21 Nov 06
You have a rechargable battery, fully charged. Then you put it in some solvent that can dissolve the entire battery to molecules. Where does the energy in the original charge go? Like you charge it up to 10 amp hours, a 2 volt cell would have the energy of 2 volts at 10 amps for one hour so that would be 20 watt hours of energy invested in the battery, which is then dissolved. What happened to that 20 watt hours of energy?
21 Nov 06
Originally posted by sonhouseI think it would end up heating the solution.
You have a rechargable battery, fully charged. Then you put it in some solvent that can dissolve the entire battery to molecules. Where does the energy in the original charge go? Like you charge it up to 10 amp hours, a 2 volt cell would have the energy of 2 volts at 10 amps for one hour so that would be 20 watt hours of energy invested in the battery, which is then dissolved. What happened to that 20 watt hours of energy?
21 Nov 06
2 edits
Originally posted by uzlessSo theoretically an experiment could be done to quantify the effect?
yup, energy converted to heat
1500 watts of heat for one hour is about 5000 btu/hrs of heat, so
20 watt hours in our example would be about 60 btu's so in that unit system it would heat up 6 pints about 10 degrees F more than if you dissolved a discharged battery?
Has anyone ever heard of just such an experiment? It's similar to the idea of a compressed spring dissolved in acid, where does the energy of compression go?
21 Nov 06
1 edit
Originally posted by sonhousepicture two springs in acid. one spring compressed, the other not compressed.
So theoretically an experiment could be done to quantify the effect?
1500 watts of heat for one hour is about 5000 btu/hrs of heat, so
20 watt hours in our example would be about 60 btu's so in that unit system it would heat up 6 pints about 10 degrees F more than if you dissolved a discharged battery?
Has anyone ever heard of just such an experiment? It to the idea of a compressed spring dissolved in acid, where does the energy of compression go?
The not compressed spring will dissolve as usual.
The compressed spring's molecules are compressed with the compression force acting upon them. The acid must overcome this compressive force in order to dissolve the spring. Energy (in the form of heat) is released during this dissolving process at a rate equal to the amount of potential energy contained in the compressed spring.
21 Nov 06
Originally posted by uzlessYes, I get that part. What I want to know if anyone ever heard of an experiment to quantify the effect.
picture two springs in acid. one spring compressed, the other not compressed.
The not compressed spring will dissolve as usual.
The compressed spring's molecules are compressed with the compression force acting upon them. The acid must overcome this compressive force in order to dissolve the spring. Energy (in the form of heat) is released during thi ...[text shortened]... g process at a rate equal to the amount of potential energy contained in the compressed spring.
22 Nov 06
1 edit
Originally posted by AThousandYoungYep, all the extraneous heat loads would have to be accounted for and subtracted from the final temp. readings to get the spring energy. I would have thought that would have been done somewhere. It seems like a simple high school level experiment. Two solutions, one with a spring unwound, put in what, nitric acid?, then measure the temp, then do it again under identical parameters with the exception of now the spring is coiled up tight. That leads to some difficulties, maybe it would have to be held together with a non-reactive cable tie or something, so the tiedown would not add extraneous data to be accounted for.
I don't know. You'd need to take into account any energy released or absorbed due to the solvation reaction as well.
You would have to be able to calculate the potential energy of the coiled spring and convert that to calories or BTU to make a prediction of the expected rise in temp.
22 Nov 06
1 edit
Originally posted by PBE6Finally, I get an answer correct! Must be the new type of beer I'm drinking
Found a reference to one on this website:
http://www.science.ca/askascientist/viewquestion.php?qID=761
(As is rather obvious, I'm not much of a mathematician, more of a theoretical conceptualist....which proves to be quite useless most of the time 😞 )