Calculate the enthalpy change on freezing 1.0 mol of water at 10.0 °C to ice at −10 °C.

Δ_fusH = 6.03 kJ mol⁻¹ at 0 °C 

C_p[H₂O(l)] = 75.3 J mol⁻¹ K⁻¹

C_p[H₂O(s)] = 36.8 J mol⁻¹ K⁻¹

Total ΔH = (1 mol at 10 °C → 1 mol at 0 °C) + (1 mol at 0 °C → 1 mol ice at 0 °C) + (1 mol ice at 0 °C → 1 mol ice at −10 °C)

= C_p[H₂O(l)] × ΔT + ΔH_freezing + C_p[H₂O(s)] × ΔT

= (75.3 J mol⁻¹ K⁻¹)(0 − 10)K + (−6.03 kJ mol⁻¹) + (36.8 JKmol⁻¹ K⁻¹)(−10K)

= −753 J mol⁻¹ − 6.03 kJ mol⁻¹ − 368 J mol⁻¹

= −0.753 kJ mol⁻¹ − 6.03 kJ mol⁻¹ − 0.368 kJ mol⁻¹

= −7.151 kJ mol⁻¹