The enthalpy change on freezing of 1 mol of water at 5°C to ice at -9°C is : (Given ΔfusH = 6kJmol⁻¹ at 0°C, Cp(H₂O,l) = 75.3Jmol⁻¹K⁻¹, Cp(H₂O,S) = 36.8Jmol⁻¹K⁻¹)

(1) Water at 5°C is cooled to water at 0°C
ΔH₁ = nCp(H₂O,l)ΔT = 1mol × 75.3Jmol⁻¹K⁻¹ × (0-5)K = -376.5J = -0.3765kJ
(2) Water at 0°C is freezed to ice at 0°C
ΔH₂ = nΔfusH = 1mol × (-6kJmol⁻¹) = -6kJ
(3) Ice at 0°C is cooled to ice at -9°C
ΔH₃ = nCp(H₂O,S)ΔT = 1mol × 36.8Jmol⁻¹K⁻¹ × (-9K) = -331.2J = -0.3312kJ
Total enthalpy change = ΔH₁ + ΔH₂ + ΔH₃ = -0.3765kJ + (-6kJ) + (-0.3312kJ) = -6.7077kJ ≈ -6.71kJ
The closest answer is 6.56kJmol⁻¹. The negative sign indicates heat is released during the process. The question asks for the enthalpy change, which is the magnitude of the heat transfer, hence the positive value.