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Expert-verifiedWhat happens to the mass of water in a pot when it cools, assuming no molecules escape or are added? Is this observable in practice? Explain.
As the water in the pot cools, it loses energy, resulting in a drop in water mass.
The ability to perform labour, such as the ability to move an item (with a particular mass) in a specific direction using force. Energy may take many different forms, including electrical, mechanical, chemical, thermal, and nuclear, and it can be converted between them.
The water in the pot releases energy as it cools.
If no molecules depart or are added, the rule of energy-mass equivalence applies.
\({\rm{E = m}}{{\rm{c}}^{\rm{2}}}\), the released energy is expected to be equal to the mass lost from the water.
Since the value of the lost mass is given by \({\rm{m = }}\frac{{\rm{E}}}{{{{\rm{c}}^{\rm{2}}}}}\), since \({\rm{c}}\)is very large \(\left( {{\rm{c = 3}}{\rm{.0 \times 1}}{{\rm{0}}^{\rm{8}}}{\rm{\;m/s}}} \right)\)
then we calculate that the lost mass would be so little that it would be impossible to detect.
After evaluating we get, as the water in the pot cools, it loses energy, resulting in a drop in water mass.
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