Consider the diffusion of some hypothetical metal Y into another hypothetical metal \(Z\) at \(950^{\circ} \mathrm{C}\); after \(10 \mathrm{~h}\) the concentration at the \(0.5 \mathrm{~mm}\) position (in metal \(Z\) ) is \(2.0 \mathrm{wt} \% \mathrm{Y}\). At what position will the concentration also be \(2.0 \mathrm{wt} \% \mathrm{Y}\) after a \(17.5-\mathrm{h}\) heat treatment again at \(950^{\circ} \mathrm{C}\) ? Assume preexponential and activation energy values of \(4.3 \times 10^{-4} \mathrm{~m}^{2} / \mathrm{s}\) and \(180,000 \mathrm{~J} / \mathrm{mol}\), respectively, for this diffusion system.

Fick's second law of diffusion is given by the equation: $$C = C_0 erf\Big(\frac{x}{2\sqrt{Dt}}\Big)$$ Where \(C\) is the concentration at the position \(x\), \(C_0\) is the initial concentration at \(x=0\), \(D\) is the diffusion coefficient, \(t\) is the time, and \(erf\) is the error function. We will also use the Arrhenius equation to find the diffusion coefficient: $$D = D_0 e^{-\frac{Q}{RT}}$$ Where \(D_0\) is the pre-exponential factor, \(Q\) is the activation energy, \(R\) is the universal gas constant, and \(T\) is the temperature in Kelvins.

First, convert the temperature from Celsius to Kelvin: $$T = 950 + 273.15 = 1223.15\ K$$ Next, calculate the diffusion coefficient using the given Arrhenius equation and values provided for \(D_0\) and \(Q\). $$D = 4.3\times10^{-4} e^{-\frac {180,000}{8.314 \times 1223.15}} \approx 3.63\times10^{-13} \frac{m^2}{s}$$

We are given that the concentration at position \(0.5\ mm\) after \(10\ h\) is \(2.0\ wt\%\). Using Fick's second law of diffusion and the diffusion coefficient we determined in Step 2, let's determine the position for the first heat treatment. Set \(C=0.02\) in the Fick's Diffusion equation: $$0.02 = C_0 erf\Big(\frac{0.5\times10^{-3}}{2\sqrt{3.63\times10^{-13} \times 10 \times 3600}}\Big)$$ Now, we need to find the value of \(C_0\).

In this step, we will solve the above equation for \(C_0\). $$C_0 = \frac{0.02}{erf\Big(\frac{0.5\times10^{-3}}{2\sqrt{3.63\times10^{-13} \times 10 \times 3600}}\Big)} \approx 0.0302$$ So, the initial concentration \(C_0\) is approximately \(0.0302\ wt\%\).

Now that we have the initial concentration value, we'll use Fick's second law of diffusion again to find the position where the concentration of metal Y is \(2.0\ wt\%\) after \(17.5\ h\) of heat treatment. Set \(C = 0.02\) and \(t = 17.5\times3600\) in Fick's equation: $$0.02 = 0.0302\ erf\Big(\frac{x}{2\sqrt{3.63\times10^{-13} \times 17.5 \times 3600}}\Big)$$ Now, we need to find the value of \(x\).

In this step, we will solve the above equation for \(x\). $$\frac{0.02}{0.0302} = erf\Big(\frac{x}{2\sqrt{3.63\times10^{-13} \times 17.5 \times 3600}}\Big)$$ Find the inverse error function to obtain: $$erf^{-1}\Big(\frac{0.02}{0.0302}\Big) = \frac{x}{2\sqrt{3.63\times10^{-13} \times 17.5 \times 3600}}$$ Solve for \(x\): $$x = 2\ erf^{-1}\Big(\frac{0.02}{0.0302}\Big)\sqrt{3.63\times10^{-13} \times 17.5 \times 3600} \approx 6.67\times10^{-4}\ m$$ Converting to \(mm\), the position is approximately: $$x \approx 0.667\ mm$$ So, after a \(17.5-\mathrm{h}\) heat treatment at \(950^{\circ} \mathrm{C}\), the concentration will also be \(2.0 \mathrm{wt} \% \mathrm{Y}\) at a position of approximately \(0.667 \mathrm{mm}\) in metal Z.