Kinetics, Diffusion, Solidification and Phase Transformations

This chapter answers the second and third paragraphs of Section XE2.2, “Thermodynamics, Kinetics and Phase Transformations”: “Reaction kinetics, rate constants, order of reactions, Arrhenius law, Fick’s laws, steady state and non-steady state solutions of diffusion equations, diffusion distance and diffusion time, applications of solutions of diffusion equations, atomistic mechanisms of diffusion, fast diffusion paths” and “Solidification of pure metals and alloys, homogeneous and heterogeneous nucleation, nucleation rate, growth, partitioning during binary solidification; diffusional solid-state phase transformations (precipitation and eutectoid), overall transformation kinetics (TTT and CCT), martensitic/displacive transformation; glass transition.” It belongs to Section XE2.2 of XE2 Materials Science, an optional section of the Engineering Sciences (XE) paper. The numericals are carburising depth and time from D(T), the effect of temperature on diffusion time, the critical nucleus radius, the Scheil liquid composition and the Avrami fraction transformed.

1. Reaction kinetics and the Arrhenius law

Thermodynamics says whether a change can happen (ΔG < 0); kinetics says how fast. The rate of a reaction is the rate of change of concentration C, and its order is the exponent of concentration in the rate law. The rate constant k collects everything except concentration.

Integrated rate laws
OrderRate lawIntegrated formHalf-lifeLinear plot
0−dC/dt = kC = C₀ − ktC₀/(2k)C against t
1−dC/dt = kCC = C₀ e^(−kt)ln 2 / kln C against t
2−dC/dt = kC²1/C = 1/C₀ + kt1/(kC₀)1/C against t

Arrhenius law: k = A exp(−Q/RT), where Q is the activation energy per mole, R = 8.314 J mol⁻¹ K⁻¹ and A the pre-exponential factor. Taking logarithms, ln k = ln A − (Q/R)(1/T): a plot of ln k against 1/T is a straight line of slope −Q/R. The same law governs diffusivity, vacancy concentration and nucleation. Between two temperatures, k₂/k₁ = exp[(Q/R)(1/T₁ − 1/T₂)]. Because T sits inside an exponent, a small rise in temperature changes the rate by a large factor, and the factor is larger for a larger Q.

⚠️ Always absolute temperature
Use kelvin in the exponent and the same Q units as R: Q in J mol⁻¹ with R = 8.314, or Q in kJ mol⁻¹ with R = 0.008314. A first-order half-life is independent of C₀; a second-order half-life is not.

2. Fick’s laws, steady state and diffusion mechanisms

Fick’s first law: the flux is proportional to the concentration gradient, J = −D dC/dx, with J in atoms (or mol) per m² per second and D in m²/s; atoms flow down the gradient. Fick’s second law: conservation of matter gives ∂C/∂t = D ∂²C/∂x² when D is constant. In steady state (∂C/∂t = 0) the flux is the same at every plane and the profile across a plate is linear, so J = D(C₁ − C₂)/Δx.

Temperature dependence: D = D₀ exp(−Q/RT). Mechanisms. In vacancy diffusion an atom jumps into a neighbouring vacant site, so it needs both a vacancy and the energy to migrate: Q = Q_formation + Q_migration, which is large (self-diffusion in close-packed metals). In interstitial diffusion a small atom (C, N, H in iron) hops between interstitial sites, needing no vacancy, so Q is lower and D is far larger than substitutional diffusivity at the same T. Fast diffusion paths are the surface, grain boundaries and dislocation cores (“pipes”): the open structure gives a lower activation energy, and typically D_surface > D_grain boundary > D_lattice. Because their Q is smaller, they dominate at low temperature and in fine-grained or heavily deformed material, while lattice diffusion takes over at high temperature.

🎯 Diffusion distance x ~ √(Dt)
The solutions of the second law depend on x and t only through x/(2√(Dt)), so the depth that reaches a given concentration scales as √(Dt). To double the depth at fixed T the time must be multiplied by 4. Equally, the time for a given depth is t ≈ x²/D up to a numerical factor: halving D doubles the time.

