1
$\begingroup$

Water at temperature $T_s$ is added to a lake at temperature $T_0 > T_s$ until the total system reaches equilibrium. What is the change in entropy of the Universe?

This is a homework question but I believe it touches on something very general. The lake is a heat bath and the water added is our small thermodynamic system. We need to calculate the change in entropy of both system and surroundings and then add them to get the change in entropy of the Universe.

First of all we know that this process is completely irreversible therefore $dS \neq \frac{dQ}{T}$. However, we can invent a thermodynamic process that is reversible and use that to get the change in entropy of the system instead since entropy is just a function of state.

$$dS_{system}=\frac{dQ_{rev}}{T}=\frac{CdT}{T}$$ $$\Delta S_{system}=C\int_{T_s}^{T_0}\frac{1}{T}dT=C\ln\frac{T_0}{T_s}$$

NB $C=$ heat capacity of system (i.e. of the water added).

Now we need to work out the change in entropy of the surroundings, and this is where I get very confused. We invented a new thermodynamic process to calculate the above change in entropy. The heat the system absorbs in this reversible process $Q_{rev}$ should not be the same as the heat absorbed in the irreversible process that actually happens. Furthermore, since the real process is irreversible $dS_{surroundings}\neq \frac{dQ}{T}$. So how do we actually calculate the change in entropy of the surroundings?

This example raises another very concerning question. If we follow the reversible path for the system and surroundings then we can sy for the surroundings simply that

$$\Delta S_{surroundings} = \frac{-Q_{rev}}{T_0}=\frac{C(T_s-T_0)}{T_0}.$$

But in this case $\Delta S_{universe} = \Delta S_{system} + \Delta S_{surroundings} \neq 0$, even though the process has been reversible and therefore the Universe must undergo no change in entropy. How do we resolve this contradiction and how do we calculate the change in entropy of the Universe the correct way?

Improve this question
$\endgroup$
1
  • $\begingroup$ When considering the reversible changes, the reversible paths of the system and the surroundings don't have to match. They can be done separately. You have determined the entropy change of the system correctly. For the entropy change of the surroundings, it must have the same initial and final states as for the irreversible process. $\endgroup$
    – Chet Miller
    Oct 10, 2020 at 16:37

1 Answer 1

Reset to default
2
$\begingroup$

The equation you wrote down for the system in the reversible case corresponds physically to using a continuous sequence of ideal reservoirs at slightly different temperatures (running from Ts to To) rather than a single reservoir at To. This equation correctly determines the entropy change for the system in the irreversible case. The equation you wrote down for the surroundings also correctly determines the entropy change for the surroundings in the irreversible case. So your final answer was correct. And the entropy changes for system and surroundings in their respective irreversible cases should not be expected to sum to zero. They actually sum to the entropy generated within the system during the irreversible process.

Improve this answer
$\endgroup$
5
  • $\begingroup$ Why is the entropy change I calculated for the reservoir valid for the irreversible case? I thought it was reversible since we used $Q_{rev}$ the heat generated by the system in the reversible case. $\endgroup$
    – Pancake_Senpai
    Oct 11, 2020 at 6:51
  • $\begingroup$ What would the reversible entropy change of the surroundings even look like if what I've written is for the irreversible case? $\endgroup$
    – Pancake_Senpai
    Oct 11, 2020 at 6:53
  • $\begingroup$ You do realize that, for the alternate reversible path(s) you are using to determine the entropy change(s), the path you apply to the system will not match the path you apply to the original surroundings, right? In the case of the system, you needed to use a new set of surroundings, consisting of not just one, but a sequence of constant temperature reservoirs. In the case of the original surroundings (the ideal reservoir at To), its entropy change between the same two end states is always $Q_{res}/T_0$. Here, $Q_{res}=\Delta U_{res}=-Q_{sys, irrev}$ $\endgroup$
    – Chet Miller
    Oct 11, 2020 at 13:10
  • $\begingroup$ Thank you! It didn't occur to me that the surrounding were physically different in both processes. That's cleared up the confusion. $\endgroup$
    – Pancake_Senpai
    Oct 11, 2020 at 19:17
  • $\begingroup$ Don't feel bad. It is very poorly explained in textbooks. $\endgroup$
    – Chet Miller
    Oct 11, 2020 at 19:41

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.