Entropy change upon mixing cold and hot water

Question

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

Calculate the entropy change (in $\pu{JK–1}$ ) 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:

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

Change in entropy = $172.2/(41.4+273) = \pu{0.55 JK-1}$

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/