3. Non-steady-state solutions and applications

Semi-infinite solid with constant surface concentration C_s (the error-function solution): the bulk starts at C₀ and the surface is held at C_s, so (C_s − C_x)/(C_s − C₀) = erf(x/(2√(Dt))). The error function runs from erf(0) = 0 to erf(∞) = 1; erf(0.477) = 0.5 and erf(1) = 0.843. This is carburising (and decarburising or doping from a gas): the profile spreads inward without limit as t grows, and the surface stays at C_s.

Thin-film (instantaneous source) solution: a fixed amount M per unit area deposited on the surface of a semi-infinite solid spreads as a Gaussian, C(x, t) = [M/√(πDt)] exp(−x²/4Dt). The surface concentration falls as t^(−1/2) and the area under the curve stays M. Used for dopant drive-in and tracer diffusion experiments.

ℹ️ Worked: carburising time
A steel with 0.2 wt% C is carburised at 1273 K with the surface held at 1.0 wt% C. Take D₀ = 2.3 × 10⁻⁵ m²/s, Q = 148 kJ/mol. D = 2.3 × 10⁻⁵ exp[−148 000/(8.314 × 1273)] = 1.94 × 10⁻¹¹ m²/s. For 0.6 wt% C at x = 0.5 mm: (1.0 − 0.6)/(1.0 − 0.2) = 0.5 = erf(z), so z = 0.477 = x/(2√(Dt)). Then t = x²/(4z²D) = (5 × 10⁻⁴)²/(4 × 0.2275 × 1.94 × 10⁻¹¹) = 1.41 × 10⁴ s = 3.9 h.
⚠️ Check the profile before using erf
The erf form needs a fixed surface concentration. A finite deposited amount needs the Gaussian instead. Always convert the concentration ratio to the erf argument before solving for x or t.

4. Nucleation and growth in solidification

A melt freezes only below the melting point T_m, by undercooling ΔT = T_m − T. A solid sphere of radius r changes G by ΔG(r) = −(4/3)πr³ΔG_v + 4πr²γ: volume gives a negative term (ΔG_v > 0 is the driving force per volume), and the solid–liquid interface of energy γ gives a positive one. The maximum, at the critical radius r* = 2γ/ΔG_v, is the barrier ΔG* = 16πγ³/(3ΔG_v²). Embryos smaller than r* shrink; larger ones grow. With ΔG_v ≈ L_vΔT/T_m (L_v the latent heat per unit volume), r* = 2γT_m/(L_vΔT) ∝ 1/ΔT and ΔG* ∝ 1/ΔT².

Heterogeneous nucleation on a mould wall or inoculant needs a smaller barrier because part of the interface is replaced: ΔG_het = S(θ) ΔG_hom with S(θ) = (2 + cos θ)(1 − cos θ)²/4, where θ is the contact angle. S = 0.5 at θ = 90°, S → 0 for a perfectly wetting substrate and S = 1 at θ = 180°. The critical radius r* is unchanged but the critical nucleus has less volume. This is why real melts freeze at small undercooling and why grain refiners work. Nucleation rate I ∝ exp(−ΔG/kT)·exp(−Q_d/kT): it rises with undercooling as ΔG collapses and then falls at very large undercooling when atomic mobility is too low, giving a maximum. Growth rate follows the same competition between driving force and mobility.

ℹ️ Worked: critical radius
Take γ = 0.20 J/m², L_v = 1.5 × 10⁹ J/m³, T_m = 1000 K, ΔT = 100 K. ΔG_v = L_vΔT/T_m = 1.5 × 10⁸ J/m³, so r* = 2γ/ΔG_v = 0.40/(1.5 × 10⁸) = 2.67 × 10⁻⁹ m = 2.67 nm. Doubling ΔT halves r* and cuts ΔG* to one quarter.

5. Partitioning during binary solidification

In an alloy the first solid and the liquid have different compositions. The equilibrium partition coefficient k₀ = C_s/C_L at the interface, read from the solidus and liquidus, is below 1 when solute lowers the melting point (the solid rejects solute into the liquid) and above 1 when it raises it. The first solid to freeze from an alloy of composition C₀ has C_s = k₀C₀.

