thermodynamics - Entropy Change for mixing water at two different temperatures

Assume that the heat capacity of water is independent of the temperature. Calculate the net entropy change when 1 mole of water at 0 degrees Celsius is mixed with 1 mole of water at 100 degrees Celsius. Assume that the heat capacity of water is (4.184 J/(K-g))(18 g/mol) = 75.3 J/(K-mol) and that the heat capacity of the calorimeter is negligible.

I was wondering if the following general methodology of my attempt was correct.

In attempting to solve this, I first found that the equilibrium temperature was (273+373)/2 = 323 K.

Then, I calculated the entropy by using the equation S = (heat capacity)*ln(T_eq/T_initial) for both 0 degrees and 100 degrees.

Afterwards, I added the two values for entropy and obtained 1.827 J/(mol-K) as the answer.

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