Entropy Change for mixing water at two different temperatures
Well, your procedure is entirely correct. There's only one problem: the units. Since the number of moles of water is specified, your final answer should not include $\text{mol}^{-1}$. And since there's exactly $1 \text{ mol}$ of both the cold and hot water, the numerical answer should be the same.

I don't know where your error arises from but I suspect it may be because you used the units of $C_{p,m}$ (molar heat capacity at constant pressure) instead of $C_p$ (heat capacity at constant pressure). The former is an intrinsic property, whereas the latter is an extrinsic one, equal to the former multiplied by the amount of substance: $C_p = n\cdot C_{p,m}$. And their relation to the $\Delta S$ of heating at constant pressure:

$$\begin{align}
\Delta S &amp;= C_p\ln{\left(\frac{T_2}{T_1} \right)} &amp; &amp;\text{Units: J K}^{-1} \\
\Delta S_m &amp;= C_{p,m}\ln{\left(\frac{T_2}{T_1} \right)} &amp; &amp;\text{Units: J K}^{-1}\text{ mol}^{-1}
\end{align}$$

I'm entering nitpicking mode again so I shall also pedantically point out that as far as this equation is concerned, entropy comes with a triangle in front of it. Meaning you write $\Delta S$ instead of $S$.