Question 1

[1 mark]

The question

A lead–tin thermal-equilibrium phase diagram is shown.

A lead-tin thermal-equilibrium phase diagram. Temperature in degrees Celsius runs up the side from 0 to 400, and % Lead (Pb) runs along the bottom from 0 to 100, on a fine grid. Two curves fall from the left edge at about 227 degrees: one meets a V at 38.1% lead and 183 degrees, the other drops steeply to a point labelled 2.5% at 183 degrees and then curves down to near the left edge at 0 degrees. From the V a curve rises to the right edge at about 327 degrees, and a second curve rises from a point labelled 81% at 183 degrees to the same place. A dashed horizontal line runs at 183 degrees from 2.5% to 81% lead, labelled 183 °C, and a dashed vertical line drops from the V at 38.1% to the axis. From the 81% point a curve falls to about 95% lead at 0 degrees. The phase fields are not labelled.

What phases would be present in an alloy containing 20% Pb at a temperature of 200 °C?

Not quite. You picked L only, but that is the field above the liquidus. At 20% Pb that starts at about 210 °C, not 200 °C.

Not quite. $\alpha+\mathrm{L}$ is the matching field on the lead-rich side of the diagram. You have $\alpha$ and $\beta$ swapped.

Not quite. $\alpha+\beta$ is the fully solid field below the 183 °C eutectic line. Your alloy is at 200 °C, above that line.

That is it. Step through the hints below to see why.

The printed diagram redrawn. Shaded: the phase field of the option you are testing. The dotted lines run from your alloy, 20% Pb at 200 °C, to both axes. Testing (D) also draws the tie line across the field at 200 °C.

At 20% Pb the L field only starts at the liquidus, at about 210 °C. Your alloy at 200 °C sits below it, outside the shaded field. Not this one.

The shaded $\alpha+\mathrm{L}$ field is the matching field on the lead-rich side, far to the right of 20% Pb. Not this one.

The shaded $\alpha+\beta$ field lies wholly below the dashed 183 °C line. Your alloy at 200 °C is above that line. Not this one.

Your alloy sits inside the shaded field, between the $\beta$ solidus at about 2% Pb and the tin-rich liquidus at about 27% Pb. This is the one.

Hint 1

This diagram counts % Lead (Pb) along the bottom, so the tin-rich side is on the left.

Convention (as in QCAA papers): $\alpha$ is the lead-rich solid and $\beta$ is the tin-rich solid.

Hint 2

At 200 °C the 20% Pb point lies between the $\beta$ solidus (about 2% Pb) and the tin-rich liquidus (about 27% Pb), above the 183 °C eutectic line.

$$2\%<20\%<27\%\ \Rightarrow\ \beta+\mathrm{L}$$

Why?

Between solidus and liquidus the alloy is part solid, part liquid, and on the tin-rich side of the eutectic the solid is $\beta$.

At 200 °C the 20% Pb point lies between the $\beta$ solidus (about 2% Pb) and the tin-rich liquidus (about 27% Pb), above the 183 °C eutectic line.

$$\beta+\mathrm{L}$$

(D)

Compare with past QCAA questions

The past QCAA questions this one was modelled on

2024 Question 1

Multiple choice · 1 mark

Plot 35% tin at 225 °C on a lead-tin phase diagram and name the phase field it lands in.

All past QCAA Engineering questions

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

Back to the index

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

Spot a mistake on this page? Report it