Fermi estimate: one gram of matter

Problem

Fermi estimate: one gram of matter. How much energy, in familiar units (J, kWh), does the rest energy of one gram of ordinary matter contain? Compare with a family’s annual consumption.

Solution

Rest energy. With $m = 1\ \text{g} = 10^{-3}$ kg and $c^2 \approx 9\cdot 10^{16}\ \text{m}^2/\text{s}^2$:

$E_0 = m c^2 = (10^{-3})(9\cdot 10^{16}) = 9\cdot 10^{13}\ \text{J}.$

In kilowatt-hours. Recalling that $1\ \text{kWh} = 3{,}6\cdot 10^6$ J:

$E_0 = \frac{9\cdot 10^{13}}{3{,}6\cdot 10^6}\ \text{kWh} \approx 2{,}5\cdot 10^7\ \text{kWh}.$

$\ev{E_0 \approx 9\cdot 10^{13}\ \text{J} \approx 2{,}5\cdot 10^7\ \text{kWh}}$

Comparison with a family. A family consumes about $3000$ kWh of electricity a year. So

$\frac{2{,}5\cdot 10^7}{3\cdot 10^3} \approx 8\cdot 10^3\ \text{years}.$

The rest energy of a single gram of matter, if fully converted, would cover a family’s electricity needs for nearly ten thousand years. This is the reason for the immense power of nuclear processes, where, however, only a tiny fraction of the mass is converted into energy.

Links

Topics: Special relativity Concepts: Mass-energy equivalence Skills: Fermi estimation