Will there be a mass defect of a water 1kg boiling into vapour? When the vapour is turned back into water, do you think there will be a mass defect or we get the same mass?
Asked
Active
Viewed 39 times
1 Answers
4
The “mass defect” is the difference in mass between a bound system and its unbound pieces. It is useful in nuclear physics because the binding energies can be quite large: the fissioning of uranium releases about one part per thousand of its mass as energy.
In atomic physics, binding energies are typically small enough that they can be ignored when computing the mass of a system. For instance, if we say that the hydrogen atom has a mass of $\rm 1\ u\approx1\ GeV$ (in energy units), we are already being too sloppy to notice the half-MeV mass of the electron; the part-per-billion correction due to the binding energy, 13 eV, is way, way smaller.
The “binding energy” in the transition between liquid and vapor is called the “latent heat” of the phase change. A typical scale is “kilojoules per mole,” or
$$ \frac{\rm 10^3\ J}{\rm mol} \times \frac{\rm 1\ eV}{\rm 1.6\times 10^{-19}\ J} \cdot \frac{\rm 1\ mol}{\rm 6\times10^{23}\ molecule} \sim 10^{-2}\rm\ eV/molecule $$
Compare with thermal energies at room temperature, $kT\approx 25\ \text{milli-eV}$. This is a part-per-trillion correction to the mass.
rob
- 89,569