# Mass difference of a charged and a drained battery cell

1. Apr 23, 2010

### Ralph Spencer

Is there a difference between the mass of a battery cell which is charged and the one which is drained? If yes, is difference of the two values proportional to the total work done by the cell (including loss due to resistance), with c^2 as the proportionality constant?

2. Apr 23, 2010

### LostConjugate

The mass should remain the same, the same number of protons/neutrons and electrons remain in the battery when it is discharged as when it is charged. Only the molecular structure is changed.

I can't think of any reason you would lose mass.

The battery performed work by reducing the potential energy of the molecules, like when hydrocarbons are burned to water and carbonoxides

3. Apr 23, 2010

### vin300

and that means loss of mass

4. Apr 23, 2010

The drained battery has less energy and will be less massive as in accordance with E=Mc squared.

5. Apr 23, 2010

### LostConjugate

Thats curious, it still has the same number of particles. Is this because the atoms in unstable molecules have higher energy states and more internal kinetic energy which translates to a more massive battery?

6. Apr 23, 2010

### vin300

having the same number of particles does not mean having the same mass, that's all the point of E=mc^2

7. Apr 23, 2010

### LostConjugate

Because the particles have relative velocity correct?

8. Apr 23, 2010

### Staff: Mentor

The potential energy of the system also contributes to (or subtracts from) the system's mass.

9. Apr 24, 2010

### Ralph Spencer

I think the mass will change, but too small to be easily detected - and may be possibly lost to error.

Am I correct?

10. Apr 24, 2010

The energy stored in the battery is easily measured and the resulting mass lost can be calculated.The mass loss is far too small to be detected by direct mass measuring techniques such as weighing.

11. Apr 25, 2010

### Ralph Spencer

Is this the correct proof:

Code (Text):

Let m₁ = Initial mass of the battery cell.
m₂ = Mass of the drained battery cell.

Now, m₃ = m₁ - m₂.

E = mc²
and, m=m₃

Therefore, E=m₃c²
⇒ m₃c² = E
⇒ m₃ = E/c²

So, as far as E is non-zero, m₃ has to be have a positive value
however small.

As m₃ is positive, m₁ > m₂.
i.e. A battery looses mass as it gets drained.

Last edited: Apr 25, 2010
12. Apr 25, 2010

### LostConjugate

The difference in mass is equal to the difference in energy divided by c^2. I think you had it right in your original post there.

13. Apr 25, 2010

### Ralph Spencer

Why so? It should be total energy divided by c²? I don't understand where the difference of energy comes from.

14. Apr 25, 2010

### rcgldr

15. Apr 26, 2010

### nucleus

For a wet lead acid battery, a battery that has a mixture of water and sulfuric acid, it does weigh less when discharged. A common way to check the battery for discharge is to measure its specific gravity, which is a measure of its electrolyte compared to same amount of pure water. Using a hydrometer, pure water has an SP of 1.00 while a charged battery may have an SP of 1.260. This means that electrolyte is 1.260 times heaver than pure water. It is the sulfuric acid that enables it to weigh more than water. In a dead (sulfated) battery, the hydrogen portion of the sulfuric acid combines with the oxygen portion of the lead oxide of the positive plate producing water.

During discharge there are different reactions occurring to positive, negative and electrolyte simultaneously. Positive plates contain lead oxide (lead and oxygen), which is a compound that can be separated. Negative plates contain sponge lead as its active material. When discharging the lead portion of the lead oxide on positive plate starts to mix with the sulfate found in the sulfuric acid, which forms lead sulfate on the positive plates. At same time lead in negative plate combines with the sulfate from sulfuric acid to form lead sulfate on negative plates. The oxygen portion of the active material on positive plates combines with the hydrogen of the sulfuric acid to form water. The water drastically reduces the strength of the electrolyte. Both the negative plates and positive plates now contain heavy concentrations of lead sulfate and water.

During recharging chemical reactions are reversed. The individual chemicals split from their compounds reform to original state.

16. Apr 26, 2010

### PhilDSP

L. Brillouin addresses this problem in the monograph "Relativity Re-examined". He essentially says that the equation $$E = mc^2$$ takes only kinetic energy into account. He derives an extra term for potential energy in a simple case. Maybe in this case it makes sense to differentiate inertial mass from potential mass which is stored in the E and B fields?

Last edited: Apr 26, 2010