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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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  • $\begingroup$ This looks great to me. I actually did a problem extremely similar to this the other night because it was part of my p-chem homework for the week. Conceptually, your answer makes sense because the change in entropy for this process is positive. $\endgroup$ – ctkw Sep 24 '15 at 23:52
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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:

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

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 $S$.

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  • $\begingroup$ The triangle is a “delta,” and it appears because you’re calculating the change in a state property (in this case, entropy). $\endgroup$ – aeismail Dec 26 '17 at 2:51

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