Heat Balance (20 marks): 21-Mat-A3 Structure and Characterization of Materials, December 2013, Q2

Question 2 — Heat Balance (20 marks)

Question text not reproduced: the examination questions are © Engineers and Geoscientists BC. Open the official past paper (linked at the top of this page) to read the question, then follow the worked solution below.

Given. One kilogram of iron is taken from 100 °C to 1600 °C, crossing three solid-state transformations and the melting point, with a heat capacity supplied for every phase along the way.

Given data
Interval or event	Temperature range	C_p or ΔH
α-Fe (BCC, ferrite)	100 → 760 °C	17.5 + 24.8 × 10⁻³T J mol⁻¹ K⁻¹
α → β transformation	760 °C	2760 J/mol
β-Fe	760 → 910 °C	37.7 J mol⁻¹ K⁻¹
β → γ transformation	910 °C	920 J/mol
γ-Fe (FCC, austenite)	910 → 1400 °C	7.7 + 19.5 × 10⁻³T J mol⁻¹ K⁻¹
γ → δ transformation	1400 °C	1180 J/mol
δ-Fe (BCC)	1400 → 1535 °C	44 J mol⁻¹ K⁻¹
Melting	1535 °C	15 680 J/mol
Liquid Fe	1535 → 1600 °C	42 J mol⁻¹ K⁻¹
Atomic mass of iron	—	55.85 g/mol

Find. The total enthalpy increase ΔH for 1 kg of iron over the stated interval, in kJ.

Figure 2.1 — The enthalpy path for one mole of iron. Sloped runs are sensible heat computed by integrating C_p; the four vertical risers are the latent heats absorbed at fixed temperature. The largest single step is melting, at 15.68 kJ/mol.

Approach. Enthalpy is a state function, so the change is the sum of the sensible heat in each phase field, obtained by integrating its own C_p between the bounding temperatures, plus the latent heat of every transformation crossed; the molar total is then scaled to one kilogram through the atomic mass.

Convert every temperature to kelvin. The heat capacities are written with $T$ in kelvin, so all limits must be converted before integrating: $T=t+273.15$, giving $$373.15,\quad 1033.15,\quad 1183.15,\quad 1673.15,\quad 1808.15,\quad 1873.15\ \text{K}$$ for 100, 760, 910, 1400, 1535 and 1600 °C respectively.
Write the general sensible-heat integral. For a phase with $C_p=a+bT$ heated from $T_1$ to $T_2$ at constant pressure, $$\Delta H=\int_{T_1}^{T_2}C_p\,dT=a\,(T_2-T_1)+\tfrac{b}{2}\left(T_2^{2}-T_1^{2}\right)$$ which reduces to $a(T_2-T_1)$ whenever $b=0$. Every sloped run in Figure 2.1 is one application of this expression.
Heat the α phase, 373.15 → 1033.15 K. With $a=17.5$ and $b=24.8\times10^{-3}$, $$\Delta H_\alpha=17.5(1033.15-373.15)+\tfrac{24.8\times10^{-3}}{2}\left(1033.15^{2}-373.15^{2}\right)$$ $$\Delta H_\alpha=11\,550+0.0124\times928\,158=11\,550+11\,509=23\,059\ \text{J/mol}$$ This one interval is nearly a third of the whole answer, because it is by far the widest temperature span.
Add the α → β latent heat and heat the β phase. At 1033.15 K the structure absorbs $\Delta H_{trf}=2760$ J/mol at constant temperature. Then, from 1033.15 to 1183.15 K with a constant $C_p=37.7$, $$\Delta H_\beta=37.7\times(1183.15-1033.15)=37.7\times150=5655\ \text{J/mol}$$
Add the β → γ latent heat and heat the γ phase. The transformation at 1183.15 K contributes 920 J/mol. Integrating the austenite capacity from 1183.15 to 1673.15 K with $a=7.7$, $b=19.5\times10^{-3}$, $$\Delta H_\gamma=7.7\times490+\tfrac{19.5\times10^{-3}}{2}\left(1673.15^{2}-1183.15^{2}\right)=3773+13\,646=17\,419\ \text{J/mol}$$
Add the γ → δ latent heat and heat the δ phase to the melting point. The 1400 °C transformation contributes 1180 J/mol, and the delta ferrite is then taken from 1673.15 to 1808.15 K: $$\Delta H_\delta=44\times(1808.15-1673.15)=44\times135=5940\ \text{J/mol}$$
Melt the iron and superheat the liquid. Fusion at 1808.15 K absorbs 15 680 J/mol — the single largest term — after which the melt is superheated the last 65 K to 1600 °C: $$\Delta H_{liq}=42\times(1873.15-1808.15)=42\times65=2730\ \text{J/mol}$$
Sum the molar path. Adding the five sensible terms and the four latent terms, $$\Delta H_m=23\,059+2760+5655+920+17\,419+1180+5940+15\,680+2730$$ $$\boxed{\Delta H_m=75\,343\ \text{J/mol}=75.34\ \text{kJ/mol}}$$ Of this, 54 803 J/mol (72.7 %) is sensible heat and 20 540 J/mol (27.3 %) is latent heat — a useful reminder that the phase changes are not a footnote in a furnace heat balance.
Scale to one kilogram. One kilogram of iron is $$n=\frac{1000\ \text{g}}{55.85\ \text{g/mol}}=17.905\ \text{mol}$$ so the enthalpy required is $$\Delta H=75\,343\times17.905=1.349\times10^{6}\ \text{J}$$ $$\boxed{\Delta H=1349\ \text{kJ per kg of iron}=1.349\ \text{MJ/kg}}$$

