ΔH = Cp.ΔT (for changing 1mol of liquid water from 5 to -0°)
=75.3 x5 = .377kj
from (0 to -5)
36.8 x-5 = - .184kj
Total enthalpy change
= .377 +(-6.03) + (-.184)
= -5.837 kjmol-1
(p.s.: Kindly verify the solution, as it has been solved by a commerce guy .)