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The charge on my smartphone was zero. I charged my phone to its maximum capacity.

How much mass $\Delta m$ did my mobile gained after charging the battery?

  • Maximum capacity of the smartphone's battery: $2000mAh$
  • Weight of the smartphone with no charge: $200g$
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closed as off-topic by tpg2114, Jon Custer, sammy gerbil, AccidentalFourierTransform, Kyle Kanos Feb 27 '17 at 11:07

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  • 1
    $\begingroup$ Electricity is just the flow of electrons. When you charge the battery, the electrons are shifted (forcefully) in opposite directions than when discharging (naturally), that's it. There is no change of mass involved according to me, but I don't know whether there is some specific mechanism or reaction that would affect the mass per se. $\endgroup$ – Kalpak Gupta Feb 24 '17 at 6:58
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    $\begingroup$ @KalpakGupta As far as the OP isn't concerned about the negligible relativistic effects, you are right. $\endgroup$ – Yashas Feb 24 '17 at 7:01
  • $\begingroup$ But where did that electrons from the charger went??? $\endgroup$ – Creepy Creature Feb 24 '17 at 7:04
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    $\begingroup$ You have a fundamental concept error. The energy stored in a battery is not measured in terms of number of electrons. In fact, they have the same number of electrons when charged and uncharged. $\endgroup$ – Yashas Feb 24 '17 at 7:05
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    $\begingroup$ @CreepyCreature: The "electricity" form the charger is used to move "electrons" in the metals in your battery from one terminal to the other. Consider a spring instead. You wind up a spring loaded toy with your hands. How much mass does the toy gain? If you say none then I ask you your question: where does the calories in your muscles go? $\endgroup$ – slebetman Feb 24 '17 at 8:05
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Your rechargeable battery certainly relies on an electrochemical reaction which gives electrical power to your device when going one way, and goes the other way when you provide the necessary amount of energy when charging.

Since you just moved electrons the other way around, there should be no change in mass.

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  • $\begingroup$ Then where did the electrons of my charger went? $\endgroup$ – Creepy Creature Feb 24 '17 at 7:08
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    $\begingroup$ The energy is the flow of electrons, not the electrons themselves. The incoming number of electrons should be equal to the outcoming number. $\endgroup$ – yketa Feb 24 '17 at 7:21
  • $\begingroup$ @CreepyCreature, It's just like how the links in a bicycle's chain carry energy from the crank set to the back wheel. The energy only flows in one direction, but the same links keep going 'round and 'round. $\endgroup$ – Solomon Slow Feb 24 '17 at 19:12
  • $\begingroup$ Now I got the point !! $\endgroup$ – Creepy Creature Feb 25 '17 at 5:06
  • $\begingroup$ Complex question have a simple answer !! $\endgroup$ – Creepy Creature Feb 25 '17 at 5:12
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There will be an increase of the mass of the battery when you charge it, though that increase is going to be undetectably small.

I would do the calculation in reverse i.e. start with a fully charged battery and calculate how much it decreases in mass when you run it down. The mass decreases because the battery does work $E$ on the electrons that flow through it and that work is related to the mass lost $\Delta m$ by Einstein's famous equation:

$$ E = \Delta m c^2 \tag{1}$$

We can do an approximate calculation using the information you provided:

Maximum capacity of the smartphone's battery: 2000mAh

That capacity means the battery can supply a current of $2$ amps for an hour, so if the battery voltage is $V$ it can supply a power of $2V$ for an hour and the total energy is therefore:

$$ E = 7200V \tag{2} $$

If we substitute this into equation (1) and rearrange we get:

$$ \Delta m = \frac{7200V}{c^2} $$

Smartphone batteries have $V\approx 4$V and this gives us:

$$ \Delta m \approx 3.2 \times 10^{-13}\,\text{kg} \approx 0.32\,\text{nanograms} $$

This is approximate because the battery voltage isn't constant as the battery discharges so our equation (2) isn't exact. However it gives us a good estimate of how much the mass decreases on discharging and increases again on recharging.

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    $\begingroup$ @MaxW: Not just chemistry. The equation concerns energy. So it also affects things like springs. In theory a spring would gain mass when you compress it even though no atoms are added to the thing. $\endgroup$ – slebetman Feb 24 '17 at 8:07
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    $\begingroup$ @MaxW At all stages of education, learning one thing means leaving out details so that the lesson is clear. When you first learn subtraction, it's always taking a smaller number from a bigger one. Negative numbers come later. When you learn square roots, it's always of positive numbers; imaginary numbers come later. Relativistic mass changes are a useless detail when learning chemistry, and would only confuse students when they are first introduced to the subject. Even professional chemists largely ignore relativity because of the tiny effect. $\endgroup$ – Mark H Feb 24 '17 at 8:12
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    $\begingroup$ @MaxW: Of course we have discussed this here already :-) See for example Conversion of mass to energy in chemical/nuclear reactions $\endgroup$ – John Rennie Feb 24 '17 at 8:27
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    $\begingroup$ Just a very small suggestion. V in this answer stands for both voltage, and unit Volt, it looks quite confusing especially to untrained eye. At the 1st glance it looks like you state that energy is 7,2kV. It would be nice if the calculation were edited to be clearer. $\endgroup$ – luk32 Feb 24 '17 at 9:42
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    $\begingroup$ While I don't disagree with the analysis, in a very real sense, the phone weight does not change from this, because the weight of the phone is not even defined to the accuracy needed to measure this. Your phone is shedding atoms from wear and gaining atoms from dust and other contamination all the time. These mass changes far exceed $0.32$ nanograms. $\endgroup$ – Paul Sinclair Feb 24 '17 at 14:23

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