How does the change in entropy of the universe vary as water is heated in different increments

I get the same answers for 2 parts of a question that I think should be different. The question is... If 0 $$^{\circ}$$ C water is heated by a 100 $$^{\circ}$$ C reservoir to thermal equilibrium, what is the total change of entropy of the universe? Instead, if it is heated by a 50 $$^{\circ}$$ C reservoir to thermal equilibrium, then a 100 $$^{\circ}$$ C reservoir, how much does the entropy of the universe change? This is what I did: (heat capacity of the mass of water is C)

$$\Delta S = \int_{273}^{373} \frac{CdT}{T}$$

$$\Delta S = C\ln\frac{373}{273}$$

Then for the second part:

$$\Delta S = \int_{273}^{323} \frac{CdT}{T} + \int_{323}^{373} \frac{CdT}{T}$$

$$\Delta S = C\ln\frac{323}{273} + C\ln\frac{373}{323}$$

$$\Delta S = C\ln\frac{373}{273}$$

I am assuming that I need to take the entropy change of the reservoir into account, but how would I do this as the temperature of the reservoir doesn't change?

• Note that the entropy change of the supersystem can be made arbitrarily close to zero, if we heat the cold water in a more complicated way, using reversible heat engine that extracts heat from the reservoir, provides useful work which is extracted away from the supersystem, and dumps the waste heat into the cold water. This can achieve equilibrium at 100 Celsius as well, but due to the reversibility, entropy of the supersystem won't increase. However, if no such engine is used and there is direct heat exchange, then the calculation you and Chet Miller describe is correct. – Ján Lalinský May 2 at 0:01

In thermodynamics, a reservoir is virtually always treated as ideal, so, in the first case, $$\Delta S_{res}=\frac{-Q}{T_{res}}=\frac{-C(373-273)}{373}$$By analogy, how do you think it would be handled for the 2nd case?