Discussion
Find the volume % of pearlite for the steel just below 723°C for 0.45 wt.% carbon steel.
Answer: B
The pearlite is about 55.1%.
The pearlite forms from the austenite present just above the eutectoid temperature, so its fraction is the austenite fraction from the lever rule:
\[ \begin{aligned} W_{p} &= \frac{0.45 - 0.02}{0.8 - 0.02} \\ &= \frac{0.43}{0.78} = 0.551 \end{aligned} \]
With nearly equal densities, this is also the volume fraction, 55.1%. It is the complement of the 44.9% proeutectoid ferrite, which is the trap option. The lever rule here must use the eutectoid composition, 0.8 wt.% C, as its end point, not the 6.67 wt.% C of cementite, because pearlite is the product of austenite of eutectoid composition.
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