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In this comment in a blog kudzu computes the energy of a marshmallow with mass $M=25 grams$ by using $E=mc^2$:

$E=Mc^2 = 2.247\times 10^{+15} Joules$

I may be wrong but this seems like a huge energy for a marshmallow. Can anybody help me put this in context.

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That's why controlled nuclear fusion would dwarf the production of any energy source to date, and why controlled matter-antimatter reactions would dwarf even that. – McGarnagle Oct 21 '12 at 18:03
@Zeynel What is the energy (kCal) in a marshmallow from the back of the packet? (I ask because I don't have a bag of 'mellows around and I am curious to see how close the digestive process is to total annihilation! :P ) – drN Oct 21 '12 at 23:28
lols to all who participated in that comment discussion, but they were flagged and not especially necessary to the question, so I've deleted those comments. – David Z Oct 22 '12 at 14:12

As @lurscher said, it seems like a huge energy because it is. But the important thing to realize is that the mass energy equivalence formula, $E = mc^2$, represents the total internal energy. This means that you would have to (quite literally) annihilate the marshmallow at the particle level in order to obtain this energy.

This is in contrast with the total usable energy that you would obtain from the marshmallow if you were to burn or digest it. In order to calculate this energy, let's refer to the caloric content in one 25 gram marshmallow.

According to Fit Day a 25 gram marshmallow is a 79.5 cal treat. In nutrition, a calorie actually refers to a kiloCalorie and $1 kCal = 4184 J$ so... $$E_{marshmallow} = 79.5 cal = 79.5 kCal = (79.5 kCal)(4184 \frac{J}{kCal}) = 332,628 J$$ As you can see, $332,628J \ll 2.247*10^{15} J$ which explains why when you eat a marshmallow, you are happy, rather than spontaneously combusting with the energy of a nuclear weapon.

To address the following question posed by @Zeynel:

What is total "internal" energy? Is there an "external" energy? Is this the thing people usually call the "rest" energy? What is it? And how do you annihilate something at "particle level"? As opposed to any other level? Can you expand on these question a little bit?

Internal energy is an interesting concept, and it refers to the total energy required to create, or conversely, destroy a system. To visualize this, lets start at the smallest level. Let's say you have a tub of water and you want to know how much internal energy it has.

Well what does it take to make one molecule of water ($H_2O$)? Two hydrogen atoms and one oxygen atom. What does it take to make a Hydrogen atom, that would be one proton and one electron. So we can say that the total energy contained in a hydrogen atom would be: $$E_{hydrogen}=E_{proton}+E_{electron}+E_{joinH}$$ Where $E_{joinH}$ is the energy required to bring the two particles together, $E_{proton}$ is the rest energy of a proton, and $E_{electron} is the rest energy of an electron. Now for the Oxygen atom, we have to bring together 24 particles, 8 neutrons, 8 protons, and 8 electrons. Thus the energy required to make an Oxygen atom is:


Where $E_{joinO}$ is the energy required to join all the particles. Now that we see the energy it takes to make an individual atom, lets see what it takes to make the water molecule. $$E_{H_2O}=2E_{hydrogen}+E_{oxygen}+E_{joinH_2O}$$ Where $E_{hydrogen}$ is the internal energy of a hydrogen atom as calculated above, $E_{oxygen}$ is the internal energy of an Oxygen atom as calculated above, and $E_{joinH_2O}$ is the energy required to bring all three atoms together. Now that we know how much energy is in a single atom, we can figure out how much energy there is in the entire tub of water, which will be given by: $$E_{tub}=N*E_{H_2O}+E_{interactions}$$ Where $N$ is the number of water molecules in the tub, and $E_{interactions}$ is the energy contained in the intermolecular interactions of the water molecules. This $E_{interactions}$ is often a good measure of the usable energy in the system since you would have to take the water molecules apart (a chemical reaction) to obtain any more energy.

Coming back to the Marshmallow, we can see that in order to obtain the entire internal energy of the system, we would have to disassemble every single atom in the marshmallow and extract it's rest energy. The only way to extract the rest energy from a particle is to, as I stated earlier, annihilate the particle. Which is done in a particle accelerator, perhaps you've heard of this one?

LHC Collision Point Figure 1: Large Hadron Collider

I hope this is somewhat clear and accurate, if anyone sees of ways to improve this answer I have opened it up as a community wiki, as I think this is an interesting question which deserves a proper answer.

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So you are saying that the internal energy of the marshmallow is the sum of the internal energies of proton, electron, neutron etc. How does removing "internal energy" from macro to micro scale explains "internal energy"? You are not saying anything more than saying that a bucket of water is the sum of water drops. What am I missing here? – Zeynel Oct 27 '12 at 22:46
@Zeynel: If I throw that bucket of water at you, after I let go the handle, the bucket of water has (extra) kinetic energy that is not apparent from the reference frame of the inside of the bucket. – RedGrittyBrick Oct 28 '12 at 10:31
Yes, the "internal energy" is the sum of the rest energies that make up the system. That is what I am saying because it is really that straightforward. I am not sure what you mean when you say "macro to micro scale." Internal energy is an inherently micro-scale phenomenon. As RedGrittyBrick brings up, there are other types of energy that can be contained in the system, but internal energy is strictly determined by particle level interactions. – Michael Leonard Oct 28 '12 at 22:16
Since "rest energy" and "internal energy" are synonyms, you are saying that "internal energy of the marshmallow is the sum of the internal energies of electron, protons etc." How does this explain the concept of internal energy? Because you are not answering the question what is internal energy? – Zeynel Oct 29 '12 at 12:34

Just to clarify other comments, the food energy content in a 7.5 gram marshmallow is 29.6 calories. Which when converted to joules is:

$$29.6 cal \times 4,184 \dfrac{j}{cal} = 123,846.4 j$$

or for a 25 gram marshmallow:

$$98.6 cal \times 4,184 \dfrac{j}{cal} = 412,821.3 j$$

Which is 10 orders of magnitude less than energy associated with its rest mass.

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