charged capacitors and energy dissipation.
  • Investigate the effects of magnetic fields on energy storage and release.
USEFUL FOR

Students of physics, engineers working with mechanical systems and electrical components, and anyone interested in the interplay between chemical reactions and energy conservation.

suhagsindur
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Case:1 I have one iron spring. I compress it gives potential energy E= ½* K*x2 & bind it by glass wire which is not melted in acid. Now I put this spring in acid & it melted completely, So where the potential energy gone?

Case:2 I put one magnet in glass vessel & hold it in vessel by some mechanical means. I put another magnet outside of vessel in such a way that there is a repulsive force between them & so both magnet possesses potential energy w.r.t. each other. Now I pour acid in vessel & magnet inside vessel is melted, So where the potential energy gone?

Case:3 I charge capacitor & break it. Both the charged plate taking outside carefully so it not get discharged. Previously the energy which is in the dielectric media, now where is that energy? If I destroy both the plate of capacitor by throwing it into acid OR furnace, How the energy will conserve which I have given to charge the capacitor?
Please for giving answer take suitable acid & material.
Please give explanation from microscopic point of view also, so I can understand better.
 
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Final temperatures of solutions should differ.

Case 3: was it easy to pull the plate apart?

Note: metal doesn't melt in acid, it dissolves.
 
Last edited by a moderator:
suhagsindur said:
Case:1 I have one iron spring. I compress it gives potential energy E= ½* K*x2 & bind it by glass wire which is not melted in acid. Now I put this spring in acid & it melted completely, So where the potential energy gone?
Borek is correct theoretically, but from a practical standpoint you will get stress corrosion cracking which will break the spring and mechanically release the remaining potential energy long before it dissolves.
 
DaleSpam said:
break the spring and mechanically release the remaining potential energy

Heating the solution :-p
 
Last edited by a moderator:
Borek said:
Final temperatures of solutions should differ.

Case 3: was it easy to pull the plate apart?

Note: metal doesn't melt in acid, it dissolves.

If I make leyden jar apparatus in which outside metal plate is there & inside water or metal plate & then charge this jar. Now it is easy to pull apart both the charged media with some care.
 
Hello, Borek, Dalespam.
Thank you for giving reply.
 
suhagsindur said:
If I make leyden jar apparatus in which outside metal plate is there & inside water or metal plate & then charge this jar. Now it is easy to pull apart both the charged media with some care.

You have missed the point. You need to apply some force to pull apart charged plate, don't you? It is not different from pulling apart two charges.
 
Last edited by a moderator:
OK, it required force to pull apart & this additional energy which I give plate to pull apart is giving higher increase in temperature then theoretical calculation in which I only take electrical energy for calculation of rise in temp.
thank you.
 

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