Two quantities of water ($M_\mathrm{r}=18$) each of mass $216\ \mathrm{g}$ are mixed together in a vessel. The temperatures of the two quantities before mixing are $303\ \mathrm{K}$ and $333\ \mathrm{K}$. The entire system is perfectly insulated and the vessel has a negligible heat capacity. Calculate the change in entropy after equilibrium has been established, and comment on your result in the light of the Second Law of Thermodynamics.
Note that $C_{\mathrm{m},p} = 75.5\ \mathrm{J\ K^{−1}\ mol^{−1}}$ for liquid water in this temperature range.

Is it correct to assume that the temperature of the water at equilibrium will be half way between the two values (i.e. $318\ \mathrm{K}$)? If so, I think the rest of the problem can be solved by using $\Delta S=\frac{C_V}T\Delta T$. Is this correct or is it more complicated?

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    $\begingroup$ I would say the equation is $dS = \frac{C_v }{T}dT$, and thus you'd better integrate it: $\Delta S=C_v \Delta (\ln T)$ $\endgroup$ – Curt F. Jun 3 '15 at 22:37
  • $\begingroup$ Good point. Is the final temperature 318K? $\endgroup$ – RobChem Jun 3 '15 at 22:39
  • $\begingroup$ Yeah, it would be. You don't need to assume it, and you shouldn't, because it's easily proven. Since the system is isolated, the heat gained by the water at 303 K must be equal to the heat lost by the water at 333 K, and since you are mixing equal amounts of the same substance, you can use $q = mc\Delta T$ to show that $T_f = 318$ K. $\endgroup$ – orthocresol Jun 4 '15 at 0:26
  • $\begingroup$ Also, your equation should read $C_p$, not $C_v$, because your process is being carried out at constant pressure: $dq = C_p\,dT$ so $dS = dq/T = C_p\, dT/T$. $\endgroup$ – orthocresol Jun 4 '15 at 0:30
  • $\begingroup$ @RobChem, not sure where we are on this question, but if the comments so far have enabled you to solve it, would you be able to write up an answer? I'd upvote it if you did. $\endgroup$ – Curt F. Jun 8 '15 at 21:56

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