Re: Mass and Energy

From: Keith P Walsh (keith.p.walsh_at_btinternet.com)
Date: 09/03/04 

Date: Fri, 3 Sep 2004 11:23:52 +0000 (UTC)

On 1 Sep 2004 09:47:47 -0700, hhc314@yahoo.com (Harry Conover) wrote:

>Keith P Walsh <keith.p.walsh@btinternet.com> wrote in message news:<uvi4j05bjh472rh47nv2dkkln9md39j2os@4ax.com>...
>>
>> "If you burn hydrogen gas in oxygen gas you create water, but you also
>> produce a lot of light and heat. Energy, or mass, has been expelled
>> from the material. So even before you make any measurements you can be
>> sure, by Einstein's rule (I.e. E=mc^2), that the water must be lighter
>> in weight than the hydrogen and oxygen that made it."
>>
>> I would suggest that anyone who is unable to recognise the veracity of
>> this statement doesn't really understand what the equation E=mc^2
>> means.
>>
>> Would anyone disagree?
>>
>> Keith P Walsh
>
>Sure Keith, I would disagree.
>
>First, whe you burn hydrogen gas in oxygen, the flame is essentially
>invisible.
>
>Second, the heat from a hydrogen/oxygen torch has a high temperature,
>but produces very little heat. In pop science demonstrations, this
>allow you to run your hand through the flame and then turn around and
>melt fused quartz.
>
>This impresses only the tourists in the crowd.
>
>Of course this is simply a demonstration of simple chemistry, and
>since the products of combustion of hydrogen and oxygen exactly equate
>to the mass of the inputs, E=MC^2 is not demonstrated.
>
>I believe you need to learn to distingush nuclear reaction from simple
>chemical reactions, as does Niger Calder (but this is only my
>opinion).
>
> Harry C.

Harry,

Thank you for your message.

I think that you have fallen foul of a common misconception, which is
that since all of the atoms comprising the molecules in the reactants
of this reaction can be identified and accounted for in the
composition of the molecules of the products, then no mass can have
been lost.

But the behavior of matter at this scale does not obey any such rule.

The amount of energy required by the various component atoms in order
for them to exist in their "reactant state" is greater than the amount
of energy that they need to exist in their "product state". The
difference in energy between these two states is equal to the amount
of energy given off in the reaction. (The combustion of hydrogen in
oxygen is an exothermic reaction.)

I'm sure that you agree with this perfectly well so far.

However, in order to appreciate the true meaning of the equation
E=mc^2, it is necessary to recognise that the energy given off in the
reaction must have contributed to the mass of the reactants before the
reaction took place. And that this amount of energy must be accounted
for by a reduction in the total rest mass of the products as compared
to that of the reactants.

In other words, if you measured the (rest) mass of the reactants
(hydrogen and oxygen) the value you obtain would include a proportion
which is equal to the energy given off in the reaction, according to
the relationship E=mc^2.

And if you then measure the (rest) mass of the products (water), this
amount of mass is missing because it has been released and dissipated
to the surroundings as (different forms of) "energy".

In short, it is not possible to derive useful energy from an
exothermal chemical reaction (or any other physical process) without
there being some sacrifice of (rest) mass from the initial state.

An important point to note is that the misconception which I described
initially carries the implication that the masses of molecular systems
are defined absolutely by the numbers of the different species of
atoms from which they are comprised.

This is not so.

The total mass of the system is also dependent upon the particular
arrangement into which the atoms are combined.

For example, two moles of H2O weigh less than the combined weight of
two moles of H2 and one mole of O2.

This is expressed by the equation:

2H2 + O2 = 2H2O + Energy

Remember that the "energy" in this equation is energy which
contributed to the rest mass of the reactants, but does not contribute
to the rest mass of the products.

And Nigel Calder and I are not the only people who understand this.

In all sincerity, and without any intention of gratuitous abuse, I
believe that the lack of understanding displayed by your contribution
to this discussion is typical of the widespread ignorance which
exists, even in "scientific" circles, regarding these matters, and
that this ignorance represents a serious obstacle to the advancement
of a broader understanding of nature.

Keith P Walsh

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