Question

If copper (which has a melting point of 1085°C) homogeneously nucleates at 849°C, calculate the critical radius given values of –1.77 × 109 J/m3 and 0.200 J/m2, respectively, for the latent heat of fusion and the surface free energy.

Answer 1

Answer:

The critical radius is -1.30 nm

Explanation:

Temperature for homogenous nucleation of copper, $T_{H} = 849^{0} C = 849 + 273 = 1122 K$

Melting point of copper, $T_{cu} = 1085^{0} C = 1085 + 273 = 1358 K$

Latent heat of fusion, $H_{f} = -1.77 * 10^{9} J/m^{3}$

Surface free energy, $\gamma = 0.200 J/m^{2}$

Critical radius, r = ?

The formula for the critical radius is given by:

$r = \frac{2 \gamma T_{cu} }{H_{f}(T_{cu} - T_{H}) }$

$r = \frac{2 * 0.2*1358 }{(-1.77 * 10^{9}) (1358 - 1122) }$

$r = \frac{543.2 }{(-1.77 * 10^{9}) 236}\\r = -1.30 * 10^{-9} m\\r = -1.30 nm$

The critical radius is -1.30 nm

Answer 2

The critical radius upon homogenous nucleation is; r_c = -1.25 nm

We want to find critical radius and it is gotten from the formula;

r_c = (2γ × T_f)/(h_f,v × ΔT)

Where;

r_c is critical radius

γ is surface free energy.

T_f is temperature of melting point of element

h_f,v is latent heat of fusion or vaporization

ΔT is change in temperature = T_f - T_h

Where T_h is the temperature at which which the element homogeneously nucleates

We are given;

γ = 0.2 J/m²

T_f = 1085 °C = 1358 K

T_h = 849 °C = 1112 K

latent heat of fusion; h_f = -1.77 × 10^(9) J/m³

Thus;

r_c = (2 × 0.2 × 1358)/(-1.77 × 10^(9) × (1358 - 1112))

r_c = -1.25 × 10^(-9) m

r_c = -1.25 nm

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