Physics help

derbski

derbski

Active member
alright, I am guessing that only college level physics folks will know how to do this, but if any of you think you can do it, give it a whirl. Basically, I just need to find out how much energy is lost in a battery, then the rest of the question is as easy as plugging that into E=mc^2. so here is the problem

A certain 1.5-V battery (whose original mass is .425 kg) has a capacity of 4.5 Ah. About how much lighter is the battery when it is discharged than when it is full [Hint: It has the same number of electrons when "discharged" as it does when it is full]

so, based on that hint, like I said, it should be as simple as finding the energy released in that battery from when it is full to when it is dead, and plugging that into E=mc^2, I can find how much mass was turned into energy.
 
wait, so is that supposed to be the energy lost in the battery, or what? also, could you tell me how you got that? thanks man, +k
 
Ha, no. I just did the electrical chapter and I couldn't remember it, so I put in the kinetic energy equation by mistake.
 
Q=MxCx△T

Q=Energy

M=Mass of object

C=Specific heat

△T=Change in temp.

plug in your numbers and solve like algebra.

is that it?
 
You don't even have enough significant figures to measure the loss in mass. The delta m in a discharged battery vs a charged one is way less than a 1/1000th of a kg. I don't feel like learning how to do circuits and shit again but if you take the energy used by a 1.5 V battery and divide by 9*10^16, I t will be 0 no mater what.
 
I figured it out, like I said, I just had to find out how much energy was used up (but, just like you, I had no idea how to do the simple circuit science behind it), so I texted chacha and found out that it was 5.06 joules, and dividing that energy by the speed of light squared gave the mass of 5.622x10^-17 kilograms. So, like you said, the loss in mass was nowhere near 1/1000th of a kilogram, but when you convert it into atomic mass units, you find out that it is 3.39x10^10 u, or the mass of 33 billion neutrons.
 
do you know the resistance of the battery? because i think you could solve for E as being V^2/R, not sure though
 
gotchya. I read the question wrong, I thought it was asking for the final weight of the battery not how much mass the battery lost. I don't remember any of that atom shit, its been a few years.
 
oh hey there, back again with a different approach. how about C=Q/V and you use the known values C and V to find Q, which you plug into E=(1/2)Q^2/C which you then plug into einstein's eqtn
 
4.5Ah = 4.5A for 1 hour

P = I * V

P = 4.5 * 1.5

P = 6.75 watts

Battery will deliver 6.75 watts for 1 hour (3600 seconds)

W = P * S

W = 6.75 * 3600

W = 24.3K Joules

doesn't matter what time you use to solve since it will cancel out
 
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