Calculating Rest Mass: A Small Mass Increase due to Energy Input

In part (a), we first must find the energy stored in the battery, which equals what the battery can supply in the form of electrical potential energy. Since ${\text{PE}}_{\text{elec}}=\text{qV}$, we have to calculate the charge $q$ in $600\mathrm{A·h}$, which is the product of the current $I$ and the time $t$. We then multiply the result by 12.0 V. We can then calculate the battery’s increase in mass using $\mathrm{\Delta }E={\text{PE}}_{\text{elec}}=(\mathrm{\Delta }m)c^2$. Part (b) is a simple ratio converted to a percentage.

1. Identify the knowns. $I\cdot t=\text{600 A}\cdot \text{h}$; $V=\text{12}\text{.}0\text{V}$; $c=3\text{.}\text{00}\times {\text{10}}^8\text{m/s}$
2. Identify the unknown. $\mathrm{\Delta }m$
3. Choose the appropriate equation. ${\text{PE}}_{\text{elec}}=(\mathrm{\Delta }m)c^2$
4. Rearrange the equation to solve for the unknown. $\mathrm{\Delta }m=\frac{{\text{PE}}_{\text{elec}}}{c^2}$
5. Plug the knowns into the equation.
   $$\begin{array}{lll}\mathrm{\Delta }m& =& \frac{{\text{PE}}_{\text{elec}}}{c^2}\\ & =& \frac{\text{qV}}{c^2}\\ & =& \frac{(It)V}{c^2}\\ & =& \frac{(\text{600 A}\cdot \text{h})(\text{12.0 V})}{(3.00\times {\text{10}}^8{)}^2}\text{.}\end{array}$$

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   Write amperes A as coulombs per second (C/s), and convert hours to seconds.

   $$\begin{array}{lll}\mathrm{\Delta }m& =& \frac{(\text{600 C/s}\cdot \text{h}\left(\frac{\text{3600 s}}{\text{1 h}}\right)(\text{12.0 J/C})}{(3.00\times {\text{10}}^8\text{m/s}{)}^2}\\ & =& \frac{(2.16\times {\text{10}}^6\text{C})(\text{12.0 J/C})}{(3.00\times {\text{10}}^8\text{m/s}{)}^2}\end{array}$$

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   Using the conversion $1\text{kg}\cdot {\text{m}}^2{\text{/s}}^2=1\text{J}$, we can write the mass as

   $\mathrm{\Delta }m=2.88\times {\text{10}}^{-\text{10}}\text{kg}\text{.}$

1. Identify the knowns. $\mathrm{\Delta }m=2.88\times {\text{10}}^{-\text{10}}\text{kg}$; $m=\text{20.0 kg}$
2. Identify the unknown. % change
3. Choose the appropriate equation.
4. Plug the knowns into the equation.
   $$\begin{array}{lll}\text{\% increase}& =& \frac{\mathrm{\Delta }m}{m}\times \text{100\%}\\ & =& \frac{2.88\times {\text{10}}^{-\text{10}}\text{kg}}{\text{20.0 kg}}\times \text{100\%}\\ & =& 1.44\times {\text{10}}^{-9}\text{\%}\text{.}\end{array}$$

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Both the actual increase in mass and the percent increase are very small, since energy is divided by $c^2$, a very large number. We would have to be able to measure the mass of the battery to a precision of a billionth of a percent, or 1 part in ${10}^{11}$, to notice this increase. It is no wonder that the mass variation is not readily observed. In fact, this change in mass is so small that we may question how you could verify it is real. The answer is found in nuclear processes in which the percentage of mass destroyed is large enough to be measured. The mass of the fuel of a nuclear reactor, for example, is measurably smaller when its energy has been used. In that case, stored energy has been released (converted mostly to heat and electricity) and the rest mass has decreased. This is also the case when you use the energy stored in a battery, except that the stored energy is much greater in nuclear processes, making the change in mass measurable in practice as well as in theory.