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} $.

  • Are you saying that you want to determine the entropy generated in each portion of the system? Or are you saying that you want to determine the entropy generated in the combined system? If it is the latter, then the answer given by @Bob D is the correct one. – Chester Miller Sep 11 at 12:21
  • @ChesterMiller I was referring to the latter but now I'm curious; how would we calculate entropy generation separately for the rock and bath individually? – Drew Sep 11 at 13:11
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    To do this, we would have to solve the transient heat conduction/convection (transport) equations (including natural convection fluid flow) for both the rock and the liquid, with matching of the temperatures and heat fluxes at the interface. The solution to this problem would involve spatial and time variations of the temperature and heat flux at the interface. To get the rate of transfer of entropy across the interface, we would have to integrate the heat flux divided by the temperature over the area of the interface. We would then integrate this with respect to time. (Continued) – Chester Miller Sep 11 at 14:15
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    This would give us the total transfer of entropy across the interface. Knowing the individual entropy change of both the rock and the fluid could then be combined with this result to calculate the individual generation of entropy for the rock and fluid. It ain't simple, but it can be done. – Chester Miller Sep 11 at 14:18
up vote 1 down vote accepted

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

  • I see. The combined system has the same entropy change as a reversible process since entropy is a property. Another question: how would we calculate entropy transfer for the rock? I don't think we could use dS = TdQ since the boundary temperature T is not constant. – Drew Sep 11 at 12:42
  • Drew, you mean dS = δQ/T, right? – Bob D Sep 11 at 12:53
  • Yes, thanks for correction. I'm confused how we would calculate the entropy temperature for the rock, however, since the boundary temperature $T$ for the rock is changing. Would we have to know that temperature as a function of time? – Drew Sep 11 at 13:19

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