91Ó°ÊÓ

The extracted heat from the water is\(Q = mL\).

The change in entropy is\(\Delta S = \frac{Q}{T}\).

The volume of the water is \(V = 1.00\;{{\rm{m}}{\rm{3}}}\).

The freezing temperature of the water is \(T = {0 \circ }{\rm{C}} = 273\;{\rm{K}}\).

You know the latent heat for freezing is \(L = 3.33 \times {105}\;{\rm{J/kg}}\) , and the density of the water is \(\rho = 1000\;{\rm{kg/}}{{\rm{m}}{\rm{3}}}\).

Now the mass of the water is \(m = \rho V\).

Then, the heat energy taken away from the water is,

\(\begin{array}{c}Q = mL\\ = \rho VL\end{array}\)

Therefore the change in entropy is,

\(\begin{array}{c}\Delta S = - \frac{Q}{T}\\ = - \frac{{\rho VL}}{T}\\ = - \frac{{\left( {1000\;{\rm{kg/}}{{\rm{m}}{\rm{3}}}} \right) \times \left( {1.00\;{{\rm{m}}{\rm{3}}}} \right) \times \left( {3.33 \times {{10}5}\;{\rm{J/kg}}} \right)}}{{273\;{\rm{K}}}}\\ = - 1.22 \times {106}\;{\rm{J/K}}\end{array}\)

Hence, the change in entropy is \( - 1.22 \times {106}\;{\rm{J/K}}\).