Answer
$\Delta S=3.8\times10^4\frac{J}{K}$
Work Step by Step
1. Loss of thermal energy as water freezes.
$\Delta S=\frac{Q}{T}=\frac{mL_w}{T}=\frac{(1000kg)(333\times10^3\frac{J}{K})}{273K}$
$=-1.22\times10^6\frac{J}{K}$
2. Loss of enthropy from ice as it cools from $0^oC$ to $-8^oC$
$T=\frac{(0-8^oC)}{2}=-4+273K=269K$
$\Delta S=\frac{Q}{T}=\frac{mc\Delta T}{T}=\frac{(1000kg)(2100)(0-8^oC)}{269K}$
$=-6.25\times10^4\frac{J}{K}$
3. Heat gain of the large mass of ice
$T=-8^oC=265K$
$\Delta S=\frac{Q}{T}=\frac{(1000kg)(333\times10^3\frac{J}{K})+(1000kg)(2100)(0-8^oC)}{265K}=1.32\times10^6\frac{J}{K}$
$\Delta S_t=-1.22\times10^6\frac{J}{K}-6.25\times10^4\frac{J}{K}+1.32\times10^6\frac{J}{K}=3.8\times10^4\frac{J}{K}$
