Question

Slow cooling of hypereutectoid steel results in formation of __________

(a) Proeutectoid cementite

(b) Posteutectoid cementite

(c) Proeutectoid ferrite

(d) Proeutectoid austenite

This question was addressed to me in class test.

Question is from Transformation of Steel and Cast Iron in division Constitution of Alloys and Phase Diagrams of Engineering Material & Metallurgy

Answer 1

Correct option is (a) Proeutectoid cementite

Explanation: Over 0.8% carbon steel is heated at around 925^oC and held at this temperature. This results in the formation of austenite. Then the steel is slowly cooled to form proeutectoid cementite which will nucleate and grow at austenite grain boundaries.