The enthalpy change on freezing of 1.0 mole of water at ${10}^{0}C$ to ice at $-{10}^{0}C$ is

[Given ${\mathrm{\Delta }}_{\mathrm{fus}}\mathrm{H}=6.03\mathrm{kj}/\mathrm{mole}\mathrm{at}{0}^0\mathrm{C}$, ${\mathrm{C}}_{\mathrm{p}}\left({\mathrm{H}}_2{\mathrm{O}}_{\left(\mathrm{l}\right)}\right)=75.3\mathrm{J}/\mathrm{mole}.\mathrm{k}$ ${\mathrm{C}}_{\mathrm{p}}\left({\mathrm{H}}_2{\mathrm{O}}_{\left(\mathrm{s}\right)}\right)=36.8\mathrm{J}/\mathrm{mole}.\mathrm{k}$

$1moleof{H}_{2}Oat{10}^{0}C\stackrel{\Delta H_1}{\to }1moleof{H}_{2}Oat{0}^{0}C\stackrel{\Delta H_2}{\to }1moleoficeat{0}^{0}C\stackrel{\Delta H_3}{\to }1moleoficeat-{10}^{0}C$$\Delta H_{Total}=\Delta H_1+\Delta H_2+\Delta H_3$

$\Delta H_1=C_p\Delta T=75.3\left(0-10\right)=-753j/mole=-0.753kj/mole$

$\Delta H_2=-6.03kj/mole$

$\Delta H_3=C_p\Delta T=36.8\left(-10-0\right)=-368j/mole=-0.368kj/mole$

$\Delta H_{Total}=-0.753-6.03-0.368=-7.151kj/mole$