1
$\begingroup$

When you heat an object it's mass increases $E=mc^2$. What I don't understand is that this energy can be converted to mass and still be thermal energy too (object still hot).

When a radioactive isotope radiates energy it's mass decreases (mass is turned in to energy). But when you heat an object thermal energy is turned in to mass but still is also left as termal energy? Hence it seems to me that the total energy/mass in the world increases, which obviously it can not.

I would be very grateful for any help explaining this.

$\endgroup$
  • $\begingroup$ Welcome to physics.SE. When you type math, you can put dollar signs before and after it so that it gets rendered properly. You may want to try that now by editing your answer. $\endgroup$ – Ben Crowell Feb 18 at 16:05
1
$\begingroup$

But when you heat an object thermal energy is turned in to mass but still is also left as termal energy?

This is a misunderstanding, a very common one, of Einstein's insight. What Einstein proposed is actually this: when internal energy of a body increases by $\Delta E$, its effective mass (measured by accelerating it, or by weighing it) also increases, by amount $\Delta E/c^2$ (similarly if energy of the body decreases). It does not matter what kind of process is changing the internal energy, the body could be accepting some energy due to radiation from remote source, or accepting work from some nearby body via contact mechanical forces.

There is no conversion of energy into mass in physics, because that would mean energy gets lost. There is also no conversion of mass into energy in physics, for the same reason. No energy ever gets created or lost in known processes; in physics, energy can only transform from one kind to another.

If a body accepts heat from the environment (whether mechanical, electromagnetic or other), this energy transforms into internal energy. Einstein realized that no matter the form the energy comes from, if it ends up in the body, then its mass changes as a manifestation of this new addition of energy.

Sometimes people talk about conversion of mass into energy, for example when talking about radioactive decay processes. This is strictly speaking incorrect for the above reasons, but common due to fact that the energy in the final form is more visible or useful (radiation, heat or electrical energy).

But the energy was always there even before the decay, it was just in the form of binding energy of nuclei, and then the decay changed this energy into the more visible form.

$\endgroup$
1
$\begingroup$

There are two common ways that people talk about $E=mc^2$. They talk about conversion between mass and energy and equivalence of mass and energy. Both can be correct as casual descriptions, but really in a careful description of the topic you end up not using either word. Here's an outline of the logical basis for all this:

  • The mass of a system is defined as $m^2=E^2-p^2$ (in units with $c=1$).

  • Mass is the same in all frames of reference. (This is different from the customary presentation from the 1950's involving "relativistic mass.")

  • The quantity $E$ is interpreted as the "mass-energy" of a system, because it's nonzero for an object at rest ($p=0$), but also increases if the object is moving.

  • Mass is not additive.

  • The mass-energy $E$ and momentum $p$ are both conserved, but mass is not.

When a radioactive isotope radiates energy it's mass decreases (mass is turned in to energy).

Let's say a nucleus emits an alpha particle. The nucleus has some mass $m$ and mass-energy $E$ before the decay. In the rest frame, $p=0$, so $m=E$ (i.e.,$E=mc^2$ in units with $c\ne1$). After the decay, we have an alpha particle and the remaining daughter nucleus. The total $E$ and $p$ are still the same, because they're conserved. Therefore by the definition of mass, the mass of the whole system is the same as before. So we haven't really converted mass to energy. If this decay process happened in a solid, then the alpha and the recoiling nucleus will now rapidly slow down, and their kinetic energy will go into heating the solid. The mass of the entire solid is still the same (because $E$ and $p$ are conserved). However, the sum of the masses of the particles has gone down, and this reduction is equal to the change in thermal energy.

But when you heat an object thermal energy is turned in to mass but still is also left as termal energy? Hence it seems to me that the total energy/mass in the world increases, which obviously it can not.

When you heat the object, you're transporting mass-energy into it from the outside. If the object's momentum doesn't change, then the increase in mass-energy means that the system's mass has also increased ($m^2=E^2-p^2$). To describe the outcome of the math, we usually say that the increase in energy is equivalent to an increase in mass. But it doesn't really matter which words you use to describe the outcome of the math.

$\endgroup$
0
$\begingroup$

$E=mc^2$ refers to one object moving with velocity v, it is the relativistic mass and would be very complicated to use it at the atomic and molecular level.

relmass

In terms of momentum, the relativistic mass is

ener

At the atomic level one uses four vectors , and the atoms and molecules vibrating in solids all have individual four vectors describing them, and the "length" of the energy momentum four vector is the mass of each atom and molecule, $m_0$. The kinetic energy part is the $(pc)^2$ in the formula above.

To get the mass of the solid one has to add vectorially the individual four vectors. The fact that the object has a classical temperature that characterizes it as "hot" will mean that the addition of the four vectors will give a larger "mass" for the solid, than the addition of the individual masses at rest, but this does not mean that that the kinetic part of the energy is absorbed anywhere. It is there and defines the temperature.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.