# Entropy change upon mixing cold and hot water

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?

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/