# Thermodynamics of a single spring in gravitational potential

I'm trying to solve this thermodynamics problem. There's a spring of elastic constant $$k$$ attached to a mass $$m$$ in it's equilibrium position

$$$$x_{0}=\frac{mg}{k}$$$$

Suddenly we add a new mass $$M$$ in the end of the spring so the spring oscillates for a while until it settles in a new equilibrium position. This happens at a constant temperature $$T$$ in contact with the enviroment.

$$$$x_f=\frac{(m+M)g}{k}$$$$

The question is: How much heat was released or absorbed by the spring during that process and how much did the entropy of the universe increase?

My problem is that I have two alternative methods for this and they give two results. Furthermore, there are certain subtleties of thermodynamics that are confusing me. Let me explain the two methods:

First Method:

The energy that the system has lost is equal to:

\begin{aligned} \Delta E &= E_f-E_i \\ &= -\frac{g^2(m+M)^2}{k}+\frac{g^2m^2}{k} \end{aligned}

since we know that this energy was lost due to friction it had to be released in the form of heat so we get

$$$$Q_{out}=-\Delta E >0$$$$

If we write down the first law of thermodynamics for this system and we use the fact that the internal energy of an harmonic oscillator is $$U=Nk_bT$$ we get

\begin{aligned} dU&=\delta Q- \delta W \\ 0 &= \delta Q - F dx \\ \rightarrow \delta Q &=kx dx \end{aligned}

where I'm using that the work done by the system is $$\delta W = kxdx$$. Integrating this we get

\begin{aligned} Q&=k \int_{x_i}^{x_f}xdx \\ &=-\Delta E \end{aligned}

which agrees with the prediction that the heat should we equal to the change in energy.

Second Method:

Instead of using the energy of a thermal harmonic oscillator let's use the mechanical energy plus something that might depend on temperature (which is not important since the temperature remains constant):

$$$$dU=kx dx+CdT$$$$

and let's think of the masses as external agents applying a force on the spring. Then the first law takes the form

\begin{aligned} dU&=\delta Q - \delta W \\kx dx &= \delta Q + (m+M)g dx \end{aligned}

Here comes an important doubt: If the system and it's enviroment are doing unequal forces, which one should we take into account when we write the work? In the first method I was using the force that the spring does, here I'm using the force that the masses do. Now we can integrate the heat to get

\begin{aligned} Q&=\int \Big[kx-(m+M)g\Big] dx \\ &= -\frac{g^2M^2}{2k} \end{aligned}

On one hand I like this equation because when we take the limit $$M\rightarrow 0$$ the heat goes straight to zero. Also, I like this method because in the first law it's clear that if we do the expansion cuasistationarily (meaning that the external force is always equal to the Hooke force) then there's no heat exchange. However, I still don't see why the heat should be less than the energy lost in the expansion.

Regarding the entropy question, I'm just not sure how to calculate it.

• is $T$ the temperature of the spring or temperature of the environment? – Bob D Sep 25 '19 at 2:20
• same they are in thermal equilibrium – P. C. Spaniel Sep 25 '19 at 2:25

In my judgment, both methods are incorrect (although your second method is closer to being correct). The form of the first law that you should be using for the universe (masses plus surroundings) is the more general form: $$\Delta E=\Delta U+\Delta(PE)+\Delta(EE)+\Delta(KE)=Q-W$$where $$\Delta U$$ is the change in internal energy of the universe (masses plus surroundings), $$\Delta (PE)$$ is the change in potential energy of the masses, $$\Delta (EE)$$ is the change in stored elastic energy of the spring, and $$\Delta (KE)$$ is the change in kinetic energy of the masses. The universe (masses plus surroundings) is isolated, so this combined system does no work W and receives no heat Q. And the change in kinetic energy of the masses is zero. So the above equation reduces to: $$\Delta U+\Delta(PE)+\Delta(EE)=0$$

The change in stored elastic energy of the spring is $$\Delta (EE)=\frac{1}{2}k(x_f^2-x_i^2)=\frac{g^2}{k}\left(mM+\frac{M^2}{2}\right)$$The change in potential energy of the masses is given by $$\Delta (PE)=-(M+m)g(x_f-x_i)=-\frac{g^2}{k}(M^2+Mm)$$So the change in internal energy of the universe between its initial and final states is given by: $$\Delta U=\frac{M^2g^2}{2k}$$ If we devised an alternative reversible process to bring about this same change in internal energy of the universe, we would have to transfer an amount of heat to the universe (from some unspecified source) equal to $$\Delta U$$. Since the temperature would be virtually unchanged, the entropy change of the universe would be $$\Delta E=\frac{\Delta U}{T}=\frac{M^2g^2}{2kT}$$

