Heat Balance (20 marks): 21-Mat-A3 Structure and Characterization of Materials, May 2015, Q7

Question 7 — 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.

Given data
Quantity	Symbol	Value
Charge mass	$m$	1 kg Cu
Initial temperature	$T_1$	20 °C (293 K)
Final temperature	$T_2$	1200 °C (1473 K)
Melting point	$T_m$	1083 °C (1356 K)
$C_p$, solid	—	$22.64+6.28\times10^{-3}T$ J K^-1 mol^-1
$C_p$, liquid	—	31.38 J K^-1 mol^-1
Latent heat of fusion	$L_f$	13,000 J mol^-1
Atomic weight	$M$	63.57 g mol^-1

Find. Total heat input $Q$ (J) to take 1 kg of copper from 20 °C to 1200 °C, assuming no heat losses.

Figure 7.1 — Heating path for the 1 kg copper charge: sensible heat in the solid, a constant-temperature plateau while it melts, then sensible heat in the liquid.

Approach. Split the path into three legs — heat the solid from 20 °C to the melting point, supply the latent heat of fusion, then heat the liquid to 1200 °C — and sum, converting the 1 kg charge to moles throughout since all the given $C_p$ and $L_f$ data are per mole.

Moles of copper. $$n=\frac{1000\ \text{g}}{63.57\ \text{g mol}^{-1}} = 15.73\ \text{mol}$$
Sensible heat, solid (293 K → 1356 K). Integrate the temperature-dependent $C_p$: $$Q_1 = n\int_{293}^{1356}\left(22.64+6.28\times10^{-3}T\right)dT = n\left[22.64\,\Delta T + \frac{6.28\times10^{-3}}{2}\left(T_m^2-T_1^2\right)\right]$$ $$Q_1 = 15.73\left[22.64(1063)+3.14\times10^{-3}(1356^2-293^2)\right]$$ $$\boxed{Q_1 \approx 465.2\ \text{kJ}}$$
Latent heat of fusion. $$Q_2 = nL_f = 15.73\times13{,}000$$ $$\boxed{Q_2 \approx 204.5\ \text{kJ}}$$
Sensible heat, liquid (1356 K → 1473 K). $C_p$ is constant here, so integration is a simple product: $$Q_3 = nC_{p,l}(T_2-T_m)=15.73\times31.38\times(1473-1356)$$ $$\boxed{Q_3 \approx 57.8\ \text{kJ}}$$
Total heat input. Sum the three legs: $$Q = Q_1+Q_2+Q_3 = 465.2+204.5+57.8$$ $$\boxed{Q \approx 727{,}400\ \text{J} \approx 727.4\ \text{kJ}}$$

Final results — Question 7
Leg	Heat (kJ)
Solid sensible heat, 20–1083 °C	465.2
Latent heat of fusion	204.5
Liquid sensible heat, 1083–1200 °C	57.8
Total heat input	≈ 727,400 J (727.4 kJ)

Topic. Sensible + latent heat balance across a solid-to-superheated-liquid heating path.

Key relations. $Q_{\text{sensible}}=n\int C_p\,dT$ (temperature-dependent $C_p$ integrates to a linear-plus-quadratic form in $T$); $Q_{\text{latent}}=nL_f$ at the phase-change temperature; total heat is additive across legs because enthalpy is a state function and the path (solid → melt → liquid) is the only physically realizable route between the given end states.

Why this works. Because $C_p$ for the solid is given as a function of $T$ rather than a constant, the sensible-heat integral must be evaluated analytically (or numerically) rather than as a simple $mc_p\Delta T$ product — using the solid's room-temperature $C_p$ value across the whole 293–1356 K range would understate $Q_1$, since $C_p$ rises with temperature. The fusion plateau is a fixed-temperature energy input with no temperature rise, which is why it appears as a flat step in the heating-curve figure rather than a sloped segment.

Common pitfall. The most common error is using $C_p$ in J K^-1 mol^-1 directly against a mass in kg without converting to moles first, or forgetting to convert °C to K before evaluating the temperature-dependent integral (only the span $\Delta T$ is insensitive to the choice of scale; the $T^2$ term in the integrated $C_p$ is not, so the melting point and final temperature must both be in kelvin before substituting into $Q_1$).

Source. D. R. Gaskell, Introduction to the Thermodynamics of Materials, Ch. 5 (heat capacity, integrated enthalpy) and Ch. 8 (latent heat of transformation).

Question 7 of 7: Heat Balance (20 marks)

Question 7 — Heat Balance (20 marks)