Entropy generated for a warm bath heating a cold rock

Question

Put a cold rock of heat capacity $C_r$ and temperature $T_{r1}$ into an insulated fluid bath of finite heat capacity $C_b$ (the bath is not a reservoir) and temperature $T_{b1}$.

Now let the entire system come to internal thermal equilibrium, thus reaching state 2 with $T_2 = T_{r2} = T_{b2}$.

How do we find the entropy generated for this obviously irreversible process?

I started with the first law of thermodynamics on the whole system:

$$E_2 - E_1 = 0 = C_r(T_2 - T_{r1}) + C_b(T_2 - T_{b1}) $$

From which the final temperature is , $$T_2 = \frac{C_r T_{r1} + C_b T_{b1}}{C_r + C_b} $$

Now that we know the initial and final temperatures, the entropy change for the bath + rock can be calculated as:

$$ S_2 - S_1 = C_r \ln\left(\frac{T_2}{T_{r1}}\right) + C_b \ln\left(\frac{T_2}{T_{b1}} \right)$$

Now to get the entropy generated, we should find the entropy transfer so that we can apply the second law of thermodynamics:

$$ S_2 - S_1 = S_\text{transfer} + S_\text{generated} $$

But what exactly is the entropy transfer in this case? I'm confused because the heat transfer is not occurring over a constant temperature boundary, so we can't write $\mathrm dS = \frac{\delta Q}{T} $.

Answer 1

Consider your rock as a closed system and the bath as the surroundings (or vice-versa). Together they can be considered as an isolated system. For such an isolated system the total entropy change would be zero if all processes were reversible. Any positive non-zero total entropy change would thus be considered entropy generated. Thus we can say that the entropy change of an isolated system formed by a closed system and its surroundings is given by

$$\Delta S_{isolated} = \Delta S_{sys} + \Delta S_{surr} = S_{gen}$$

For your example, if the heat transfers were reversible, you would have

$$S_2 – S_1 = 0$$

but since the transfers are irreversible you have

$$S_1 - S_2 =S_{gen} = C_r \ln\frac{T_2}{T_{r1}} + C_b \ln\frac{T_2}{T_{b1}}$$

Hope this helps