Final results — Question 2
Contribution	Value (J/mol)
Sensible heat, α-Fe (100 → 760 °C)	23 059
Latent heat, α → β	2 760
Sensible heat, β-Fe (760 → 910 °C)	5 655
Latent heat, β → γ	920
Sensible heat, γ-Fe (910 → 1400 °C)	17 419
Latent heat, γ → δ	1 180
Sensible heat, δ-Fe (1400 → 1535 °C)	5 940
Latent heat of fusion at 1535 °C	15 680
Sensible heat, liquid Fe (1535 → 1600 °C)	2 730
Total per mole	75 343 (75.34 kJ/mol)
Total per kilogram (17.905 mol)	1 349 000 J = 1349 kJ/kg

Check: temperature-scale convention. The heat capacities are quoted per kelvin with $T$ appearing explicitly, so the integration limits must be absolute temperatures. Using $T=t+273.15$ rather than the older $t+273$ shifts the total by about 6 J/mol (less than 0.01 %), so the answer is insensitive to the choice; what is not insensitive is integrating in degrees Celsius, which would understate $\Delta H_\alpha$ alone by roughly 9 kJ/mol.

Topic. Enthalpy path calculations across phase transformations — the arithmetic behind every furnace heat balance.

Key relations. $\Delta H=\int C_p\,dT$ within a phase; $\Delta H=\Delta H_{trf}$ at a transformation, taken at constant temperature; Kirchhoff's construction, in which the total is path-independent so the intervals may be summed in any order; $n=m/M$ to move between molar and mass bases.

Why this works. Enthalpy is a state function, which means the change between 100 °C and 1600 °C depends only on those two end states and not on how the metal got there. That licenses the standard device of replacing the real, continuous heating with a chain of idealised steps — heat at constant structure, transform at constant temperature, heat again — whose individual enthalpies are all tabulated. Each transformation appears as a vertical riser in Figure 2.1 because the latent heat is absorbed with no temperature rise: the energy goes into rearranging the lattice, not into raising the mean atomic vibration. Summing the risers and the ramps reconstructs the true enthalpy difference exactly.

Common pitfall. The most common error is integrating $C_p=a+bT$ with the limits in degrees Celsius; because the $bT$ term is linear in absolute temperature, this silently removes a 273 K offset from every interval and produces a badly low answer. A close second is forgetting one of the four latent heats — the β-Fe field is only 150 K wide and its two bounding transformations are easy to skip — which would lose 3.7 kJ/mol here and, in an industrial balance, badly undersize the burner.

Source. Gaskell, Introduction to the Thermodynamics of Materials, Ch. 6 (heat capacity, enthalpy and entropy); Rosenqvist, Principles of Extractive Metallurgy, Ch. 4 (heat balances in metallurgical reactors).

Question 2 of 7: Heat Balance (20 marks)

Question 2 — Heat Balance (20 marks)