Engineering · Mock exam 2026 All mock exam questions
Mock exam · Section 2 · Written response

Question 13

[6 marks] Simple familiar
The question

Two identical bars of eutectoid steel (0.83% carbon) are heated to 800 °C. Bar 1 is then quenched in water. Bar 2 is cooled slowly inside the furnace.

a)

Explain how the cooling rate affects the microstructure and the mechanical properties of each bar.

[4 marks] b)

Bar 1 is to be used as a punch in a metal fabrication workshop. Explain why Bar 1 is tempered before it is used.

[2 marks]
Watch the situation first

Same steel, same start, two cooling rates

Both bars start as austenite at 800 °C. Watch each cooling curve and the microstructure it leaves behind, then watch Bar 1 go back into the heat to be tempered.

Both bars are held at 800 °C, above 723 °C, so both are austenite Bar 1 is quenched. The carbon has no time to diffuse out Bar 2 cools slowly and transforms at 723 °C into coarse pearlite Martensite is very hard but brittle. Pearlite is softer and tougher Bar 1 is reheated to below 723 °C to temper it Tempered martensite is tougher, for a small loss of hardness 0 200 400 600 800 723 Temperature (°C) Time (not to scale) Bar 1 Bar 2 transforms tempering, below 723 °C Bar 1, quenched then temperedBar 2, furnace cooled Bar 1: austenite Bar 1: martensite Bar 1: tempered martensite Bar 2: austenite Bar 2: coarse pearlite very hard but brittle tougher, a little less hard softer, more ductile, tougher

The cooling rate decides the structure. Fast cooling traps the carbon as martensite. Slow cooling lets it form layers of pearlite.

Now for the engineering

Pick a cooling route, then read the structure it gives

Each route starts from austenite at 800 °C. Choose one to see its cooling curve, and the microstructure and properties it leaves.

0 200 400 600 800 723 Time (not to scale) Temperature (°C) tempering, below 723 °C quench furnace cool
Water quench Furnace cool 723 °C
Microstructure at room temperature

Marking guide · 6 marks

Work through the model response one mark at a time

Four marks in a) for a structure and a property for each bar, and two in b) for what tempering does and why the punch needs it. Try each sentence yourself before you reveal it.

a) Mark 1Part a) · 1 mark
Model response

Water quenching cools Bar 1 so rapidly that the carbon does not have time to diffuse out of the austenite, so the austenite transforms into martensite, with carbon trapped in a distorted iron lattice.

Why?

You name martensite for the quenched bar and you give the reason: there is no time for the carbon to diffuse out of the austenite. The name without the reason is only half of what the marker looks for.

Marking

For the first mark, accept ‘carbon is trapped in the iron lattice’ or ‘there is no time for pearlite to form’.

Marking

Do not accept pearlite, bainite or cementite as the structure of Bar 1.

Marker: identifies that water quenching (rapid cooling) forms martensite, because the carbon does not have time to diffuse out of the austenite

a) Mark 2Part a) · 1 mark
Model response

Martensite is very hard but brittle, so Bar 1 resists wear but cracks under impact.

Why?

You state that martensite is very hard but brittle. The question asks about mechanical properties as well as microstructure, so each bar needs its properties, and here both halves count: hard, and brittle.

Marker: identifies that martensite is very hard but brittle

a) Mark 3Part a) · 1 mark
Model response

Furnace cooling is slow, so at 723 °C the austenite has time to transform into coarse pearlite (alternating layers of ferrite and cementite).

Why?

You link the slow furnace cooling to the austenite transforming into coarse pearlite. Slow cooling is the cause and pearlite is the effect, so say both.

Marking

For Bar 2, accept pearlite, or layers of ferrite and cementite. ‘Coarse’ is not required.

Marking

A sketch of the microstructures may support the response but is not required.

Marker: identifies that furnace cooling (slow cooling) allows the austenite to transform into coarse pearlite

a) Mark 4Part a) · 1 mark
Model response

Coarse pearlite is softer and more ductile and tougher than martensite, so Bar 2 can bend and absorb impact without cracking.

Why?

You compare pearlite with martensite: softer, more ductile, tougher. The comparison with Bar 1 is what earns the mark, so write “than martensite”.

Marking

Or other suitable response consistent with a reasonable understanding.

Marker: identifies that coarse pearlite is softer and more ductile (or tougher) than martensite

b) Mark 1Part b) · 1 mark
Model response

Untempered martensite is too brittle for a punch, which is struck repeatedly. Bar 1 is reheated to a temperature below 723 °C (about 200 to 300 °C), so some carbon leaves the distorted lattice and tempered martensite forms.

