CBSE Class 11-science Answered

The enthalpy change on freezing from 10 degree celcius to -10 degree celcius may be expressed as:

                             ΔH₁                          ΔH₂                            ΔH₃

Liquid (10^οC) Liquid (0^οC) Solid (0^οC) Solid (-10^οC)

ΔH₁= n C_p [H₂O(l)] x ΔT = 1 x 75.03 x 10 = 0.7503 kJ mol^-1

ΔH₂= n (-Δ_fusH^o) = -1 x 6.03 = -6.03 kJ mol^-1

ΔH₃= n C_p [H₂O(s)] x ΔT = -1 x 36.8 x 10 = - 0.368 kJ mol^-1

ΔH = ΔH_{1 +}  ΔH_{2 +}  ΔH₃

ΔH = 0.7503 -6.03 - 0.368 = -5.6477 kJ mol^-1