Question

Calculate the enthalpy change on freezing of $1.0mol$ of water at ${10.0}^{\circ }C$ to ice at $-{10.0}^{\circ }C$
${△}_{fus}H=6.03KJ$ $mo{l}^{-1}$ at $0^{\circ }C$

$C_p\left[H_{2}O\right(1\left)\right]=75.3Jmo{l}^{-1}{K}^{-1}$

$C_p\left[H_{2}O\right(s\left)\right]=36.8Jmo{l}^{-1}{K}^{-1}$

Answer 1

The process is represented below.

$Liquid\left({10}^{0}C\right)\stackrel{\Delta H1}{\to }Liquid\left(0^{0}C\right)\stackrel{\Delta H2}{\to }solid\left(0^{0}C\right)\stackrel{\Delta H3}{\to }Solid(-{10}^{0}C$

$\Delta H1=n{C}_P\left[H_{2}O\right(l\left)\right]\times \Delta T=1\times 75.3\times 10=753J/mol=0.753kJ/mol$

$\Delta H2=n(-{\Delta }_{fus}{H}^0)=-1\times 6.03=-6.03kJ/mol$

$\Delta H3=n{C}_P\left[H_{2}O\right(s\left)\right]\times \Delta T=1\times 36.8\times 10=368J/mol=0.368kJ/mol$

$\Delta H1+\Delta H2+\Delta H3=0.753-6.03-0.368=-5645kJ/mol$