Why?

You say that Bar 1 is reheated below 723 °C (about 200 to 300 °C) and that tempered martensite forms. Both the temperature limit and the new structure are in the mark.

Marking

A specific tempering temperature is not required. Accept any stated temperature from 150–650 °C.

Marker: identifies that tempering reheats the quenched steel to a temperature below 723 °C to form tempered martensite

b) Mark 2Part b) · 1 mark
Model response

This increases toughness (reduces brittleness) for a small loss of hardness, so the punch stays hard enough to cut metal but withstands repeated impact without shattering.

Why?

You say toughness goes up and brittleness goes down for a small loss of hardness, and you link it to the punch surviving impact. The link to the punch is what makes it an explanation.

Marking

Accept ‘relieves internal stresses’ as part of the second mark only when it is linked to reduced brittleness or increased toughness.

Marking

Allow FT marks for errors in part a).

Marking

Or other suitable response consistent with a reasonable understanding.

Marker: explains appropriately using wording that indicates tempering increases toughness (reduces brittleness) for a small loss of hardness, so the punch can withstand impact without shattering

The model response in full

a) Water quenching cools Bar 1 so rapidly that the carbon does not have time to diffuse out of the austenite, so the austenite transforms into martensite, with carbon trapped in a distorted iron lattice.a) Mark 1 Martensite is very hard but brittle, so Bar 1 resists wear but cracks under impact.a) Mark 2

Furnace cooling is slow, so at 723 °C the austenite has time to transform into coarse pearlite (alternating layers of ferrite and cementite).a) Mark 3 Coarse pearlite is softer and more ductile and tougher than martensite, so Bar 2 can bend and absorb impact without cracking.a) Mark 4

b) Untempered martensite is too brittle for a punch, which is struck repeatedly. Bar 1 is reheated to a temperature below 723 °C (about 200 to 300 °C), so some carbon leaves the distorted lattice and tempered martensite forms.b) Mark 1 This increases toughness (reduces brittleness) for a small loss of hardness, so the punch stays hard enough to cut metal but withstands repeated impact without shattering.b) Mark 2

Putting it all together

Each phrase of the question gave you something

“eutectoid steel (0.83% carbon)”
Slow cooled, the whole bar becomes pearlite, with no extra ferrite or cementite. That makes the two structures easy to compare.
“heated to 800 °C”
Above 723 °C, so both bars start as austenite. Every structure in your answer forms from that austenite.
“Bar 1 is then quenched in water”
Rapid cooling. The carbon has no time to diffuse out, so martensite forms. Very hard but brittle.
“Bar 2 is cooled slowly inside the furnace”
Slow cooling. At 723 °C the austenite has time to transform into coarse pearlite. Softer, more ductile and tougher than martensite.
“the microstructure and the mechanical properties of each bar”
Four marks: a structure and a property for each bar. Compare the properties of the two bars directly.
“to be used as a punch”
A punch is struck repeatedly, so it must stay hard enough to cut metal without shattering.
“Explain why Bar 1 is tempered”
Reheat below 723 °C to form tempered martensite. Toughness goes up for a small loss of hardness, so the punch survives impact.
Common errors

If I had to guess where you might lose marks…

  • Naming pearlite, bainite or cementite for the quenched bar. Quenching leaves no time for pearlite to form. The only structure that earns the first mark for Bar 1 is martensite.
  • Describing properties for only one bar, or calling pearlite “weaker”. Each bar needs its own properties. For Bar 2, compare ductility or toughness with martensite, not just strength.
  • Saying tempering “softens” or “anneals” the punch, or reheating above 723 °C. Above 723 °C the steel re-forms austenite and undoes the quench. Tempering stays below 723 °C and costs only a little hardness.
  • Writing “relieves internal stresses” on its own in b). It only counts toward the second mark when you link it to reduced brittleness or increased toughness.
Compare with past QCAA questions

New for 2026: QCAA has not examined this yet

No past QCAA question tests heat treatment, so compare the structure of the question instead.

Similar structure
2024 Question 14
Short response · 4 marks
Explain how the low carbon content of mild steel makes it suit seamless tubes, with two properties.
2023 Question 15
Short response · 8 marks
Sketch and label the ferrite and layered pearlite in medium carbon steel, then explain two properties that suit train rails.

All past QCAA Engineering questions

Only mark this done when you could do it without help. Reading the solution does not count.

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A Tangent Tuition mock exam question. It is not a past QCAA question, and Tangent Tuition is not affiliated with the QCAA.