b
In order to calculate the enthalpy change for \(H_2O\) at \(5\,^oC\) to ice at \(-\,5\,^oC\) , we need to calculated the enthalpy change of all the transformation involved in the process.

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

\( \Rightarrow \,{T_2}\, = \,304.92\,\,K\)

\({T_1}\, = \,300\,\,K,\,\)   \({T_2}\, = 304.92\,\,K\)

\(\Delta T\, = \,\,{T_1}\, - {T_2}\, = 4.92\,\,K\)