Answer
$r^{*} = 1.10 \times 10^{-9} m = 1.1 nm$
Work Step by Step
Required:
If ice homogeneously nucleates at -40°C, calculate the critical radius given values of $-3.1 \times 10^{8} J/m^{3} $ and $25 \times 10^{-3} J/m^{2}$, respectively, for the latent heat of fusion and the surface free energy.
Solution:
Using Equation 10.6:
$r^{*} = (-\frac{2γT_{m}}{ΔH_{f}})(\frac{1}{T_{m} - T}) = [-\frac{(2)(25 \times 10^{-3} J/m^{2})(0 + 273 K)}{(-3.1 \times 10^{8} J/m^{3})}) (\frac{1}{0 °C - (-40 °C) })] = 1.10 \times 10^{-9} m = 1.1 nm$
