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As I understand when a rubber band is stretched adiabatically (which I am assuming means no change in entropy and so no heat flow from surroundings) its polymers naturally are straightened and thus their entropy would be reduced, so in order to keep entropy constant, the rubber band's temperature must increase. So how exactly does the work done on the rubber band become manifested as randomised thermal energy?

My initial guess is that the polymers 'lining up better' equates to them being in a lower potential energy state and so kinetic energy (randomised into thermal energy) is released in this process but I am not sure. Or possibly the work done just transfers kinetic energy directly to the particles in the rubber?

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Rubber is pretty damn complex!

On a macroscopic level, you do work to the rubber and it heats. When the rubber does work back, it cools. Just like when you compress a gas and then the gas expands.

The microscopic situation with the gas is simple (assuming you just play with a piston in a cylinder of gas) : the molecules bounce from a moving wall and either loose some energy (when the piston goes away) or gain energy (when the piston compresses the gas).

Rubber has long entangled and mostly spiral molecules and their thermal motion keeps them in some equilibrium shape. When you stretch the rubber, you tend to unwind some molecular spirals and impose more tension over the chemical bonds. This increases their vibrational energy and this energy is quickly "thermalized" over the whole material.

A (very rough) analogy: a rope, hanged more or less horizontally between two points. You quickly move one of these points to straighten the rope and it starts to oscilate.

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  • $\begingroup$ The Feynman video referenced in another comment mentions the polymer chains essentially sort of 'knocking' the surroundings smaller particles too when they straighten out, is that a different complimentary effect? $\endgroup$
    – Alex Gower
    Oct 5 '20 at 14:03
  • $\begingroup$ Although I guess your method would also cause this indirectly too $\endgroup$
    – Alex Gower
    Oct 5 '20 at 14:04
  • $\begingroup$ @AlexGower hey, I am not Feynman at any rate, but both his and my own version are a gross oversimplifications on what happens in the rubber. I (but I am not Feynman, remember) would use these surrounding particles in order to illustrate the viscous loses in the rubber. $\endgroup$
    – fraxinus
    Oct 5 '20 at 14:13
  • $\begingroup$ Fair enough! As a final point, Feynman explains the force causing the rubber to contract with increased temperature as the smaller surrounding atoms to the polymer chain 'bombarding it more and causing more kinks in it'. Do you have any alternative simplified explanation to that too? $\endgroup$
    – Alex Gower
    Oct 5 '20 at 14:34
  • $\begingroup$ The chain's own atoms can do the same, as well as other chains' atoms. $\endgroup$
    – fraxinus
    Oct 6 '20 at 5:53
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When the rubber band is stretched it heats up because of the viscous friction of the chain molecules slides on each other. On the other hand, it is not because of adiabatic expansion because there is no expansion in the rubber band and it is an incompressible material (it deforms on stretching but volume change is 0).

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  • $\begingroup$ But what about for a theroetical adiabatic expansion? $\endgroup$
    – Alex Gower
    Oct 5 '20 at 13:44
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    $\begingroup$ Viscous friction is not your friend when you have to explain the cooling when the rubber is relaxed. $\endgroup$
    – fraxinus
    Oct 5 '20 at 13:44
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    $\begingroup$ I know yt vids are not up to par, but it is Feynman, after all, and the title is "rubber bands": youtube.com/watch?v=XRxAn2DRzgI $\endgroup$
    – JEB
    Oct 5 '20 at 13:55
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    $\begingroup$ It's definitely not viscous friction because, when you release the tension, it adiabatically cools back down.. $\endgroup$ Oct 5 '20 at 14:00

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