# Entropy exchange of a free fall

I have a problem in which the tell me that you drop a bag of 50 kg of sand from 10 meters high, and you have to caltulate the entropy difference of the sand, asuming that the speific heat of the sand is so high that its temperature (298K) doesn't change. My result is 16.5 J/K, and my problem is with the sign. The book doensn't give an explanation but just an answer and it's positive, but I'm thinking that in the impact the heat must flow out of the sand, as it's loosing its kinetic energy, so the entropy would decrease, and the surroundings entropy would increase; what am I missing, or what is really going on here?

-
It's a self-delusion to force oneself into thinking that the temperature doesn't change - and "therefore" the entropy carried by the temperature of the sand doesn't change. In reality, if the specific heat of the sand is high, it just means that it has many degrees of freedom that may carry entropy, so a small temperature change corresponds to a high entropy change. In reality, the entropy change won't depend on the heat capacity much and the entropy change of the bag can never be neglected. –  Luboš Motl Feb 4 '13 at 17:59
I don't understand your last line (entropy change of the bag can never be neglected) I'm not doing that, I'm calculating it, I'm not supposing the entropy doesn't change because I'm calculating it. The problem explicitly says to suppose the temperature doesn't change from the impact, just to make calculations easier, what doesn't have anything to do with my doubt about the sign of the entropy, as supposing a change of the temperature to estimate that sign generates the same problem I'm in. –  MyUserIsThis Feb 4 '13 at 18:30

In reality the sandbag is going transfer energy via drag to air, work on the ground during impact (by denting the ground), and heat transfer to the ground etc., but I am going to assume your textbook wants the problem to be simple. So, let us consider that the sand bag falls to the ground and stops, has no drag in the air and energy transfer to the ground.

So coming to a halt with no energy transfer to air or ground, all the potential energy must become internal energy of the sand. Consider the sand bag to be a homogeneous thermodynamic system, its internal state changes as per the Gibbs equation \begin{align*} TdS = dU+PdV-\mu dN \end{align*} No change in composition or volume ($dV=0, dN=0$). That leaves, \begin{align*} \int_{initial}^{final}dS = \int_{initial}^{final}\frac{dU}{T} \end{align*} Now sand is not ideal gas! So technically changes in internal energy can occur with no change in temperature. But regardless of that fact, one can also say that the specific heat is high the temp change is negligible and pull the temperature out of the integral, \begin{align*} \Delta S = \frac{\Delta U}{T} \end{align*} All this entropy change will be positive (entropy generated by the internal dissipation of potential energy to internal energy in the molecules comprising sand). I am guessing this should work out numerically to your answer. If not maybe we are missing something, that I can't see.

-
Thanks, with this I "intuitively" saw the thing ""all the potential energy must become internal energy of the sand"". What my textbook wanted me to do is, as we're ignoring every drag or energy transferred to the ground, to consider all the potential energy as the heat transmited to the sand bag and just do $dS=dQ/T\Rightarrow \Delta S=mgh/T$ to calculate the entropy difference. I understand that this is no realistic at all. –  MyUserIsThis Feb 4 '13 at 23:33
Yes, a lot of ppl (and sadly even texts) confuse internal energy and heat transfer. Heat is energy transferred out of a system in a diffusive process, but energy is stored in matter as internal energy (and external energy, e.g., potential forms) not heat. Glad I could help –  Sankaran Feb 4 '13 at 23:46
A good answer, however I believe it is incomplete. The final formula is true for positive or negative values of entropy change - it is a valid statement of the first law of thermodynamics - but you also need to use the second law to prove that the change in entropy (and internal energy) must be positive. Also, the product g*h must be negative for a consistent coordinate system. My answer was incomplete too, so I will extend it below. –  Mark Rovetta Feb 5 '13 at 5:44
If you say so, but the second law is trivial in this case: $dS=\delta S_{gen}\geq0$ and the change being positive is evident. –  Sankaran Feb 5 '13 at 6:18

If the bag of sand falls and impacts the ground while doing no work and exchanging no heat with the surroundings, the system is thermodynamically isolated. The Helmholtz free energy of this system is:

\begin{align*} A = U-TS \end{align*}

Taking the differential

\begin{align*} dA = dU-TdS-SdT \end{align*}

It's a condition of the problem that the temperature remains constant and so $dT=0$. For any spontaneous change in the state of the system, we know also that $dA\leq0$ must be true after impact. Then substituting these

\begin{align*} 0\geq dU-TdS \end{align*}

After rearranging and integrating

\begin{align*} \Delta S\geq \frac{\Delta U}{T} \end{align*}

The change in the internal energy is due both to the loss of gravitational potential energy, and possibly any added changes due to the creation of crystal defects or fractures in the sand grains.

\begin{align*} \Delta U = mgh+\Delta U_{fractures} \end{align*}

For example, $\Delta U_{fractures}$ would be non-zero if fractures released residual strain energy that was locked inside of the lattices of crystalline sand grains. If no residual strain energy was released, then 16.5 J/K would be the minimum possible entropy change (driven only by the gravitational potential energy change.)

There are a number of ways additional entropy could manifest itself in the bag of sand without increasing the temperature. The entropy would then be latent rather than sensible.

For example, 16.5 J/K is a very small change in the entropy of a 50 kg bag of silica sand. The bulk specific entropy (J/(gK)) of a silica crystal is less than the specific entropy of the surface of a silica crystal (J/Kcm^2). If the impact pulverizes the sand into dust of a smaller average grain-size, this would increase the surface area of the sand, and slightly increase the entropy of the sand in the bag.

Another possibility, if this is a bag of wet sand, is the impact could cause some silica to dissolve into the pore fluid. The entropy of mixing of a solute into solution is always positive.

-