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