How much more does a fully charged MS weigh?

I think the answer is 4.641 mg assuming a nominal battery voltage of 375 volts and an 85kWh car.

snapped from the net:
Convert kilowatt hours to electron volts | energy conversion
1 kwh = 2.247x10^23 e/kwh
Electron - Wikipedia, the free encyclopedia
weigth of e = 9.109x10^-31 kg/e

facts:
Telsa charged with 85 kwh

total weigth of charge then become
85 kwh * 2.3x10^25 e/kwh * 9.109x10^-31 kg/e = 1.740x10^-3 kg

so generally speeking 1.7 g lighter when not charged comparet to a fully charged, but remeber a Tesla is never fully discharged so in practice it would "only" be around 1.5 gram

Fascinating stuff. And an ICE car with 17 (US) gallons on board would weigh an additional 103.24 Lbs, or 46829 grams.

That is all good and peachy but those electrons just move to the other side of the battery. You need to use E=MC^2 to determine the mass differential. We did it before I'll look for the link.

or you can just put 85kWh into WolframAlpha
85kWh in g - Wolfram|Alpha

Ends up being 0.000003405 grams! So not grams but mircograms.

maybe some point in futre your could fit the 85 kwh pack in a nut shell - or even load the car once in a lifetime in a larger pack and drive almost for ever.

think about old fasion floppy discs (80 kb) and now you got a memory microSD card witch holds 64Gb data - back 20 years ago this was pure fantasy
at same time size have shrinked a factor 100 - that is total a improvement of aprox 100.000.000 times.

Im not saying its going to happen to battery storage, just saying it might happen at some point as demant of better battery are pressent and huge amount is invested into this field.

Like a wind-up toy with a spring, after the toy is wound up it weighs the same as before, except it has potential energy now that can be discharged. Of course, e=mc^2 applies (mass energy equivalence) so that stored energy has a very small weight as ElSupreme identifies.

your right - electron has not been filled, just moved.
E = mc^2
m = E / c^2

1 [kwh] = 3.6x10^6 [J]
c = 300.000.000 [m/s]
1 Tesla charge [Tesla*] = 85 [kwh]
1 [kg] = 1 [J] / (1 [m^2]/ 1 [s^2])

m [kg] = (85 [kwh/Tesla*] * 3.6x10^6 [J/kwh])/300.000.000 ^2 [m/s]
m [kg] = 3.06x10^8 [J/Tesla*] / 9x10^16 [m^2/s^2]
m [kg] = 0.34x10^-8 [J/Tesla*] / [m^2/s^2]
m [kg] = 0.34x10^-8 [kg/Tesla*]
m = 0.0034 [mg/Tesla*]
3.4 microgram

Actually, I'm just thinking that a battery may not be potential energy, it may be chemical energy. But the concept still applies.

My P85 on the scales.... 4,690....

NielsChr said:

your right - electron has not been filled, just moved.
E = mc^2
m = E / c^2

1 [kwh] = 3.6x10^6 [J]
c = 300.000.000 [m/s]
1 Tesla charge [Tesla*] = 85 [kwh]
1 [kg] = 1 [J] / (1 [m^2]/ 1 [s^2])

m [kg] = (85 [kwh/Tesla*] * 3.6x10^6 [J/kwh])/300.000.000 ^2 [m/s]
m [kg] = 3.06x10^8 [J/Tesla*] / 9x10^16 [m^2/s^2]
m [kg] = 0.34x10^-8 [J/Tesla*] / [m^2/s^2]
m [kg] = 0.34x10^-8 [kg/Tesla*]
m = 0.0034 [mg/Tesla*]
3.4 microgram

Click to expand...

Extra credit to Niels for showing his work.

fiksegts said:

My P85 on the scales.... 4,690....

Click to expand...

But was it fully charged at the time?

Note that having a positive potential (charged) battery doesn't necessarily mean you added electrons. It just means that you added some electrons to one side and took away some from the other side of the barrier, causing a potential difference between the two. So aside from whatever asymmetries that exist in a LiON cell (how much there is, I don't know), there should be no difference in weight

I nominate this thread for Question of the Year.

JohnQ said:

Extra credit to Niels for showing his work.

Click to expand...

With a name like Niels, would you expect anything less??

ElSupreme said:

That is all good and peachy but those electrons just move to the other side of the battery. You need to use E=MC^2 to determine the mass differential. We did it before I'll look for the link.

or you can just put 85kWh into WolframAlpha
85kWh in g - Wolfram|Alpha

Ends up being 0.000003405 grams! So not grams but mircograms.

Click to expand...

I'm pretty impressed how many people on this board know enough science to answer this question correctly. Very different than the previous car club I was a part of!

I'm really curious how people on other EV forums would do with this question. Is anyone willing to post a similar question on the Chevy Volt forum, the Nissan LEAF forum, the Fisker Karma forum, etc. and report the results? (you would obviously change the name of the EV to match the forum)

wraithnot said:

I'm pretty impressed how many people on this board know enough science to answer this question correctly. Very different than the previous car club I was a part of!

I'm really curious how people on other EV forums would do with this question. Is anyone willing to post a similar question on the Chevy Volt forum, the Nissan LEAF forum, the Fisker Karma forum, etc. and report the results? (you would obviously change the name of the EV to match the forum)

Click to expand...

We could keep score based on how many responses before someone get the right answer. In our case, ElSupreme nailed it on the fifth response. Of course, this will only work if no one gives away the answer on the other forums