Let's see if I can make it clearer.

For simplicity I'm going to call the ~2mol of 0C water "A" and the ~1mol of 100C water "B".

From my previous post, I'm saying the following is true:

S

_{total} = S

_{A} + S

_{B}This equation follows from what I said earlier. Taking A and B, putting them next to each other until they come to the same temperature, then mixing them is equivalent to taking the cold A and hot B and mixing them straight away (and I'm conjecturing that mixing water with water has no associated change in entropy)

In this equation S

_{X} is the entropy change from heating or cooling, which I'm saying is n*C

_{v}*ln(T

_{f}/T

_{i})

So, the total entropy should be:

S

_{total} = n

_{A}*C

_{v}*ln(T

_{f}/T

_{A}) + n

_{B}*C

_{v}*ln(T

_{f}/T

_{B})

Where does T

_{f} come from? Well, you should have already done stuff on thermal equilibrium, the only heat being transferred is from B to A, from the 100C water to the 0C water. So to get Tf set q

_{A}=q

_{B} and solve for the final temperature.

Just so you know where I got that n*C

_{v}*ln(Tf/Ti), I remembered the definition of entropy and integrated in the following way (this forum really needs LaTeX support!):

So you might want to double-check that this is the right equation (I may have messed up my maths)