Chapter 9, Problem 9.11P

Interpretation Introduction

Interpretation:

For ${10}^{16}$ spheres of copper, the total interfacial surface energy having a critical radius $r^{*}$ needs to be determined.

Concept introduction:

The total energy is related to the interface of the surface and the grain boundaries in the alloy or metals which holds great importance in today's metallurgical practice because it controls the problem of brittleness in the metal.

Answer to Problem 9.11P

The total interfacial surface energy for ${10}^{16}$ spheres of copper is 35 mJ.

Explanation of Solution

Given Information:

From the copper properties table, the values of freezing temperature, latent heat of fusion surface energy and maximum intercooling for the selected materials are as follows:

  $\begin{array}{l}\Delta T_m={1085}^0\text{C},{\sigma }_{SL}=177\times {10}^{-7}\frac{\text{J}}{{\text{cm}}^{\text{2}}}\\ \Delta Hf=1628\frac{\text{J}}{{\text{cm}}^{\text{3}}}\end{array}$

Calculation:

The formula for the critical radius $r^{*}$ is as follows:

  $r^{*}=\frac{2\sigma sc{T}_m}{\Delta H_f\Delta T}$

Here,

  $\begin{array}{l}{\sigma }_{SL}=\text{Surface energy of solid-liquid interface}\hfill \\ T_m=\text{Freezing temperature}\hfill \\ \Delta Hf=\text{Heat of fusion}\hfill \\ \Delta T=\text{Typical under for homogenous nucleation}\hfill \end{array}$

By putting the values in the equation (1),

  $\begin{array}{l}{r}^{*}=\frac{2\times 177\times {10}^{-7}\times (1085+273)\text{ K}}{1628\times 236}\\ r^{*}=12.51\times {10}^{-8}\text{cm}\end{array}$

Obtaining the relation for total interfacial surface energy of copper for ${10}^{16}$ spheres,

  ${\sigma }_{total(SL)}=n\times ({\sigma }_{SL})\times A_{spheres}$ --------------------- (2)

Here,

  $\begin{array}{l}n=\text{Number of spheres}\hfill \\ A_{spheres}=\text{Area of sphere}\hfill \end{array}$

From the formula of area of sphere:

  $A_{\text{spheres}}=4\pi r^2$

Calculate the area of a sphere as follows:

  $A_{\text{spheres}}=4\pi r^{*}{}^2$

Here, $r^{*}=$ the critical radius of the sphere

Substituting the value $r^{*}$ from equation (1)

  ${\text{A}}_{\text{spheres}}{\text{=4π}}^{\text{2}}{\text{(12}{\text{.51×10}}^{\text{-8}}\text{)}}^{\text{2}}$

  $\text{=1}{\text{.966×10}}^{\text{-13}}{\text{cm}}^{\text{2}}$

After getting the value of the area ofa sphere, the total interfacial surface energy of copper for ${10}^{16}$ spheres cam be calculated using the following values.

  $\begin{array}{l}n={10}^{16},A=1.966\times {10}^{-13}\\ {\sigma }_{SL}=177\times {10}^{-7}\end{array}$

Put the above values in equation (2)

  $\begin{array}{l}{\text{σ}}_{\text{total(SL)}}{\text{=10}}^{\text{16}}{\text{×177×10}}^{\text{-7}}\text{×1}{\text{.966×10}}^{\text{-13}}\\ \text{=0}{\text{.035×10}}^{\text{-3}}\text{J}\\ {\text{σ}}_{\text{total(SL)}}\text{=35mJ}\end{array}$

Hence, the total interfacial surface energy of copper for ${10}^{16}$ spheres is 35 mJ.

Conclusion

Thus, from the properties of the copper table,one gets the total interfacial surface energy of copper for ${10}^{16}$ spheres as 35 mJ.