---
title: "Slow cooling of 1.2% C steel"
subject: "Material Science"
chapter: 6
key: "semester_3/material_science:hypereutectoid-steel-12-carbon"
source: https://www.bctengineer.com/q/semester_3/material_science/hypereutectoid-steel-12-carbon
site: BCT Engineering
---

# Slow cooling of 1.2% C steel

Material Science · Chapter 6: Iron – Iron Carbide diagram and Heat Treatment of Steels

- Asked in: year not given, 6 marks
- Video: [Eutectoid, Hypoeutectoid and Hypereutectoid steels](https://www.youtube.com/watch?v=S7GH0FH0wtI) — Introduction to Materials Science and Engineering
- Answer checked by Claude Opus 5.5 (high effort) on 2026-10-14
- Page: https://www.bctengineer.com/q/semester_3/material_science/hypereutectoid-steel-12-carbon
- Whole chapter: https://www.bctengineer.com/md/semester_3/material_science/chapter-6

## Question

A hypereutectoid steel contains 1.2 wt% carbon. Explain its microstructure development on slow cooling from the austenite region, and calculate (a) the mass fraction of pro-eutectoid cementite and pearlite just below the eutectoid temperature, (b) the amount of each in 1 kg of steel, and (c) the total mass of ferrite and cementite.

## Short answer

Slow cooling yields a network of pro-eutectoid cementite and pearlite. Mass fractions: 7.45% pro-eutectoid cementite, 92.55% pearlite. In 1 kg: 74.5 g cementite, 925.5 g pearlite; total ferrite 822.8 g, cementite 177.2 g.

## Answer

#### Microstructure development
1. At high temperature the alloy is austenite of 1.2% C.
2. On cooling to the A$_{cm}$ line (about 900 °C for 1.2% C), austenite becomes saturated with carbon and **pro-eutectoid cementite** precipitates, mostly as a network along the austenite grain boundaries.
3. The remaining austenite loses carbon and reaches 0.76% C at 727 °C.
4. At 727 °C it transforms to **pearlite**.
5. Room-temperature structure: cementite network + pearlite. The network makes the steel hard and brittle, which is why such steels are spheroidise-annealed.

#### Lever rule (just below 727 °C)
Pro-eutectoid cementite (tie line 0.76 to 6.67):

$$
W_{Fe_3C'} = \frac{1.20 - 0.76}{6.67 - 0.76} = \frac{0.44}{5.91} = 0.0745
$$

Pearlite:

$$
W_{P} = \frac{6.67 - 1.20}{6.67 - 0.76} = \frac{5.47}{5.91} = 0.9255
$$

#### (b) In 1 kg
Pro-eutectoid cementite = 0.0745 kg = 74.5 g; pearlite = 0.9255 kg = 925.5 g.

#### (c) Total phases
$$
W_{Fe_3C} = \frac{1.20 - 0.022}{6.67 - 0.022} = \frac{1.178}{6.648} = 0.1772 \quad (177.2\ \text{g})
$$

$$
W_{\alpha} = 1 - 0.1772 = 0.8228 \quad (822.8\ \text{g})
$$

**Answer: pro-eutectoid cementite 7.45%, pearlite 92.55%; per kg: 74.5 g cementite network + 925.5 g pearlite; total ferrite 822.8 g and total cementite 177.2 g.**

## Check yourself

The answer's key points as multiple-choice questions, with the correct option and why.

1. On cooling 1.2% C austenite below the Acm line, what precipitates first?
   A. Pro-eutectoid ferrite
   B. Martensite
   C. Ledeburite
   D. Pro-eutectoid cementite, mainly along grain boundaries
   Answer: D. Austenite becomes saturated in carbon and rejects cementite, which forms a network at grain boundaries. (Key point: Pro-eutectoid phase)

2. For pro-eutectoid cementite in 1.2% C steel just below 727 °C, which tie line is used?
   A. 0.022 to 6.67
   B. 0.76 to 6.67
   C. 0.022 to 0.76
   D. 2.14 to 6.67
   Answer: B. Just below the eutectoid the two constituents are pearlite (0.76% C) and cementite (6.67% C), so W = (1.20 - 0.76)/(6.67 - 0.76) = 0.0745. (Key point: Lever rule tie line)

3. How much pearlite is present in 1 kg of this steel?
   A. 74.5 g
   B. 177.2 g
   C. 822.8 g
   D. 925.5 g
   Answer: D. Pearlite fraction = (6.67 - 1.20)/5.91 = 0.9255, i.e. 925.5 g. (Key point: Mass in 1 kg)

4. Why are such steels spheroidise-annealed?
   A. To form martensite
   B. To convert the brittle cementite network into spheres
   C. To increase carbon content
   D. To make austenite stable at room temperature
   Answer: B. The grain-boundary cementite network makes the steel hard and brittle; spheroidising breaks it into spheres. (Key point: Harmful network)

## Related in Material Science

- [Slow cooling of 0.4% C steel](https://www.bctengineer.com/question.md?key=semester_3%2Fmaterial_science%3Aslow-cooling-04-carbon-steel) — Material Science, chapter 6
- [Cooling of 3% C cast iron](https://www.bctengineer.com/question.md?key=semester_3%2Fmaterial_science%3Ahypoeutectic-cast-iron-3-carbon) — Material Science, chapter 6

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