recrystallization at \(350^{\circ} \mathrm{C}\) of a previously deformed aluminum are tabulated here. Assuming that the kinetics of this process obey the Avrami relationship, determine the fraction recrystallized after a total time of \(116.8 \mathrm{~min}\). \begin{tabular}{cc} \hline Fraction Recrystallized & Time (min) \\ \hline \(0.30\) & \(95.2\) \\ \hline \(0.80\) & \(126.6\) \\ \hline \end{tabular}

We are given two data points: Fraction recrystallized: \(X_{1}=0.30\) at \(t_{1}=95.2\) min Fraction recrystallized: \(X_{2}=0.80\) at \(t_{2}=126.6\) min From the Avrami relationship, we have two equations: 1. \(0.30 = 1 - \exp(-k\cdot 95.2^{n})\) 2. \(0.80 = 1 - \exp(-k\cdot 126.6^{n})\) We will now take the natural logarithm of both sides of the equations to get linear equations: 1. \(\ln (1 - 0.30) = \ln (\exp(-k\cdot 95.2^{n})) = -k\cdot 95.2^{n}\) 2. \(\ln (1 - 0.80) = \ln (\exp(-k\cdot 126.6^{n})) = -k\cdot 126.6^{n}\) Dividing equation 1 by equation 2, we get: \(\frac{-k\cdot 95.2^{n}}{-k\cdot 126.6^{n}} = \frac{\ln(1 - 0.30)}{\ln(1 - 0.80)} \Rightarrow \frac{95.2^{n}}{126.6^{n}} = \frac{\ln(1 - 0.30)}{\ln(1 - 0.80)}\) Now take the logarithm of both sides again: \(n\cdot \ln{95.2} - n\cdot \ln{126.6} = \ln (\frac{\ln(1 - 0.30)}{\ln(1 - 0.80)})\) Now we can solve for \(n\): \(n = \frac{\ln (\frac{\ln(1 - 0.30)}{\ln(1 - 0.80)})}{\ln{95.2}-\ln{126.6}} \approx 1.33\) Now that we have found the value of \(n\), we can find the value of \(k\) using equation 1: \(k\cdot 95.2^{1.33} = -\ln (1 - 0.30)\) \(k \approx \frac{-\ln (1 - 0.30)}{95.2^{1.33}} \approx 6.16 \times 10^{-4} \, \mathrm{min}^{-n}\)

Now that we have the values of \(n\) and \(k\), we can use the Avrami relationship to find the fraction recrystallized fo \(t=116.8\) min: \(X = 1 - \exp(-k\cdot t^{n})\) \(X = 1 - \exp(-6.16 \times 10^{-4}\cdot (116.8)^{1.33})\) \(X \approx 0.48\) The fraction recrystallized after a total time of \(116.8\) minutes is approximately \(0.48\).