Two limiting solidification models
ModelAssumptionLiquid composition at solid fraction f_s
Equilibrium (lever rule)Complete diffusion in solid and liquidC_L = C₀ / [1 − f_s(1 − k₀)]
Scheil (non-equilibrium)No diffusion in solid, complete mixing in liquid, local equilibrium at the interfaceC_L = C₀(1 − f_s)^(k₀ − 1)

For k₀ < 1 the Scheil liquid becomes enriched without limit as f_s → 1, so the last liquid reaches the eutectic composition even in alloys that equilibrium cooling would freeze as single-phase solid; the result is coring (composition varies across each grain, lowest solute at the centre) and interdendritic eutectic. Real cooling lies between the two limits, closer to Scheil when cooling is fast and solid-state diffusion slow.

ℹ️ Worked: Scheil liquid
Take C₀ = 1.5 wt% solute, k₀ = 0.4, f_s = 0.75. Scheil: C_L = 1.5 × (0.25)^(−0.6) = 1.5 × 2.297 = 3.45 wt%. Lever rule: C_L = 1.5/(1 − 0.75 × 0.6) = 2.73 wt%. The Scheil liquid is richer because the solid that froze earlier cannot take solute back.

6. Solid-state transformations: precipitation, eutectoid, TTT and CCT

Precipitation from a supersaturated solid solution (quenched from the single-phase field and aged inside the two-phase field) proceeds by nucleation of a second phase, often on grain boundaries or dislocations, then diffusional growth; low-energy coherent interfaces make nucleation easier than incoherent ones. The eutectoid reaction γ → α + Fe₃C needs carbon to redistribute over the lamella spacing, so it is diffusional: the interlamellar spacing of pearlite shrinks as the transformation temperature falls, because larger undercooling means a larger driving force and shorter diffusion distances. Finer pearlite is harder.

Overall kinetics (nucleation plus growth) follow the Johnson–Mehl–Avrami equation: fraction transformed f = 1 − exp(−k tⁿ), an S-shaped curve with a slow start, a fast middle and a slow finish. The exponent n (typically 1 to 4) reflects the nucleation and growth mode, and k contains the temperature dependence through nucleation and growth rates. The time for half transformation is t₀.₅ = (ln 2/k)^(1/n). A TTT (time–temperature–transformation) diagram plots the start and finish of an isothermal transformation against log time: it is C-shaped because driving force rises and diffusivity falls as T drops. For a eutectoid steel, austenite transforms to pearlite above the nose and to bainite below it, and below the martensite start temperature M_s the diffusionless martensite forms. A CCT diagram is for continuous cooling and lies below and to the right of the TTT curve; cooling faster than the critical rate that just misses the nose avoids pearlite and leaves martensite.

ℹ️ Worked: Avrami fraction
Take n = 3 and k = 8 × 10⁻⁴ s⁻³. At t = 10 s, k tⁿ = 8 × 10⁻⁴ × 1000 = 0.8, so f = 1 − e^(−0.8) = 1 − 0.449 = 0.551, i.e. 55.1 %.

7. Martensitic transformation and the glass transition

The martensitic (displacive) transformation is a cooperative shear of the lattice: atoms move less than one interatomic distance, there is no diffusion, so the product has the same composition as the parent (in steel, carbon stays trapped in the new body-centred tetragonal lattice, a supersaturated and very hard phase). It is very fast, athermal (the amount formed depends on how far T is below M_s, not on time; it ends at M_f) and produces a plate or lath shape and a surface relief. Tempering later relieves the brittleness.

The glass transition occurs when a supercooled liquid, cooled fast enough to avoid crystallising, becomes so viscous that it falls out of equilibrium. At the glass transition temperature T_g the volume–temperature and enthalpy–temperature curves change slope and the heat capacity drops in a step, but there is no latent heat and no jump in volume, so it is not a first-order transition. T_g is a kinetic quantity: it moves to a higher temperature when the cooling rate is increased, and by convention the viscosity at T_g is about 10¹² Pa·s. Glasses are non-crystalline, lack long-range order and keep a residual entropy; metallic glasses need very rapid quenching, silicate and many polymer melts form glass easily.

