Past exam of the materials science course of the University of Cambridge / 2018 / Paper 1 / 4 / b / ii / Solution

For eutectoid pearlite, apply the lever rule between ferrite and cementite at the transformation temperature. Using standard approximate Fe–C values $C_e=0.76$, $C_{\alpha }=0.022$ and $C_{\mathrm{cem}}=6.67$ in weight percent carbon,These are mass fractions. To compare with an unbiased geometric line fraction, convert using the phase densities:With representative densities ${\rho }_{\mathrm{cem}}\simeq 7.69$ and ${\rho }_{\alpha }\simeq 7.87\mathrm{Mg}{\mathrm{m}}^{-3}$, this gives about $11.3vol\%$ cementite and $88.7vol\%$ ferrite. Because the densities are similar, the mass and volume fractions differ only modestly. The phase-diagram values here are conventional approximations, not a claim that the exact 2018 Data Book has been recovered.