Compute the mass fraction of eutectoid cementite in an iron-carbon alloy that contains \(1.00 \mathrm{wt} \% \mathrm{C}\).

From an iron-carbon phase diagram, we know that the eutectoid composition is \(0.76 \mathrm{wt} \% \mathrm{C}\). This is the carbon content in steel at which a eutectoid phase transformation occurs, where austenite is transformed into pearlite (ferrite and cementite).

In the Fe-C phase diagram, proeutectoid cementite (formed before eutectoid transformation) is present above 0.76 wt% C. Below 0.76 wt% C, it is proeutectoid ferrite (formed above 0.02 wt% C). Eutectoid cementite is the cementite that forms as part of the pearlite structure within the eutectoid transformation range. Given alloy = \(1.00 \mathrm{wt} \% \mathrm{C}\) has cementite formation since it is above 0.76 wt% C.

We can use the Lever Rule to determine the mass fraction of eutectoid cementite (\(\alpha_{eut. cementite}\)) in the alloy. The Lever Rule equation is: \(\alpha_{eut. cementite} = \cfrac{C_{overall} - C_{\alpha}}{C_{\theta} - C_{\alpha}}\) Where: \(C_{overall}\) = Overall alloy composition (1.00 wt% C) \(C_{\alpha}\) = Eutectoid ferrite composition (0.76 wt% C) \(C_{\theta}\) = Eutectoid cementite composition (6.70 wt% C) Plugging in the values, we get: \(\alpha_{eut. cementite} = \cfrac{1.00 - 0.76}{6.70 - 0.76} = \cfrac{0.24}{5.94} = 0.0404\) To convert this fraction into a percentage, multiply by 100: Mass fraction of eutectoid cementite = \(0.0404 \times 100 = 4.04\%\) So, the mass fraction of eutectoid cementite in the given iron-carbon alloy containing 1.00 wt% C is approximately 4.04%.