Mass-Energy Equivalence (E = mc²)

Multiply the mass in kilograms by the square of the speed of light: $c^{2} = (\frac{3\times{}10^{8} m}{s})^{2} =\frac{ 9\times{}10^{16} m^{2}}{s^{2}}$. Example: one gram of matter (0.001 kg) corresponds to $E = 0.001 \times{} 9\times{}10^{16} = 9\times{}10^{13} J$, equal to 25 GWh, the annual electricity consumption of about 7,000 households. The most common mistake is forgetting the square and multiplying only by $3\times{}10^{8}$, missing a factor of 300 million. Conversely, $m =\frac{ E}{c^{2}}$ gives the mass change for an energy release. In nuclear and particle physics one often converts to electron volts: $1 eV = 1.602\times{}10^{-19} J$.

If you weigh the building blocks of an atomic nucleus individually and compare with the finished nucleus, mass is missing: the bound nucleus is lighter than the sum of its protons and neutrons. This difference is the mass defect, and it was released as binding energy during assembly, according to $\Delta{}E = \Delta{}m\cdot{}c^{2}$. In the fission of uranium-235 the defect is about 0.1% of the initial mass, in the fusion of hydrogen to helium as much as 0.7%, which is why fusion yields more per kilogram of fuel. Sample calculation: a defect of $0.2 u \approx{} 3.32\times{}10^{-28} kg$ corresponds to $E = 3.32\times{}10^{-28} \times{} 9\times{}10^{16} \approx{} 3\times{}10^{-11} J \approx{} 186 MeV$, the typical scale of a single fission event.

Yes, and massively so: the Sun radiates a power of about $3.8\times{}10^{26} W$. By $\Delta{}m =\frac{ E}{c^{2}}$ it loses $\Delta{}m =\frac{ 3.8\times{}10^{26} }{ 9\times{}10^{16}} \approx{} 4.2\times{}10^{9} kg$ per second, a good four million tonnes every second. The energy comes from nuclear fusion in the solar core: four hydrogen nuclei fuse in several steps into one helium nucleus that is 0.7% lighter than the initial particles; exactly this mass defect is released as radiation. Despite the huge numbers the loss is insignificant for the Sun: at a mass of $2\times{}10^{30} kg$ it loses only about 0.007% of its mass to radiation even in a billion years.

In principle yes, in practice almost never. Complete conversion happens only in annihilation: when a particle meets its antiparticle, for instance an electron meets a positron, the entire rest mass becomes radiation energy, used in PET diagnostics, where the two 511 keV photons reveal the annihilation site. Fission and fusion, by contrast, convert only the tiny mass defect (0.1 to 0.7%), and chemical reactions a trillion times less still. The rest of the mass persists as particles because conservation laws (such as baryon number) forbid the complete conversion of ordinary matter. $E = mc^{2}$ is thus a statement of equivalence; it does not say the conversion is technically feasible at will.

Yes. Every energy change of a system also changes its mass by $\Delta{}m =\frac{ \Delta{}E}{c^{2}}$, only the effect is immeasurably small outside nuclear physics. Combustion example: one kilogram of petrol releases about $4.3\times{}10^{7} J$; the products are thus lighter by $\Delta{}m =\frac{ 4.3\times{}10^{7}}{9\times{}10^{16}} \approx{} 4.8\times{}10^{-10} kg$, half a microgram, far below any balance precision. A heated body or a stretched spring is likewise minimally heavier than in its low-energy state. That chemists may work with Lavoisier conservation of mass is due only to the tininess of the effect. Only in nuclear processes, where millions of times more energy is converted per particle, does the mass difference become measurable and technically usable.