# Energy Conservation & E=M*C2

1. Mar 18, 2010

### suhagsindur

This is question about equation E=m*c2. Why this equation does not required any condition of matter?
Case 1: I take two mole (4 Kg)H2 & one mole(32 Kg)O2 & make water & get energy.
2H2 (4 Kg Hydrogen)+O2 (32 Kg Oxygen)=2H2O(36 Kg of water)+ 572 KJ Energy
Now I convert this water into energy by equation E=m*c2.
Total energy I get is QTotal= (36*c2) + 572 KJ Energy.

»»» Now instead of I convert hydrogen & oxygen into water, I directly convert both gases into energy so total Energy I getting is,
QTotal= m*c2 =(36*c2) Only.
This energy is lesser as compared to first one by amount of “572 KJ Energy”.

Case 2: I have one Kg of boiled water at 100 degree centigrade.
Now I extract 418.7 KJ energy from the water & make it of zero degrees centigrade.
Water at 100 C - 418.7 KJ Energy =Water at 0 C.
Now I convert this zero degree water into energy,
Total energy I getting is,
QTotal=1*c2+418.7 KJ energy

»»» Now I directly convert boiling water into energy,
I getting energy is
QTotal=m*c2=1*c2.
This energy is lesser as compared to first one by amount of “418.7 KJ Energy”.

Why it is occurs? Why energy librated is different for the same initial & final condition of mass?

2. Mar 18, 2010

### Staff: Mentor

Actually, you will get only 35.9999999999936 kg of water. Everything else then works out fine. This difference in mass between the products and the reactants is called the mass deficit and is closely related to the binding energy. See: http://en.wikipedia.org/wiki/Binding_energy. The mass deficit occurs for all interactions, although as you can see above it is usually pretty small and only noticeable for very strong interactions, like those involving the strong nuclear force.

3. Mar 18, 2010

### suhagsindur

This means that boiling water possesses higher mass then cold water by extremely small amount which is neglected for practical purpose.