Heat Balance (20 marks): 21-Mat-A3 Structure and Characterization of Materials, December 2019, 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.

Note on the given mass. 55.85 kg of iron is exactly 1000 mol (the atomic weight is 55.85 g/mol), which is very likely a deliberately chosen round conversion factor rather than a coincidence — it is used below to keep the final scaling step a bare multiplication by 1000.

Given. 55.85 kg of iron are taken from 160 °C to 1735 °C, crossing three solid-state transformations and the melting point (1735 °C is 200 °C above melting, so the liquid is also superheated), 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)	160 → 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 → 1735 °C	42 J mol⁻¹ K⁻¹
Atomic mass / mass of iron	—	55.85 g/mol; 55.85 kg (= 1000 mol)

Find. The total enthalpy increase ΔH for 55.85 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 160 °C to 1735 °C, which now includes the liquid superheat above the 1535 °C melting point.

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 — including fusion, since the end temperature lies above the melting point; the molar total is then scaled to 1000 mol.

Convert every temperature to kelvin. $T=t+273.15$ gives $$433.15,\quad 1033.15,\quad 1183.15,\quad 1673.15,\quad 1808.15,\quad 2008.15\ \text{K}$$ for 160, 760, 910, 1400, 1535 and 1735 °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, 433.15 → 1033.15 K. With $a=17.5$, $b=24.8\times10^{-3}$, $$\Delta H_\alpha=17.5(1033.15-433.15)+\tfrac{24.8\times10^{-3}}{2}\left(1033.15^{2}-433.15^{2}\right)$$ $$\Delta H_\alpha=10\,500+0.0124\times\left(1\,067\,398-187{,}619\right)=10\,500+10\,909=21\,409\ \text{J/mol}$$
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 1735 °C. Fusion at 1808.15 K absorbs 15 680 J/mol, after which the melt is superheated the last 200 K to 2008.15 K: $$\Delta H_{liq}=42\times(2008.15-1808.15)=42\times200=8400\ \text{J/mol}$$
Sum the molar path. $$\Delta H_m=21\,409+2760+5655+920+17\,419+1180+5940+15\,680+8400$$ $$\boxed{\Delta H_m=79\,363\ \text{J/mol}=79.36\ \text{kJ/mol}}$$ Of this, 58 823 J/mol (74.1 %) is sensible heat and 20 540 J/mol (25.9 %) is latent heat.
Scale to 55.85 kilograms. Because the mass equals the atomic weight exactly scaled by 1000, $$n=\frac{55\,850\ \text{g}}{55.85\ \text{g/mol}}=1000\ \text{mol}$$ $$\Delta H=79\,363\times1000=7.9363\times10^{7}\ \text{J}$$ $$\boxed{\Delta H=79\,363\ \text{kJ}=79.36\ \text{MJ for 55.85 kg of iron}}$$

Final results — Question 6
Contribution	Value (J/mol)
Sensible heat, α-Fe (160 → 760 °C)	21 409
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 → 1735 °C)	8 400
Total per mole	79 363 (79.36 kJ/mol)
Total for 55.85 kg (1000 mol)	79 363 000 J = 79 363 kJ = 79.36 MJ

Check: liquid superheat above melting. Because 1735 °C exceeds the 1535 °C melting point by 200 °C, the path must include BOTH the full 15 680 J/mol fusion enthalpy AND a genuine liquid sensible-heat term ($42\times200=8400$ J/mol) — a solution that stops at the melting point, or that treats 1735 °C as if it were still solid, would understate the answer by more than 24 000 J/mol.

Topic. Sensible and latent heat balances across allotropic transformations and melting.

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. Iron is a particularly instructive example because it changes crystal structure three times before it even melts (BCC α → FCC γ → BCC δ), and an end temperature above the melting point adds a fourth event, fusion itself.

Common pitfall. The most common error is applying one phase's $C_p$ expression across a transformation boundary instead of switching functions and adding the latent heat at that temperature. The second is stopping the integration at the melting point and forgetting the liquid superheat term, or conversely forgetting to add the fusion enthalpy itself.

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)