You know the latent heat for freezing is \(L = 3.33 \times {105}\;{\rm{J/kg}}\) , the density of the water is \(\rho = 1000\;{\rm{kg/}}{{\rm{m}}{\rm{3}}}\), and the specific heat of ice is \({c_{\rm{i}}} = 2100\;{\rm{J/kg}} \cdot {\rm{K}}\)
Now the mass of the water is \(m = \rho V\).
Then, the heat energy is taken away from the water to become the ice \({0 \circ }{\rm{C}}\),
\(\begin{array}{c}Q = mL\\ = \rho VL\end{array}\)
Therefore the change in entropy during freezing,
\(\begin{array}{c}\Delta {S_1} = - \frac{Q}{{{T_1}}}\\ = - \frac{{\rho VL}}{{{T_1}}}\\ = - \frac{{\left( {1000\;{\rm{kg/}}{{\rm{m}}{\rm{3}}}} \right) \times \left( {1.00\;{{\rm{m}}{\rm{3}}}} \right) \times \left( {3.33 \times {{10}5}\;{\rm{J/kg}}} \right)}}{{273\;{\rm{K}}}}\\ = - 1.22 \times {106}\;{\rm{J/K}}\end{array}\)
Now the change in entropy due to lowering the temperature of the ice is,
\(\begin{array}{c}\Delta {S_1} = m{c_{\rm{i}}}\ln \frac{{{T_2}}}{{{T_1}}}\\ = \rho V{c_{\rm{i}}}\ln \frac{{{T_2}}}{{{T_1}}}\\ = \left[ {\left( {1000\;{\rm{kg/}}{{\rm{m}}{\rm{3}}}} \right) \times \left( {1.00\;{{\rm{m}}{\rm{3}}}} \right) \times \left( {2100\;{\rm{J/kg}} \cdot {\rm{K}}} \right)} \right]\ln \frac{{265\;{\rm{K}}}}{{273\;{\rm{K}}}}\\ = - 6.25 \times {104}\;{\rm{J/K}}\end{array}\)
Therefore the total change in entropy is,
\(\begin{array}{c}\Delta S = \Delta {S_1} + \Delta {S_2}\\ = \left( { - 1.22 \times {{10}6}\;{\rm{J/K}}} \right) + \left( { - 6.25 \times {{10}4}\;{\rm{J/K}}} \right)\\ = - 1.28 \times {106}\;{\rm{J/K}}\end{array}\)
Hence, the change in entropy is \(1.28 \times {106}\;{\rm{J/K}}\).
