I have been searching for a definitive answer to this question but I have been unable to find it. I understand that when you stretch a rubber band, it gives off heat - an exothermic process - but at the same time you are adding kinetic energy to the rubber band to make this happen, and its entropy decreases. Contrarily, when you release the rubber band, it absorbs heat, applies kinetic energy to the rubber band, and increases its entropy. My question is, does the rubber band in the stretched state have higher or lower potential energy than the rubber band in the relaxed state? Thank you.

  • What do you think? If the stretched band has less stored energy than in the relaxed state, this means that you have to do work (expend effort) to unstretch the rubber band. Does that agree with your observations? – sammy gerbil Apr 18 '17 at 23:39
  • I've read that the energy absorbed from the environment is doing the work here, that is, if the environment was cold, the rubber band would not contract. According to the wikipedia article on rubber elasticity, the rubber band gains no potential energy as a result of stretching it. – John Panettiere Apr 19 '17 at 0:03
  • I think the key question is whether the heat released when stretching the rubber band exceeds the kinetic energy invested in stretching it, but I'm at the limit of my ability to answer this question. – John Panettiere Apr 19 '17 at 0:05
up vote 3 down vote accepted

Ideal elastomers do not gain potential energy when stretched (i.e., their stiffness is entirely entropic), but real elastomers do.

A little background: The force $f$ needed to stretch a strip of solid material slowly is $$f=\left(\frac{\partial U}{\partial l}\right)_T-T\left(\frac{\partial S}{\partial l}\right)_T\tag{1}$$ where $U$ is the internal energy, $l$ is the length, $T$ is the temperature, and $S$ is the entropy.

(You can get this from noting that you can add energy to the strip by heating it, pressurizing it, or stretching it, among other ways. We can write this in differential form as $$dU=T\,dS-p\,dV+f\,dl\tag{1a}$$ where $V$ is the volume. The volume of condensed matter doesn't change much with pressure, so $dV\approx 0$. Since we're operating slowly, let's assume constant temperature as we take the derivative with respect to $l$ to obtain Eq. (1).)

Materials such as metals, ceramics, and strongly crosslinked polymers obtain their stiffness from the $(\partial U/\partial l)_T$ term, as their entropy doesn't increase by much upon an elastic strain of, say, 0.1%, which is nearly all that their bonds can sustain. Elastomers are different, as your research up to this point has indicated. Without crosslinks, there is little preventing their long, kinked molecules from extending; thus, they gain their stiffness almost entirely from the $-T(\partial S/\partial l)_T$ term. (Note that the Wikipedia article you mentioned only states that fully entropic stiffness is a "good approximation").

Let's continue the derivation, using a Maxwell relation to turn $-(\partial S/\partial l)_T$ into $(\partial f/\partial T)_l$: $$-\left(\frac{\partial S}{\partial l}\right)_T=\left(\frac{\partial G}{\partial T\,\partial l}\right)=\left(\frac{\partial G}{\partial l\,\partial T}\right)=\left(\frac{\partial f}{\partial T}\right)_l\tag{2}$$ Thus,$$f=\left(\frac{\partial U}{\partial l}\right)_T+T\left(\frac{\partial f}{\partial T}\right)_l\tag{3}$$ Let's experimentally employ the second term to measure the relative influence of the enthalpic and entropy stiffnesses. We stretch an elastomer to a given length and measure how the resisting force changes for small changes in temperature. If we do that across a large range of temperatures, then for the ideal elastomer we obtain a line that intersects $f=0$ at $T=0$. So just as you noted, the ideal elastomer requires finite temperature to pull back against a tensile force.

For this reason, ideal elastomers have been compared to ideal gases. Just as an ideal gas pushes back against compression because of strongly temperature-dependent entropic effects, the ideal elastomer pulls back against elongation in an analogous way.

Again, however, real elastomers exhibit some enthalpic stiffness because of entanglement and other interactions. Chanda's Introduction to Polymer Science and Chemistry reports that for polybutadiene, for example, the enthalpic term $(\partial U/\partial l)_T$ contributes about 10-20% to the stiffness. Thus, the energy that goes into stretching a real elastomer is only partially converted into a temperature increase.

  • Thank you for your prompt and thorough reply. I've marked your answer as the correct answer, I apparently do not have enough reputation to up-vote it yet though. – John Panettiere Apr 19 '17 at 11:41

Your Answer


By clicking "Post Your Answer", you acknowledge that you have read our updated terms of service, privacy policy and cookie policy, and that your continued use of the website is subject to these policies.

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