According to the $E=mc^2$ equation, will an object whose thermal energy (temperature) rises also weigh more? And by the same token, will the mass of an object decrease as its temperature approaches the absolute zero?
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2possible duplicate of Is the molecule of hot water heavier than that of cold water? – TZDZ Dec 12 '14 at 08:01
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@TZDZ the linked question discusses the mass of a single molecule, where the movement is just kinetic energy. This question discusses thermal energy in a larger object in which it is generally not viewed as kinetic energy but internal thermal energy. – Richard Dec 12 '14 at 09:11
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1possible duplicate of Does a cooling mass lose mass as it radiates? – pentane Dec 12 '14 at 10:05
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All internal energy such as thermal, rotational, and internal potential energy contributes to the rest mass of an object. In fact, the vast majority of the mass of an atom is due to the internal energy between quarks that make up the nucleus rather than the rest mass of the quarks themselves.
So, yes, a hot object has greater rest mass and would weigh more when measured, if a scale were sensitive enough. Also, an object near absolute zero would also weigh less though, again, I don't know whether the difference is measurable in practical terms.
Richard
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1Speaking of weight... shouldn't you account for the increased buoyancy from thermal expansion? – John Dvorak Dec 12 '14 at 12:58
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@JanDvorak - :-) I would choose to weigh in a vacuum with all the equipment close to absolute zero, or possibly with a balance that would be attracted by its gravitational force thus negating buoyancy effects and not requiring the object to move a scale. Could even measure the deflection of a laser passing close to the object, though the effect would be smaller than plank length. But if we are going for impossible weighing tasks - might as well go all the way! – Richard Dec 12 '14 at 14:43
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Yes, when we heat an object its mass increases. The complete equation is $$E^2=(pc)^2 + (mc^2)^2$$
And from this equation if a system has zero momentum (p=0), then it has energy $E=mc^2$
When you heat an object the molecules or atoms begin to vibrate, rotate with more kinetic energy. But this doesnt increase the momentum of the system (the object is a system of many particles). Though the kinetic energy of the particles that make the system increases, the system is at rest (when you heat the object the object doesn't start moving). Hence, from the equation I wrote above the energy you provide get added as mass of object.
Another example:
A perfectly reflecting wall massless box has mass when photons are randomly moving inside the box.
Kyle Kanos
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Paul
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I think you must always take the momentum equation with the Lorentz factor $\gamma$ in relativistic calculations instead of the purely classical expression for it. – Gaurav Dec 12 '14 at 12:35
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You can also use the equation i used above... why i got the downvote please explain.. – Paul Dec 12 '14 at 12:41
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The correct equation is $E=(pc)^2+m^2 c^4$ instead of $E=(pc)^2+m^2 c^2$. – Gaurav Dec 12 '14 at 13:15
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Both equation E=mc^2 and the equation i wrote above both are correct... but first one doesnt tell you the whole story.. – Paul Dec 12 '14 at 13:36