 Solution

At $227^{\circ }\mathrm{C}\simeq 500\mathrm{K}$, $1000/T\simeq 2$. Reading the carbon line gives $D_C\sim 2\times 10^{-15}{\mathrm{m}}^2{\mathrm{s}}^{-1}$ to graphical accuracy. With $t=60\mathrm{s}$, the characteristic diffusion length isIf “distance” means a one-coordinate root-mean-square displacement, use $\sqrt{2Dt}\sim 0.49\mu \mathrm{m}$; the three-dimensional radial root-mean-square is $\sqrt{6Dt}\sim 0.85\mu \mathrm{m}$. The graph does not justify precision between these conventions. Equal diffusion distance in equal time requires equal diffusion coefficients. The copper line reaches this diffusivity around $1000/T\simeq 0.96$, giving $T_{\mathrm{Cu}}\sim 1.0\times 10^3\mathrm{K}$, about $770^{\circ }\mathrm{C}$. Carbon moves interstitially through ferrite, whereas copper self-diffusion is substitutional and vacancy-mediated, with a much higher effective activation barrier.