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Physics help

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derbski
Posts: 3173 - Karma: 62
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.
May 16 2010 6:05PM
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Benson23
Posts: 1713 - Karma: 15
i have enough of my own homework.
May 16 2010 6:16PM
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Jamestown
Posts: 4002 - Karma: 51
.5mv^2
May 16 2010 6:19PM
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derbski
Posts: 3173 - Karma: 62
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
May 16 2010 6:20PM
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Alex_G
Posts: 11902 - Karma: 1,019
ns is where i go to avoid doing my own homework
May 16 2010 6:21PM
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Jamestown
Posts: 4002 - Karma: 51
Ha, no. I just did the electrical chapter and I couldn't remember it, so I put in the kinetic energy equation by mistake.
May 16 2010 7:26PM
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saucybiscuits
Posts: 2554 - Karma: 173
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?
May 16 2010 8:29PM
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Edew19
Posts: 365 - Karma: 13
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.
May 16 2010 8:58PM
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derbski
Posts: 3173 - Karma: 62
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.
May 16 2010 9:35PM
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cbeebzthasteeze
Posts: 1862 - Karma: 85
The answer is obvious. 42.
May 16 2010 10:00PM
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TheButterHashira
Posts: 9691 - Karma: 1,702
That is so wrong. That is the equation you use for calorimetry.
May 16 2010 10:05PM
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pianoman
Posts: 2503 - Karma: 779
do you know the resistance of the battery? because i think you could solve for E as being V^2/R, not sure though
May 16 2010 10:40PM
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Edew19
Posts: 365 - Karma: 13
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.
May 16 2010 10:43PM
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pianoman
Posts: 2503 - Karma: 779
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
May 16 2010 11:18PM
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Mac.
Posts: 953 - Karma: 24
Power dissipated = VI

So current times voltage, then try to convert that over?
May 17 2010 8:59AM
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I.killCommies
Posts: 186 - Karma: 10
wow i remember my asst track coach telling me that joke like 12 years ago
May 17 2010 1:06PM
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cbeebzthasteeze
Posts: 1862 - Karma: 85
Cool story, coolstory.
May 17 2010 8:00PM
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^nate^
Posts: 1263 - Karma: 560
i'm quite positive this is the correct approach.
May 17 2010 8:13PM
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Bepo
Posts: 8195 - Karma: 17
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
May 17 2010 8:34PM
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