Question

If ice homogeneously nucleates at -44.6°C, calculate the critical radius given values of -3.1 × 10^8 J/m3 and 25 × 10^-3 J/m^2, respectively, for the latent heat of fusion and the surface free energy.

Answer 1

Answer:the critical radius for the homogeneous nucleating ice is 0.986 nm

Step-by-step explanation:

Using the formulae below to calculate the critical radius for homogeneous nucleation,

We Have that

Critical radius ( r *) = [2γ Tm/ΔHf]{ 1/ Tm- T]

where γ = surface free energy =25 × 10^-3 J/m^2

Tm= solidification temperature at equilibrium =273K

Hf= latent heat of fusion = -3.1 × 10^8 J/m3

Temperature , T = -44.6°C

Critical radius ( r *) = (2 X 25 × 10^-3 J/m^2 x 273)/ (-3.1 × 10^8 J/m3 ) X ( 1/ 273 - ( -44.6 +273)

4.40x 10^-8X ( 1/273-228.4)

4.40x 10^-8 X 1/44.6

4.40x 10^-8 x 0.0224

9.86x 10 ^-10m =0.986 x 10 ^-9m = 0.986nm