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.
[Physics] Potential energy in stretched vs unstretched rubber bands
elasticityentropypotential energy
Related Solutions
A simple example of a constant external force being applied to a spring-mass system is the force of attraction $mg$ on a mass $m$ is a gravitational field of strength $g$.
Release the mass at the end of an unstretched spring, then when the spring has been stretched by an amount $x$ the work done by the external force (gravitational attraction) is $mgx$.
The elastic potential energy stored in the spring is $\frac 12 kx^2$ where $k$ is the spring constant.
The difference between these two quantities, $mgx - \frac 12 kx^2$, is the increase in kinetic energy of the mass.
Eventually the constant external force will be smaller than the force exerted by the spring and the mass will slow down and finally stop.
This will happen when $mgx_{\rm stop} = \frac 12 k x_{\rm stop}^2$
At this position all the work done by the external force is stored as elastic potential energy.
This example is no more than the oscillation of a mass at the end of a spring but noting that $x$ is the total extension of the spring and not the extension of the spring from the static equilibrium position.
It sounds like you're describing two separate effects:
Elastomers can have a negative thermal expansion coefficient, i.e., $$\alpha=\frac{1}{L}\left(\frac{\partial L}{\partial T}\right)_F<0,$$ because the higher-entropy state (corresponding to a higher temperature) is one in which the polymer chains move around more, which has the consequence of shortening their end-to-end length. When characterizing this material property, the force $F$ is kept constant.
Most materials (including elastomers) generally get more compliant when they are heated; thus, their stiffness $$k=\left(\frac{\partial F}{\partial L}\right)_T,$$ which is characterized at a constant temperature, decreases with increasing temperature:
In polymers, the origin is the melting of weak bonds between the chains. However, exceptions can exist; for example, the stiffness of a purely entropic system increases with temperature, so once melting is complete, we might see an increase in the elastic modulus with heating:
(Hosford, Mechanical Behavior of Materials)
Therefore, if you hang a weight on a rubber band (to achieve a change in length of $\Delta L$) and heat it up, the weight will probably rise because of (1). If you then remove the weight at the same higher temperature, the rubber band may contract by more or less than $\Delta L$ as mediated by the mechanisms discussed in (2). If you then cool the system down to the original temperature, the rubber band will return to its original length, $\Delta L$ above the original position of the weight, if it is not already there and if the entire loading time was short enough that irreversible flow was negligible.
Put another way, since you're asking about elasticity in the context of a hot and a cold rubber band loaded by the same weight, I should emphasize that one can't directly measure a system's stiffness by keeping the force constant and observing the displacement when changing other things. One measures the stiffness by changing the force while watching the displacement (or vice versa) while keeping other things constant.
Best Answer
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.