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?