⚠️ Martensite is not on the equilibrium diagram
Martensite is a metastable phase: it does not appear on the Fe–Fe₃C diagram, and it forms by shear without a change of composition. T_g is not a fixed constant of a material; it depends on cooling rate.

Key takeaways

  • Rate constants follow k = A exp(−Q/RT): ln k against 1/T has slope −Q/R; a first-order half-life is ln 2/k and independent of C₀.
  • Fick: J = −D dC/dx and ∂C/∂t = D ∂²C/∂x². Constant-surface problems use erf, a finite deposited amount uses the Gaussian; depth scales as √(Dt).
  • Interstitial diffusion is faster than vacancy diffusion; surfaces, grain boundaries and dislocation cores are fast paths that matter at low temperature.
  • Nucleation: r* = 2γ/ΔG_v ∝ 1/ΔT, ΔG* ∝ 1/ΔT²; heterogeneous nucleation multiplies ΔG* by S(θ) < 1 and leaves r* unchanged; the rate has a maximum with undercooling.
  • Scheil: C_L = C₀(1 − f_s)^(k₀ − 1); it gives more solute enrichment than the lever rule and explains coring and interdendritic eutectic.
  • Avrami f = 1 − exp(−k tⁿ); TTT is isothermal and C-shaped, CCT lies below and right of it; martensite is diffusionless and athermal; T_g depends on cooling rate.

Practice questions (17)

Attempt each one before opening the answer. Every explanation names the tempting wrong option as well as the right one, because that is where marks are lost.

  1. A reaction is first order with rate constant k = 0.0231 min⁻¹ (ln C₀/C = kt). The time needed for 75 % of the reactant to be consumed is closest to

    1. 30 min
    2. 45 min
    3. 60 min
    4. 120 min
    Show answer

    Answer: C — 60 min

    For 75 % consumed, C₀/C = 4, so t = ln 4/k = 1.386/0.0231 = 60 min, which is two half-lives (ln 2/k = 30 min each). 30 min is the time for 50 % only. 45 min is only about 65 % consumed and 120 min (four half-lives) about 94 % consumed.

  2. The diffusivity of an atom follows D = D₀ exp(−Q/RT). A plot of ln D (vertical) against 1/T (horizontal) is a straight line whose slope is

    1. +Q/R, so the line rises as the temperature is raised
    2. −Q, so the slope is the activation energy in joules per mole
    3. −R/Q, so a steeper line means a smaller activation energy
    4. −Q/R, so Q is found from the slope
    Show answer

    Answer: D — −Q/R, so Q is found from the slope

    Taking logs, ln D = ln D₀ − (Q/R)(1/T), so the slope is −Q/R (negative, since 1/T falls as T rises) and Q = −R × slope. A positive slope would hold for an axis of −1/T. A slope of −Q would ignore R and have the wrong units, and −R/Q is the reciprocal.

  3. For diffusion into a semi-infinite solid held at constant surface concentration at fixed temperature, the time required to reach a given concentration at twice the depth is

    1. twice the original time, because depth is proportional to time
    2. four times the original time, because depth varies as √(Dt)
    3. √2 times the original time, because time varies as √x
    4. eight times the original time, because diffusion is three-dimensional
    Show answer

    Answer: B — four times the original time, because depth varies as √(Dt)

    A given concentration corresponds to a fixed value of x/(2√(Dt)), so x² ∝ Dt and t ∝ x². Doubling x needs four times the time. Linear scaling with time is the common slip; the √2 option reverses the dependence, and the eight-fold option has no basis in the one-dimensional solution.

