Entropy change upon mixing cold and hot water

Question

A closed, well-insulated container is filled with $454~\mathrm{g}$ of water at $94.4~^\circ\mathrm{C}$. To the hot water, $200~\mathrm{g}$ of water ice at exactly $0~^\circ\mathrm{C}$ is added. The mixture reaches an equilibrium temperature of $41.1~^\circ\mathrm{C}$. Assume the molar heat capacity is constant and all the processes are at constant pressure. The standard enthalpy of fusion for water at $0~^\circ\mathrm{C}$ is $6.008~\mathrm{kJ~mol^{–1}}$. The constant-pressure heat capacity for water is $75.291~\mathrm{J~K^{–1}~mol^{–1}}$. Water has a molecular weight of $18.015~\mathrm{g~mol^{–1}}$.

Calculate the entropy change (in $\mathrm{J~K^{–1}}$) for the system that happened because of this mixing.

I know the entropy change equals to $q/t$ because $q$ equals to the enthalpy exchange in the system as it is constant pressure, so what I did was:

$$q=(454/18.015)\times 75.291\times (41.1-94.4)+(200/18.015)\times 75.291\times(41.1)+(200/18.015)\times6008 = 172.2~\mathrm{J}$$

$$\text{change in entropy} = 172.2/(41.4+273) = 0.55~\mathrm{J~K^{-1}}$$

That is apparently incorrect, what have I done wrong?

Answer 1

The change in entropy doesn't "equals to" q/T. For a specified closed system, the change in entropy between an initial state and a final state is given by $$\Delta S=\int{\frac{dq_{rev}}{T}}$$ where the subscript "rev" refers to a reversible path between the initial and final states. The process you described is not reversible. You need to devise and employ a reversible path to get the change in entropy.

Here is a hint: For the system comprised of the 454 gm of water initially at 94.4 C, the change in entropy is $$75.291\frac{454}{18.015}\ln{[(273.2+41.1)/(273.2+94.4)]}$$ If you would like a cookbook recipe for how to determine the change in entropy for a system experiencing an irreversible process, see the following link: https://www.physicsforums.com/insights/grandpa-chets-entropy-recipe/