weight of 85kwh of electrons, a somewhat non-sensical question

[winfield100]

I realize this is a somewhat non-sensical question, but I am trying to vaguely find the "weight" of 85kWh of electrons. "It takes 6 x 10^18 electrons to equal a coulomb. A kilowatt hour is a measure of power. An amp is equal to one coulomb/second. A watt is volts x amps. SO, if you had a specific voltage, you could figure amps & watts, and from there how many electrons. "
As an example ""
120V, a kilowatt hour is an 8.33 amp device running for an hour
8.33 amps/sec = 8.33 coulombs/sec
8.33 x 60 x 60 = 29,998 coulombs/hour
29,998 x 6 x 10^18 = 1.8 x 10^23 electrons. Hmm, not even a mole? Sounds low, but I believe the maths right.
An electron's mass is 9.11×10−31 kg
9.11 x 10^-31 x 1.8 x 10^23 = 1.64 x 10^-7 kg or .16 micrograms ""

However, I don't know the amps and volts the Tesla battery runs at and physics class was 40 years ago.
I am also uncertain if the weight of the electrons "vanishes" as they are "used" or just move around but doubt that.
('gallons of sunshine' do have "weight/mass" since they will push a solar sail but this is not exactly related to the question)
The question is in realtion to someone posting that 6 gallons of gasoline weigh far less than an 85kWh battery and I am trying to point out they are not the same, the electrons are the fuel as is the gasoline,
thanks for bearing with me

 

[lphe]

If you are trying to measure the mass of 85 kWh of energy, you should be using E = mc^2. 85 kWh = 3600*1000*85 Joules = 3.06e8 Joules. The speed of light is 3e8 m/s.

So mass = 3.06e8/(3e8)^2 = 3.4e-9 kg.

 

[winfield100]

If you are trying to measure the mass of 85 kWh of energy, you should be using E = mc^2. 85 kWh = 3600*1000*85 Joules = 3.06e8 Joules. The speed of light is 3e8 m/s.

So mass = 3.06e8/(3e8)^2 = 3.4e-9 kg.

so 3.4e-6gm or 3.4e-3mg or 3.4 micrograms?
(glancing out at my PV array that helps fuel my PHEV)
(i do find it a bit tough that the equation is that simple, but it's been a long time..)

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If you are trying to measure the mass of 85 kWh of energy, you should be using E = mc^2. 85 kWh = 3600*1000*85 Joules = 3.06e8 Joules. The speed of light is 3e8 m/s.

So mass = 3.06e8/(3e8)^2 = 3.4e-9 kg.

so if i understand, about every hour an 85kW Tesla would use vaguely 3.4 micrograms of electrons and in vaguely 4-5 hours would use 14-17 micrograms of fuel (electrons)?

 

[lphe]

First, you do not actually get 85 kWh of usable energy out of the battery. From what I have seen posted at this forum, it takes about 77 kWh to fully charge an empty battery and you can get about 75 kWh out of a fully charged battery. Some energy is lost in the form of heat and other undesirable chemical reactions. So the battery is about 75/77 = 97% efficient. The usable energy capacity of the battery is 75 kWh.

I would not state that the Tesla uses 3.4 micrograms of electrons. It does not convert electrons to energy. Instead, chemical reactions take place which change the binding energy of the electrons to the atoms (if you remember Chemistry class discussing the energy released when breaking molecular bonds). It is this change in binding energies that releases energy that can be used by the car to perform work. The change in mass of the battery is due to this change in binding energies.

 

[winfield100]

yes, thats kinda why I started the thread with "..nonsensical question" it's a false equivalvency to equate the weight of 6 gallons of gas to the weight of an 85kW battery. the electrons don't "vanish" the energy in bonds makes more sense (heck or even valence levels). thank you for your kind answer

 

[Danal]

I would not state that the Tesla uses 3.4 micrograms of electrons. It does not convert electrons to energy. Instead, chemical reactions take place which change the binding energy of the electrons to the atoms. It is this change in binding energies that releases energy that can be used by the car to perform work. The change in mass of the battery is due to this change in binding energies.

Agreed. There are no more or less electrons in the fully charged or discharged pack. They just flow out the - terminal and back into the + (Ben Franklin got it backwards. Really). Within the pack, the electrons are just in different places, and at different binding energies, when at different State of Charge.




Another bit of techno-trivia about electrons:

Electricity concepts are often taught by the "Water in a Pipe" metaphor. Voltage = Pressure, Amps=Flow, Resistance = a pinch in the pipe, and so forth. This is great and leads to a lot of insight... but one place where it is VERY wrong is with regard to the "amount of flow". Our life experience with a water hoses and such leads us to believe that if we could see electrons, they'd be squirting along just like water. In fact, amps that represent a LOT of current in the real world move the electrons a VERY SMALL physical distance.

Final result: When a P85D punches at about 500kW (round numbers) this is 400V at 1250A (very round numbers). During the 3.1 seconds it takes to go 0 to 60, the electrons move through the High Voltage cables about 0.14 inches (less than 1/4 inch). This assumes 1 CM diameter HV cables. Does anyone know the actual specs? If they are bigger than this, the actual physical distance will be less, and so forth.

Anyway... as compared to our experiences with water, electrons don't move much. They make their power with tremendous "pressure".

Danal




How to calculate:

• Power about 671HP or 500kW, about 400V at 1250A
• Value for electric current: I = 1250 ampere
• Wire diameter: D = 1 cm, radius R=.5cm
• Mobile electrons per cc (for copper, if 1 per atom): Q = 8.5*10^+22
• Charge per electron: e = 1.6*10^-19

The equation:

cm/sec
= ________I_______
Q * e * R^2 * pi

 

[winfield100]

it is quite enjoyable that I can get fairly lucid technical answers to a somewhat nonsensical question. "500kW (round numbers) this is 400V at 1250A" this helps me understand why the 100% instant torque. (and why my question was "dumb")

 

[SteveS0353]

it is quite enjoyable that I can get fairly lucid technical answers to a somewhat nonsensical question. "500kW (round numbers) this is 400V at 1250A" this helps me understand why the 100% instant torque. (and why my question was "dumb")

As with most advanced technology, "early adopters" are typically a pretty savvy bunch. I've noticed that there are a great many very knowledgeable and experienced folks participating on this forum. There are no "dumb" questions, and every technical question I've seen asked, has eventually elicited a useful answer from someone here.

 

[obrien28]

The OP seems to have the right idea, I remember reading this post from Tesla's blog back in the Roadster days (2008) which measured the weight of the electrons used, coming out to 3.089mg or the weight of a small sand grain. After checking online for the mass of an electron it would appear the author got the weight wrong by over 3 orders of magnitude 10^-28 instead of 10^-31. Maybe he made some other assumptions which increased the mass, but he doesn’t go over them in the blog post. Still, the number is vanishingly small compared to the weight of a gallon of gasoline (energy per unit mass is quite another story though)