Heat Balance (20 marks): 21-Mat-A3 Structure and Characterization of Materials, May 2018, 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.

Given. 2 kg of copper heated from 25 °C to 1100 °C, crossing its 1083 °C melting point, with separate solid and liquid heat capacities and a single latent heat of fusion.

Given data
Quantity	Symbol	Value
Mass of copper	$m$	2 kg (2000 g)
Atomic weight	$M$	63.57 g/mol
Start / end temperature	$T_1,\,T_2$	25 °C, 1100 °C
Melting point	$T_m$	1083 °C
$C_p$, solid Cu	$C_{p,s}$	$22.64+6.28\times10^{-3}T$ J/(mol·K)
$C_p$, liquid Cu	$C_{p,l}$	31.38 J/(mol·K) (constant)
Latent heat of fusion	$L_f$	13 000 J/mol

Find. The total heat input (in J) needed to take the 2 kg copper charge from 25 °C to 1100 °C.

Figure 6.1 — The three-stage heating path (schematic; segment widths are illustrative, not to scale): sensible heat in the solid, a constant-temperature melting plateau, then sensible heat in the liquid.

Approach. Split the path into three legs at the melting point — sensible heat in the solid, latent heat of fusion, sensible heat in the liquid — sum them to a molar enthalpy, then scale by the number of moles in the 2 kg charge.

Convert to absolute temperature. $T_1=25+273.15=298.15$ K, $T_m=1083+273.15=1356.15$ K, $T_2=1100+273.15=1373.15$ K.
Sensible heat in the solid, $T_1\to T_m$. Integrating the temperature-dependent $C_p$, $$\Delta H_{solid}=\int_{T_1}^{T_m}(22.64+6.28\times10^{-3}T)\,dT =22.64(T_m-T_1)+\frac{6.28\times10^{-3}}{2}(T_m^2-T_1^2)$$ $$=22.64(1058.0)+3.14\times10^{-3}(1\,839\,142.8-88\,893.4)=23\,953.1+5495.8$$ $$\Delta H_{solid}=29\,448.9\ \text{J/mol}$$
Latent heat of fusion. Given directly: $$\Delta H_{fus}=13\,000\ \text{J/mol}$$
Sensible heat in the liquid, $T_m\to T_2$. With constant $C_{p,l}$, $$\Delta H_{liquid}=C_{p,l}(T_2-T_m)=31.38\times(1373.15-1356.15)=31.38\times17.0$$ $$\Delta H_{liquid}=533.5\ \text{J/mol}$$
Sum to a molar enthalpy and pivot to the boxed result. $$\Delta H_{molar}=\Delta H_{solid}+\Delta H_{fus}+\Delta H_{liquid}=29\,448.9+13\,000+533.5$$ $$\boxed{\Delta H_{molar}=42\,982\ \text{J/mol}}$$ Fusion alone accounts for $13\,000/42\,982=30.2\ \%$ of the total — a reminder that the isothermal latent step is not a minor correction next to sensible heating, even though it spans no temperature change at all.
Scale to the 2 kg charge. The number of moles is $$n=\frac{m}{M}=\frac{2000\ \text{g}}{63.57\ \text{g/mol}}=31.46\ \text{mol}$$ so the total heat input is $$Q=n\,\Delta H_{molar}=31.46\times42\,982$$ $$\boxed{Q\approx1.352\times10^{6}\ \text{J}=1352\ \text{kJ}=1.352\ \text{MJ}}$$

Final results — Question 6
Quantity	Result
Sensible heat, solid Cu (per mol)	29 448.9 J/mol
Latent heat of fusion (per mol)	13 000 J/mol (30.2 % of the molar total)
Sensible heat, liquid Cu (per mol)	533.5 J/mol
Molar enthalpy change	42 982 J/mol
Moles in 2 kg charge	31.46 mol
Total heat input, $Q$	1.352 × 10⁶ J (1352 kJ)

Topic. Sensible and latent heat in a metal heating balance.

Key relations. $\Delta H=\int C_p\,dT$ per phase; $\Delta H_{fus}$ added as a discrete isothermal term at $T_m$; $n=m/M$; $Q=n\,\Delta H_{molar}$.

Why this works. Enthalpy is a state function, so the total heat needed to go from 25 °C solid copper to 1100 °C liquid copper can be built up leg by leg — sensible heat does not "know about" the phase change ahead of it, and the fusion term does not "know about" temperature at all, it only knows the material crossed its melting point. Adding the three legs, rather than trying to write one single $C_p$ expression across the whole range, is the only way to handle a discontinuous property (here, the fact that $C_p$ itself jumps from the polynomial solid form to a constant 31.38 at $T_m$, on top of the separate latent-heat spike).

Common pitfall. The most common error is integrating the solid $C_p$ expression straight through to 1100 °C and forgetting the fusion term altogether, which understates $Q$ by roughly 30 %. The second is using Celsius directly in the $\int T\,dT$ term of the quadratic $C_p$ integral — the $T^2$ dependence means the integration must be carried out in kelvin, even though $(T_2-T_1)$ differences are the same in either scale.

Source. Gaskell, Introduction to the Thermodynamics of Materials, Ch. 5 (heat capacity and enthalpy of phase change); Habashi, Textbook of Pyrometallurgy, Ch. 3 (heat balances).

Question 6 of 7: Heat Balance (20 marks)

Question 6 — Heat Balance (20 marks)