A steel sample with 1.5 wt.% carbon (no other alloying elements present) is slowly cooled from 1100 °C to just below the eutectoid temperature (723 °C). A part of the iron-cementite phase diagram is shown in the figure. The ratio of the pro-eutectoid cementite content to the total cementite content in the microstructure that develops just below the eutectoid temperature is ________. (Rounded off to two decimal places)

The correct answer is 0.54.

1. Understanding the Phase Diagram and Key Data Points:
By analyzing the provided iron-cementite phase diagram, we can identify several critical data points at the eutectoid temperature (723 °C):

• Alloy Composition (C₀): The vertical dashed line shows the steel sample composition is 1.5 wt.% C.
• Eutectoid Carbon Composition (C_e): The eutectoid point on the diagram corresponds to 0.8 wt.% C.
• Ferrite Solubility Limit (C_α): The maximum carbon solubility in ferrite (α) at 723 °C is 0.035 wt.% C.
• Cementite Carbon Composition (C_Fe3C): The carbon content of cementite (Fe₃C) at the right-hand boundary is 6.7 wt.% C.

2. Calculating the Mass Fraction of Pro-eutectoid Cementite (m₁):
Pro-eutectoid cementite forms as the steel is cooled through the austenite (γ) + cementite (Fe₃C) region, just above the eutectoid temperature (723 °C). At this point, the phases in equilibrium are austenite of eutectoid composition (0.8 wt.% C) and pro-eutectoid cementite (6.7 wt.% C).
Using the lever rule:

$$m_1=\frac{C_0-C_e}{C_{{\mathrm{Fe}}_{3}C}-C_e}$$

Substituting the values from the diagram:

$$m_1=\frac{1.5-0.8}{6.7-0.8}=\frac{0.7}{5.9}\approx 0.1186$$

3. Calculating the Mass Fraction of Total Cementite (m₂):
Just below the eutectoid temperature (723 °C), the austenite undergoes the eutectoid reaction to transform completely into ferrite (α) and eutectoid cementite. The total cementite present in the microstructure includes both the pro-eutectoid cementite and the eutectoid cementite. The two equilibrium phases just below 723 °C are ferrite (with 0.035 wt.% C) and cementite (6.7 wt.% C).
Using the lever rule for total cementite:

$$m_2=\frac{C_0-C_{\alpha }}{C_{{\mathrm{Fe}}_{3}C}-C_{\alpha }}$$

Substituting the values:

$$m_2=\frac{1.5-0.035}{6.7-0.035}=\frac{1.465}{6.665}\approx 0.2198$$

4. Calculating the Ratio:
Now, we find the ratio of the pro-eutectoid cementite content to the total cementite content:

$$\text{Ratio}=\frac{m_1}{m_2}=\frac{0.1186}{0.2198}\approx 0.5396$$

Rounding off to two decimal places gives 0.54.