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Chapter 10: Problem 3

If copper (which has a melting point of \(1085^{\circ} \mathrm{C}\) ) homogeneously nucleates at \(849^{\circ} \mathrm{C}\), calculate the critical radius given values of \(-1.77 \times 10^{9} \mathrm{~J} / \mathrm{m}^{3}\) and \(0.200 \mathrm{~J} / \mathrm{m}^{2}\), respectively, for the latent heat of fusion and the surface free energy.

Short Answer

Expert verified
The critical radius for copper when it homogeneously nucleates at 849°C is approximately 2.85 x 10^-8 m.

Step by step solution

01

Calculate the change in Gibbs free energy per unit volume

First, we need to find the change in Gibbs free energy per unit volume, \(\Delta G_{\mathrm{v}}\), using the given latent heat of fusion, \(L\), the melting point, and the nucleation temperature. Given: - Latent heat of fusion, \(L = -1.77 \times 10^{9} \mathrm{~J} / \mathrm{m}^{3}\) - Melting point of copper, \(T_m = 1085^{\circ} \mathrm{C} = 1085 + 273 = 1358 \mathrm{K}\) - Nucleation temperature, \(T = 849^{\circ} \mathrm{C} = 849 + 273 = 1122 \mathrm{K}\) Now, calculate \(\Delta G_{\mathrm{v}}\): $$ \Delta G_{\mathrm{v}} = \frac{L}{T_m - T} = \frac{-1.77 \times 10^{9} \mathrm{~J} / \mathrm{m}^{3}}{1358 \mathrm{K} - 1122 \mathrm{K}} = -1.402 \times 10^{7} \mathrm{~J} / \mathrm{m}^{3} \mathrm{K} $$
02

Calculate the critical radius

Now that we have the change in Gibbs free energy per unit volume, \(\Delta G_{\mathrm{v}}\), we can calculate the critical radius, \(r^{*}\), using the given surface free energy, \(\gamma\): $$ r^{*} = - \frac{2\gamma}{\Delta G_{\mathrm{v}}} = - \frac{2 \times 0.200 \mathrm{~J} / \mathrm{m}^{2}}{-1.402 \times 10^{7} \mathrm{~J} / \mathrm{m}^{3} \mathrm{K}} = 2.85 \times 10^{-8} \mathrm{m} $$ So, the critical radius for copper when it homogeneously nucleates at \(849^{\circ} \mathrm{C}\) is approximately \(2.85 \times 10^{-8} \mathrm{m}\).

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Most popular questions from this chapter

The fraction recrystallized-time data for the recrystallization at \(600^{\circ} \mathrm{C}\) of a previously deformed steel are tabulated here. Assuming that the kinetics of this process obey the Avrami relationship, determine the fraction recrystallized after a total time of \(22.8 \mathrm{~min} .\) $$ \begin{array}{cc} \hline \text { Fraction Recrystallized } & \text { Time (min) } \\ \hline 0.20 & 13.1 \\ 0.70 & 29.1 \\ \hline \end{array} $$

Figure \(10.40\) shows the continuous cooling transformation diagram for a \(1.13 \mathrm{wt} \%\) C iron-carbon alloy. Make a copy of this figure and then sketch and label continuous cooling curves to yield the following microstructures: (a) Fine pearlite and proeutectoid cementite (b) Martensite (c) Martensite and proeutectoid cementite (d) Coarse pearlite and proeutectoid cementite (e) Martensite, fine pearlite, and proeutectoid cementite

Name the microstructural products of eutectoid iron-carbon alloy \((0.76 \mathrm{wt} \% \mathrm{C})\) specimens that are first completely transformed to austenite, then cooled to room temperature at the following rates: (a) \(200^{\circ} \mathrm{C} / \mathrm{s}\), (b) \(100^{\circ} \mathrm{C} / \mathrm{s}\), and (c) \(20^{\circ} \mathrm{C} / \mathrm{s}\).

Compute the rate of some reaction that obeys Avrami kinetics, assuming that the constants \(n\) and \(k\) have values of \(3.0\) and \(7 \times 10^{-3}\), respectively, for time expressed in seconds.

Make a copy of the isothermal transformation diagram for a \(0.45 \mathrm{wt} \% \mathrm{C}\) iron-carbon alloy (Figure \(10.39\) ), and then sketch and label on this diagram the time-temperature paths to produce the following microstructures: (a) \(42 \%\) proeutectoid ferrite and \(58 \%\) coarse pearlite (b) \(50 \%\) fine pearlite and \(50 \%\) bainite (c) \(100 \%\) martensite (d) \(50 \%\) martensite and \(50 \%\) austenite
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