Energy conservation in Acid

  • #1
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.
 

Answers and Replies

  • #2
Borek
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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.
 
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  • #3
Dale
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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?
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.
 
  • #4
Borek
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break the spring and mechanically release the remaining potential energy

Heating the solution :tongue2:
 
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  • #5
suhagsindur
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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.

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.
 
  • #6
suhagsindur
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Hello, Borek, Dalespam.
Thank you for giving reply.
 
  • #7
Borek
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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.
 
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  • #8
suhagsindur
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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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