Entropy Change - which formula to use?

Question

I am looking at a system of two solids A and B, that is isolated.

A is at a higher temperature; thus, we have heat transfer from A to B.

If I wished to calculate the entropy change: $$\Delta S_{tot} = \Delta S_A +\Delta S_B$$ where $$\Delta S_{AorB} = m_{AorB} \cdot c_v \ln(T_2/T_1)$$ would this formula apply ? This formula is the result of integrating $dS = Q/T = c_vm_A dT/T$ for a constant volume process. However,this formula is for an internally reversible process (where no irreversibilities occur within boundaries), but for each system A and B, there is indeed irreversibilities (heat transfer) so I am not sure how this equation applies at all.

Answer 1

Your equation is correct. To carry out the changes reversibly (in order to determine the entropy change of the system), you need to separate the two solids and subject each of them separately and reversibly to the same changes in temperature in different alternate processes. Those alternate reversible processes lead to the same formulas you stated.

Answer 2

Entropy is a state function. That means that it doesn't matter which ''path'' in the phase diagram you take (even if there is no ''path'' when it's irreversible): it only depends on the initial and final states. Therefore if you take as ''system variables'' $T$ and $V$, your function $S(T,V)$ will be the only thing you will need.

That said, the expression for the entropy you deduced is for $V$ constant. \begin{align*} \Delta S &= S(T_2, V_2) - S(T_1, V_1) \\ &= S(T_2, V_2) - S(T_1, V_2) + S(T_1, V_2) - S(T_1,V_1) \\ \Delta S &= \Delta S\rvert_{V=\text{const}} + \Delta S\rvert_{T=\text{const}} \end{align*} In your case, you only have an expression for the term with constant $V$.

Edit: My answer applies for all systems in general but I might want to point out a few things for your case. The second term is $0$ because it's a solid (their volume doesn't vary). Also that expression is, a priori, for each part of the system, but as the system preserves it's boundary conditions for each part of the system, the entropy is additive. So your total $\Delta S$ will be $$ \Delta S = \Delta S_{1}\rvert_{V=\text{const}} + \Delta S_{2}\rvert_{V=\text{const}}$$ where $S_1$ and $S_2$ are the entropies of each solid.

Answer 3

However, this formula is for an internally reversible process (where no irreversibilities occur within boundaries), but for each system A and B, there is indeed irreversibilities (heat transfer) so I am not sure how this equation applies at all.

It is correct that the heat transfer process in your example is irreversible. But to calculate the entropy change you can devise any convenient reversible process that connects the initial and final states and apply the equation. You can do this because entropy of each solid is a state function that does not depend on the process connecting the equilibrium states. So your equation is correct. For an excellent primer on calculating entropy change check out @Chet Miller entropy recipe here: https://www.physicsforums.com/insights/grandpa-chets-entropy-recipe/

To apply the equation, imagine you separately place each solid in contact with an infinite series of thermal reservoirs that begin at the initial temperature of the solid and end at the final temperature of each of the solids, $T_F$. The temperature of each reservoir in the series differing infinitesimally from the solid and from the previous thermal reservoir. This assures the heat transfer for each solid is reversible, i.e., each solid is always in thermal equilibrium.

Since the initial temperatures of the two solids $T_A$ and $T_B$ are different, the actual heat transfer process is irreversible and the principle of entropy increase should apply, that is

$$\Delta S_{tot}=\Delta S_{A}+\Delta S_{B}>0$$

Example:

For simplicity, consider the case where the two solids are identical except for their initial temperature. Then the final temperature is the mean of the initial temperatures. Let $T_A$ = 600 K and $T_B$ = 300 K. The final temperature is then 450 K. Then from the equation we get

$\Delta S_{A}$ = mc ln$\frac{450 K}{600 K}$ = - 0.2876 mc

$\Delta S_{B}$ = mc ln$\frac{450 K}{300 K}$ = + 0.4055 mc

$\Delta S_{tot}$ = + 0.1179 mc

Which equals the entropy generated by the irreversible heat transfer. We should expect that the smaller the temperature difference, the less entropy generated. For example, let $T_A$ = 400 K and $T_B$ = 300 K. The final temperature is then 350 K and we obtain.

$\Delta S_{A}$ = mc ln$\frac{350 K}{400 K}$ = - 0.1335 mc

$\Delta S_{B}$ = mc ln$\frac{350 K}{300 K}$ = + 0.1542 mc

$\Delta S_{tot}$ = + 0.0206 mc

In the limit, where $T_{A}=T_B$, the process is reversible and $\Delta S_{tot}$ = 0.

Hope this helps.