$(a)$ Energy change of $1\,mol$, $H_2O\,(l)$, at $5\,^oC$

$\to \,1\,mol$, $H_2O\,(l)$ , $0\,^oC$

$(b)$ Energy change of $1\,mol$, $H_2O\,(l)$, at $0\,^oC$

$\to \,1\,mol$, $H_2O\,(s)$ (ice) , $0\,^oC$

$(c)$ Energy change of $1\,mol$, Ice $(s)$, at $0\,^oC$

$\to \,1\,mol$, Ice $(s)$ , $-5\,^oC$

Total $\Delta H$

$ = \,{C_P}\,[{H_2}O\,(l)]\,\,\Delta T\,\, + \,\Delta H$ freezing $ + \,\,{C_P}\,[{H_2}O\,(s)]\,\,\Delta T$

$ = \,(75.3\,\,J\,\,mo{l^{ - 1}}\,{K^{ - 1}})\,( - 5)\,K\, + ( - \,6\, \times \,{10^3}\,\,J\,mo{l^{ - 1}}\,{K^{ - 1}})$ $+ \,(36.8\,\,J\,mo{l^{ - 1}}\,{K^{ - 1}})\,( - 5)\,K$

$\Delta H\,\, = \,\, - \,6.56\,\,kJ\,mo{l^{ - 1}}$ (exothermic process)

So, $\Delta H\,\, = \,\,6.56\,\,kJ\,mo{l^{ - 1}}$