  4. A fixed quantity M per unit area of a tracer is deposited on the surface of a semi-infinite solid and diffuses in, giving C(x, t) = [M/√(πDt)] exp(−x²/4Dt). If the diffusion time is increased by a factor of 4 at the same temperature, the concentration at the surface (x = 0)

    1. falls to one quarter of its value
    2. falls to one half of its value
    3. stays equal to the value of the constant surface concentration
    4. rises to twice its value as the film spreads
    Show answer

    Answer: B — falls to one half of its value

    At x = 0, C = M/√(πDt) ∝ t^(−1/2). A factor 4 in time halves the surface concentration. Quarter would follow if C ∝ 1/t. A surface held constant belongs to the erf problem, not to a fixed deposit, and the total amount M is conserved while the peak falls.

  5. Which of the following statements about atomistic mechanisms and fast diffusion paths are correct?

    1. Substitutional self-diffusion by the vacancy mechanism needs a vacancy next to the jumping atom, so Q includes the vacancy formation energy
    2. Carbon in iron diffuses by hopping between interstitial sites and, at the same temperature, much faster than iron diffuses in iron
    3. Grain-boundary diffusion has a higher activation energy than lattice diffusion, so it matters most at high temperature
    4. Reducing the grain size of a polycrystal increases the share of the total flux carried by grain boundaries at low temperature
    Show answer

    Answer: A — Substitutional self-diffusion by the vacancy mechanism needs a vacancy next to the jumping atom, so Q includes the vacancy formation energy; B — Carbon in iron diffuses by hopping between interstitial sites and, at the same temperature, much faster than iron diffuses in iron; D — Reducing the grain size of a polycrystal increases the share of the total flux carried by grain boundaries at low temperature

    Vacancy diffusion has Q = Q_f + Q_m, interstitial atoms need no vacancy and so move much faster, and finer grains mean more boundary area, which raises the boundary share at low temperature. The grain-boundary statement is reversed: boundaries have a lower activation energy than the lattice, so they dominate at low temperature and are overtaken by lattice diffusion at high temperature.

  6. A steel with 0.2 wt% C is carburised at 1273 K with the surface held at 1.0 wt% C. Take D = D₀ exp(−Q/RT) with D₀ = 2.3 × 10⁻⁵ m²/s, Q = 148 kJ/mol and R = 8.314 J mol⁻¹ K⁻¹; use the error-function solution for a semi-infinite solid, with erf(z) = 0.5 at z = 0.477. The time needed for the carbon content to reach 0.6 wt% at a depth of 0.5 mm is ____ hours (to one decimal place).

    Numerical answer — type the value.

    Show answer

    Answer: 3.9

    D = 2.3 × 10⁻⁵ exp[−148 000/(8.314 × 1273)] = 1.94 × 10⁻¹¹ m²/s. (C_s − C_x)/(C_s − C₀) = 0.4/0.8 = 0.5 = erf(z), z = 0.477 = x/(2√(Dt)). t = x²/(4z²D) = 2.5 × 10⁻⁷/(4 × 0.2275 × 1.94 × 10⁻¹¹) = 1.41 × 10⁴ s = 3.9 h. Using the ratio 0.6/1.0 instead of 0.4/0.8 would be wrong.

  7. A diffusion treatment is done either at 900 °C (1173 K) or at 1000 °C (1273 K) to reach the same diffusion depth. The activation energy is Q = 140 kJ/mol, with R = 8.314 J mol⁻¹ K⁻¹ and D₀ unchanged. The treatment time at 900 °C is longer than at 1000 °C by a factor of ____ (to two decimal places).

    Numerical answer — type the value.

    Show answer

    Answer: 3.09

    For the same depth, x² ∝ Dt, so t₁₁₇₃/t₁₂₇₃ = D₁₂₇₃/D₁₁₇₃ = exp[(Q/R)(1/1173 − 1/1273)] = exp[16 839 × 6.697 × 10⁻⁵] = exp(1.1277) = 3.09. Using Celsius temperatures in the exponent would give a completely different number.

