Re: Energy and Mass

On 11 Feb, 15:31, "Jon Slaughter" <Jon_Slaugh...@xxxxxxxxxxx> wrote:

One more example is say the forming of water

2H2 + O2 = 2H2O

By measuring the masses of 2H2 and O2 invididually one would get X and then
2H20 should way X, but does it? is it even close?- Hide quoted text -


John,

All matter consists of an arrangement of energy.

More directly, all matter IS energy.

This idea was first proposed by Albert Einstein when he introduced the
equation:

E=mc^2

What this equation means is that the total amount of energy, E,
comprising any amount of matter with mass m is equal to the mass
multiplied by the speed of light squared.

To give a random example, the pencil on my desk has a mass of 5g. If I
convert this quantity of mass (0.005kg) into a quantity of energy
using E=mc^2 with an approximate value of 3x10^8m/s for the speed of
light, I get an approximate (but pretty accurate) value for the total
energy of the pencil of:

4.5X10^13 Joules (or 4.5x10^10kJ)

My pencil isn't moving at a very high speed or anything. (At least not
relative to me it isn't).

It's just sitting there on my desk.

Now that's a lot of energy. But I would have difficulty converting any
significant proportion of this energy into any other useful form such
as heat or light. That's because the great majority of that energy
goes into making up the atoms of carbon, hydrogen, oxygen and whatever
other atoms go into making up my pencil, and I have no practical means
of converting these atoms into heat and light.

Having said that, I could convert a tiny portion of this total energy
into small amounts of heat and light by burning the wooden part of the
pencil. And in order for the conservation equation for this combustion
to balance I would have to conclude that the products of the
combustion (including any ash, water vapour, carbondioxide, etc.)
would, in total, have a mass which is slightly less than my pencil
originally had, by an amount equivalent to the energy given off -
according to E=mc^2.

Consider your equation for the combustion of hydrogen:

2H2 + O2 = 2H2O

From the point of view of mass/energy conservation this equation is
not complete. It does not balance. The reaction it describes is an
EXOTHERMIC reaction; that is, a net amount of energy is released and
dissipated to the surroundings. This amount of energy contributed to
the mass of the 2H2 and/or O2 before combustion, but does not
contribute to the mass of the H2O afterwards.

Assuming that the equation represents molar masses, a chemistry
textbook will give the amount of energy derived from this reaction as
572kJ.

This is equivalent to a mass of 572000/c^2, = 6.355x10^-13, an
extremely small mass. Nevertheless the rest mass of the products of
the reaction (just water) must be less than the total of the rest
masses of the reactants (hydrogen and oxygen) by this amount.

Be careful. In my experience some people don't believe this. In fact,
it appears that some "scientists" are convinced that the equation
E=mc^2 has nothing to do with chemical reactions.

They are wrong.

Others may argue that the difference in mass is so small that it can
be ignored. Well, for practical purposes (for example in the design of
chemical processing plant) this may be true. But from the point of
view of understanding the nature of chemical reactions it is a
fundamental truth that the mass of the products of an exothermic
reaction must be less than the mass of the reactants by an amount
which is equivalent to the amount of energy released, according to the
relationship E=mc^2. And, conversely, it is a fundamental truth that
the mass of the products of an endothermic reaction must be greater
than the mass of the reactants by an amount which is equivalent to the
amount of energy absorbed, according to the relationship E=mc^2.

For the combustion of hydrogen, the correct equation of conservation
is:

2H2 + O2 = 2H2O + Ed

Where Ed is the amount of energy released and dissipated to the
surroundings by the reaction. The quantities may be expressed in terms
of energy or mass.

Best regards,

Keith P Walsh

PS, I've just noticed that André Michaud got his answer in before I
posted mine. I think André's response is consistent with mine. I'd be
interested to hear if he agrees with me.














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