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The binding energy of nucleus is calculated as- Mass defect = (Total mass of nucleons-Mass of the nucleus) And after that $E=mc^2$ is used for calculating the binding energy.

Hot water is heavier than cold water because hot water has additional kinetic energy which adds to the mass of the water. So in the similar way the mass of the nucleus that is measured should also include the mass of the binding energy of the nucleus and there shouldn't be any mass defect measured. But why does this not happen?

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The density of water as a function of temperature has a maximum around 4 degrees Celsius, so "hot" water is actually lighter than "cold" water. The mass defect is a very small quantity for most binding energies (as you devide by $c^2$).

The mass defect of the proton and the neutron, each consisting of 3 quarks, is about 99% of their rest mass, the mass defect due to the binding energy for most nuclei is on the order of a few percent, while the mass defect due to electronic binding energy is on the order of $10^{-8}$ (one millionth of a percent). You can imagine that vibrational, rotational and translational energy have even smaller mass defects that are very hard to measure.

Let us - just for fun - assume that we could measure a 0.1% increase in the density of 1L water solely due to an increase in kinetic energy of the water molecules. Ignoring the normal temperature behavior of the density of water, this corresponds to an energy of

$\Delta E=0.1\% \cdot m_{\text{H}_2\text{O}} \cdot n \cdot c^2 \\\qquad=0.001\cdot 18 \text{ g mol}^{-1}\cdot 55.6\text{ mol L}^{-1}\cdot 9\cdot 10^{16} \text{m}^2\text{ s}^{-2}\\\qquad\approx 9\cdot10^{13}\,\text{J L}^{-1}$

Converting this kinetic energy to a temperature we get something on the order of $10^{13}$ $^\circ\!$C!

You can also try to calculate the increase in mass for water at room temperature and will see this is a very tiny fraction of the rest mass.

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