  8. In homogeneous nucleation of a solid from a melt, the undercooling ΔT is doubled. Taking ΔG_v ≈ L_vΔT/T_m, the critical radius r* and the nucleation barrier ΔG* change to

    1. r* halves and ΔG* falls to one quarter
    2. r* halves and ΔG* halves
    3. r* falls to one quarter and ΔG* falls to one eighth
    4. r* is unchanged and ΔG* halves
    Show answer

    Answer: A — r* halves and ΔG* falls to one quarter

    r* = 2γ/ΔG_v ∝ 1/ΔT, so it halves, and ΔG* = 16πγ³/(3ΔG_v²) ∝ 1/ΔT², so it falls to one quarter. Making ΔG* proportional to 1/ΔT is the usual slip. The critical radius certainly changes with ΔT in homogeneous nucleation; it is only unchanged when going from homogeneous to heterogeneous at the same ΔT.

  9. A pure metal melts at T_m = 1000 K. Take the solid–liquid interfacial energy γ = 0.20 J/m² and the latent heat of fusion per unit volume L_v = 1.5 × 10⁹ J/m³, with ΔG_v = L_vΔT/T_m. For an undercooling of ΔT = 100 K the critical nucleus radius for homogeneous nucleation is ____ nm (to two decimal places).

    Numerical answer — type the value.

    Show answer

    Answer: 2.67

    ΔG_v = 1.5 × 10⁹ × 100/1000 = 1.5 × 10⁸ J/m³. r* = 2γ/ΔG_v = 0.40/(1.5 × 10⁸) = 2.667 × 10⁻⁹ m = 2.67 nm. Omitting the factor 2 would give 1.33 nm.

  10. A nucleus forms as a spherical cap on a flat mould wall with contact angle θ = 90°. Using S(θ) = (2 + cos θ)(1 − cos θ)²/4, the ratio ΔG_het/ΔG_hom and the critical radius r* of the cap are

    1. 0.25, and r* is half of the homogeneous value
    2. 0.5, and r* is the same as the homogeneous value
    3. 0.75, and r* is the same as the homogeneous value
    4. 1.0, because a flat wall does not change the interface area
    Show answer

    Answer: B — 0.5, and r* is the same as the homogeneous value

    At θ = 90°, S = (2 + 0)(1 − 0)²/4 = 0.5. The barrier is halved because the cap has half the volume of the full sphere, but its curvature is still set by 2γ/ΔG_v, so r* is unchanged. S = 1 is the case θ = 180° (no wetting). A quarter would arise from squaring S.

  11. Below the melting point the homogeneous nucleation rate of a melt is found to increase with undercooling at first and then fall at very large undercooling. The reason for the fall is that

    1. the critical radius becomes infinite once the undercooling is large enough
    2. the barrier ΔG* rises steeply as the undercooling increases further
    3. the latent heat released always raises the melt temperature back to T_m
    4. atomic mobility, factor exp(−Q_d/kT), becomes too small
    Show answer

    Answer: D — atomic mobility, factor exp(−Q_d/kT), becomes too small

    The rate has one factor exp(−ΔG/kT), which rises with undercooling, and one factor exp(−Q_d/kT) for atomic transport across the interface, which falls as T falls. At large undercooling the second wins and the rate passes through a maximum. The barrier ΔG ∝ 1/ΔT² actually falls with undercooling, and r* decreases rather than becoming infinite.

  12. Which of the following statements about partitioning during binary solidification are correct, for an alloy with equilibrium partition coefficient k₀ < 1?

    1. The first solid to form from an alloy of composition C₀ has composition k₀C₀
    2. In the Scheil model the solid does not diffuse and the liquid is completely mixed
    3. At the same solid fraction the Scheil liquid is leaner in solute than the lever-rule liquid
    4. Scheil cooling can produce interdendritic eutectic in an alloy that is single-phase solid at equilibrium
    Show answer

    Answer: A — The first solid to form from an alloy of composition C₀ has composition k₀C₀; B — In the Scheil model the solid does not diffuse and the liquid is completely mixed; D — Scheil cooling can produce interdendritic eutectic in an alloy that is single-phase solid at equilibrium

    k₀C₀ is the first solid, and Scheil assumes no solid diffusion with a well-mixed liquid. Because earlier solid cannot return solute, the Scheil liquid is richer in solute than the lever-rule liquid, which is why it can reach the eutectic composition even for an alloy whose equilibrium state is single-phase solid. The statement that it is leaner is therefore wrong.

