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?