• The second method appears to yield a consistent result with your answer, and looks correct to me. So does your answer. It would be interesting if you could specifically point to what is wrong in the proposed second method. – user8736288 Sep 25 '19 at 15:22
• The amount of heat transfer between the spring/mass and its surroundings is way less than this. Plus, mechanistically, all the entropy change is due to viscous dissipation of mechanical energy (resulting from air drag) in the surrounding air boundary layer adjacent to the mass. This is the region that first increases in temperature. A small amount of heat then flows from the air boundary layer to the mass, and the rest is transported by convection and conduction throughout the surrounding air. So the heat flow from the mass to the surroundings is very small, and actually negative. – Chet Miller Sep 25 '19 at 16:36
• Notice that I didn't say anything about heat flow in my answer. The heat flow in the actual process is irrelevant to the actual entropy change, which must be established by devising an alternative reversible process, say, by separating the mass and the surrounding air, and transferring reversible heat to each of them separately, consistent with their changes in internal energy in the actual process (i.e., between their initial and final thermodynamic equilibrium states). – Chet Miller Sep 25 '19 at 16:42
• I understand you "devise" a reversible process in order to obtain $dS= \delta Q_{rev} / T$. If we were to write something from the standpoint of the system "mass+ spring": the work done by the conservative gravity forces should go into $\Delta Ep$ in the lhs. I would still write $\Delta U$ as the variation of strain energy. The quantity computed then corresponds to the sum of the work done by non conservative (drag) forces + some heat exchange $Q$, but this cannot further be specified, without knowing the specifics of the dissipation mechanisms. Would that be correct? Many thanks. – user8736288 Sep 25 '19 at 19:00
• It is much simpler than that. Since entropy is a physical property of the materials comprising the system (i.e, the mass and the air in the room), all you need to do for the reversible process is to take these from their initial thermodynamic equilibrium state to their final thermodynamic equilibrium state by putting them in contact with a constant temperature reservoir at a slightly higher temperature than the initial temperature until the total heat transferred matches the change in their internal energy that they experienced in the irreversible damping. No mechanical effects are necessary – Chet Miller Sep 25 '19 at 21:54

Here is an alternate approach to the same problem.

Call the masses and spring the "system," and call everything else the "surroundings." If we do a force balance on the mass and spring, we obtain: $$(M+m)g-kx-F_D=(M+m)\frac{d^2x}{dt^2}$$ where $$F_D$$ is the (damping) air drag force exerted by the surroundings on the system. Note that $$F_D$$ always has the same sign as the downward velocity of the masses. If we multiply this equation by the velocity dx/dt, and integrate between time equal to zero and infinite time (when the masses have stopped moving), we obtain $$-\Delta (PE)-\Delta (EE)-W_D=0$$So the work done by the surroundings on the system to damp out the motion of the masses is $$W_D=-\Delta (PE)-\Delta (EE)=-\frac{M^2g^2}{2k}$$Therefore, the air drag work done by the system on its surroundings is minus this: $$W=-W_D=\Delta (PE)+\Delta (EE)=\frac{M^2g^2}{2k}\tag{1}$$If we now apply the first law of thermodynamics to the spring-mass system, we obtain:$$\Delta U_{syst}+\Delta (PE)+\Delta (EE)=Q-W\tag{2}$$where here $$\Delta U_{syst}$$ represents the change in internal energy of just the spring-mass system (not like in the previous development where it represented the internal energy change for the universe). If we combine Eqns. 1 and 2, we now obtain: $$\Delta U_{syst}=Q$$where Q represents the typically tiny amount of heat flow from the surroundings to the system. This heat flow is typically tiny because both the system and surroundings experience the same equilibrium temperature change, and the mass times heat capacity of the system is typically much smaller than the mass times heat capacity of the surroundings.

If we now apply the first law of thermodynamics to the surroundings, we obtain: $$\Delta U_{surr}=-Q-W_D=-Q+\frac{M^2g^2}{2k}$$So the internal energy change of the universe is $$\Delta U=\Delta U_{syst}+\Delta U_{surr}=\frac{M^2g^2}{2k}$$