  13. An alloy of C₀ = 1.5 wt% solute with equilibrium partition coefficient k₀ = 0.4 solidifies under Scheil conditions (no solid diffusion, complete liquid mixing). When 75 % of the mass has solidified (f_s = 0.75) the liquid contains ____ wt% solute (to two decimal places).

    Numerical answer — type the value.

    Show answer

    Answer: 3.45

    Scheil: C_L = C₀(1 − f_s)^(k₀ − 1) = 1.5 × 0.25^(−0.6) = 1.5 × 2.297 = 3.45 wt%. Using the lever rule, C₀/[1 − f_s(1 − k₀)], would give 2.73 wt%.

  14. In a plain eutectoid steel, pearlite formed by isothermal transformation at a temperature just below 727 °C is compared with pearlite formed at a temperature near the nose of the TTT curve. The interlamellar spacing near the nose is

    1. larger, because the slower diffusion near the nose forces the lamellae to coarsen over long times
    2. the same at both temperatures, since the eutectoid composition alone fixes the spacing
    3. smaller, because larger undercooling gives more driving force and shorter diffusion paths
    4. zero, since bainite completely replaces pearlite as the product near the nose
    Show answer

    Answer: C — smaller, because larger undercooling gives more driving force and shorter diffusion paths

    The spacing falls as undercooling grows: the driving force rises and the carbon only has to move over short distances, so near the nose the pearlite is fine. Slower diffusion near the nose would not coarsen the structure; it is the high-temperature pearlite that is coarse. The spacing is not set by composition alone, and bainite forms below the nose, not at it.

  15. Which one of the following correctly describes the martensitic transformation in a plain carbon steel?

    1. It occurs by long-range carbon diffusion, and its amount grows steadily with holding time at a fixed temperature in the nose region
    2. It forms by a diffusionless shear, so the product has the composition of the parent austenite
    3. It is the equilibrium product of the eutectoid reaction and so appears as a field on the Fe–Fe₃C diagram
    4. It is a soft, ductile phase because the carbon atoms leave the lattice during the shear
    Show answer

    Answer: B — It forms by a diffusionless shear, so the product has the composition of the parent austenite

    Martensite forms by a diffusionless, athermal shear, so carbon stays trapped, giving a supersaturated, hard, body-centred tetragonal phase of the same composition as austenite. Diffusional growth with time dependence describes pearlite or bainite. It is a metastable phase absent from the equilibrium diagram, and because the carbon stays in the lattice it is hard rather than soft.

  16. An isothermal transformation follows the Avrami equation f = 1 − exp(−k tⁿ) with n = 3 and k = 8 × 10⁻⁴ s⁻³, where t is in seconds. The percentage of the material transformed after 10 s is ____ (to one decimal place).

    Numerical answer — type the value.

    Show answer

    Answer: 55.1

    k tⁿ = 8 × 10⁻⁴ × 10³ = 0.8; f = 1 − e^(−0.8) = 1 − 0.4493 = 0.5507, that is 55.1 %. Forgetting the exponential and quoting k tⁿ = 0.8 would give 80 %.

  17. Which of the following statements about the glass transition and continuous-cooling transformation are correct?

    1. On cooling a glass-forming liquid faster, the measured glass transition temperature T_g moves to a higher value
    2. At T_g there is a latent heat and a sudden jump in specific volume, as at a melting point
    3. A CCT curve for a given steel lies below and to the right of its TTT curve
    4. A TTT curve is C-shaped because driving force rises and diffusivity falls as the temperature is lowered
    Show answer

    Answer: A — On cooling a glass-forming liquid faster, the measured glass transition temperature T_g moves to a higher value; C — A CCT curve for a given steel lies below and to the right of its TTT curve; D — A TTT curve is C-shaped because driving force rises and diffusivity falls as the temperature is lowered

    T_g is kinetic and shifts upward with faster cooling; CCT is displaced down and to the right of TTT; the C shape is the competition between driving force and diffusivity. The statement about latent heat and a volume jump is wrong: at T_g the volume and enthalpy are continuous and only their slopes (expansion coefficient and heat capacity) change.