= Solution
At $227^\circ\mathrm C\simeq500\,\mathrm K$, $1000/T\simeq2$. Reading the <carbon> line gives $D_C\sim2\times10^{-15}\,\mathrm{m^2s^{-1}}$ to graphical accuracy. With $t=60\,\mathrm s$, the characteristic <diffusion length> is
$$
\boxed{\sqrt{Dt}\sim0.35\,\mathrm{\mu m}}.
$$
If “distance” means a one-coordinate root-mean-square displacement, use $\sqrt{2Dt}\sim0.49\,\mathrm{\mu m}$; the three-dimensional radial root-mean-square is $\sqrt{6Dt}\sim0.85\,\mathrm{\mu m}$. 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 $1000/T\simeq0.96$, giving \b[$T_{\rm Cu}\sim1.0\times10^3\,\mathrm K$, 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.
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