Heat Balance (20 marks): 21-Mat-A3 Structure and Characterization of Materials, December 2016, Q6

Question 6 — 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.

Data set. The C_p expressions and transformation enthalpies are the iron thermodynamic data used for this heat balance; the mass is 5 kg and the temperature endpoints are 60–1635 °C. The β and γ sensible-heat terms below cover exactly the 760–910 °C and 910–1400 °C spans.

Given. Five kilograms of iron are taken from 60 °C to 1635 °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)	60 → 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 → 1635 °C	42 J mol⁻¹ K⁻¹
Atomic mass of iron	—	55.85 g/mol
Mass of iron	—	5 kg

Find. The total enthalpy increase ΔH for 5 kg of iron over the stated interval, in kJ and MJ.

Figure 6.1 — The enthalpy path for iron, phase by phase. Each box is a sensible-heat integral of its own C_p; the labelled arrows between boxes are the fixed-temperature latent heats crossed on the way from 60 °C to 1635 °C.

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 5 kg through the atomic mass.

Convert every temperature to kelvin. $T=t+273.15$ gives $$333.15,\quad 1033.15,\quad 1183.15,\quad 1673.15,\quad 1808.15,\quad 1908.15\ \text{K}$$ for 60, 760, 910, 1400, 1535 and 1635 °C respectively.
Write the general sensible-heat integral. For $C_p=a+bT$ heated from $T_1$ to $T_2$, $$\Delta H=\int_{T_1}^{T_2}C_p\,dT=a\,(T_2-T_1)+\tfrac{b}{2}\left(T_2^{2}-T_1^{2}\right)$$ reducing to $a(T_2-T_1)$ when $b=0$.
Heat the α phase, 333.15 → 1033.15 K. With $a=17.5$, $b=24.8\times10^{-3}$, $$\Delta H_\alpha=17.5(1033.15-333.15)+\tfrac{24.8\times10^{-3}}{2}\left(1033.15^{2}-333.15^{2}\right)$$ $$\Delta H_\alpha=12\,250+0.0124\times944{,}316=12\,250+11\,859=24\,109\ \text{J/mol}$$ This is the widest temperature span of the path (700 K) and the single largest term.
Add the α → β latent heat and heat the β phase. At 1033.15 K, 2760 J/mol is absorbed. From 1033.15 to 1183.15 K at 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. 920 J/mol at 1183.15 K, then integrating $a=7.7$, $b=19.5\times10^{-3}$ from 1183.15 to 1673.15 K, $$\Delta H_\gamma=7.7\times490+\tfrac{19.5\times10^{-3}}{2}\left(1673.15^{2}-1183.15^{2}\right)=3773+13\,646=17\,419\ \text{J/mol}$$
Add the γ → δ latent heat and heat δ-Fe to the melting point. 1180 J/mol at 1673.15 K, then $\Delta H_\delta=44\times(1808.15-1673.15)=44\times135=5940\ \text{J/mol}$.
Melt the iron and superheat the liquid to 1635 °C. Fusion at 1808.15 K absorbs 15 680 J/mol, after which the melt is superheated the last 100 K to 1908.15 K: $$\Delta H_{liq}=42\times(1908.15-1808.15)=42\times100=4200\ \text{J/mol}$$
Sum the molar path. $$\Delta H_m=24\,109+2760+5655+920+17\,419+1180+5940+15\,680+4200$$ $$\boxed{\Delta H_m=77\,863\ \text{J/mol}=77.86\ \text{kJ/mol}}$$ Of this, 57 324 J/mol (73.6 %) is sensible heat and 20 540 J/mol (26.4 %) is latent heat.
Scale to 5 kilograms. $$n=\frac{5000\ \text{g}}{55.85\ \text{g/mol}}=89.53\ \text{mol}$$ $$\Delta H=77\,863\times89.53=6.971\times10^{6}\ \text{J}$$ $$\boxed{\Delta H=6971\ \text{kJ}=6.971\ \text{MJ for 5 kg of iron}}$$

Final results — Question 6
Contribution	Value (J/mol)
Sensible heat, α-Fe (60 → 760 °C)	24 109
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 → 1635 °C)	4 200
Total per mole	77 863 (77.86 kJ/mol)
Total for 5 kg (89.53 mol)	6 971 000 J = 6971 kJ = 6.971 MJ

Check: temperature-scale convention. Using $T=t+273.15$ rather than $t+273$ shifts the molar total by well under 0.01 %, so the answer is insensitive to that choice. What is not insensitive is integrating in Celsius rather than kelvin, which would understate $\Delta H_\alpha$ alone by roughly 9 kJ/mol given the 700 K span it covers.

Topic. Sensible and latent heat balances across allotropic transformations.

Key relations. $\Delta H=\int C_p\,dT=a(T_2-T_1)+\tfrac{b}{2}(T_2^2-T_1^2)$; enthalpy as a path-independent state function, so the total is the sum of sequential sensible and latent terms regardless of how the path is broken up.

Why this works. Because $H$ is a state function, the calculation can be split at every transformation temperature without approximation — each phase field is integrated with its own $C_p$, and each latent heat is added exactly once at the temperature it occurs, with no need to know anything about the path taken to get there. Iron is a particularly instructive example because it changes crystal structure three times before it even melts (BCC α → FCC γ → BCC δ), and the terms that depend only on transformation-temperature spacing (β and γ) are unchanged by moving the overall start and end points, while the terms touching the endpoints (α and the liquid) change with them.

Common pitfall. The most common error is applying one of the phase's $C_p$ expressions across a transformation boundary instead of switching functions and adding the latent heat at that temperature. The second is treating the latent heats as negligible; here they are 26 % of the total, which is not a rounding error.

Source. Gaskell, Introduction to the Thermodynamics of Materials, Ch. 5 (heat capacity and enthalpy) and Ch. 8 (phase transformations); Habashi, Textbook of Pyrometallurgy, Ch. 2 (heat balances).

Question 6 of 7: Heat Balance (20 marks)

Question 6 — Heat Balance (20 marks)