Entropy and Enthalpy of Heating Through Two Phase Transitions: 21-Mat-A1 Thermodynamics, December 2015, Q3

Question 3: Entropy and Enthalpy of Heating Through Two Phase Transitions (20 marks)

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Quantity	Value
$C_p$(solid A)	118.4 J mol⁻¹ K⁻¹
$C_p$(liquid A)	134.8 J mol⁻¹ K⁻¹
$C_p$(vapour A)	82.4 J mol⁻¹ K⁻¹
$\Delta H_{fus}$ at $T_{fus}=5.4\,{}^{\circ}\text{C}$ (278.55 K)	9.9 kJ mol⁻¹
$\Delta H_{vap}$ at $T_{vap}=25\,{}^{\circ}\text{C}$ (298.15 K)	33.9 kJ mol⁻¹

Find. $\Delta S$ and $\Delta H$ for each leg of the path solid ($-25\,{}^{\circ}\text{C}$) → fusion (5.4 $\,{}^{\circ}\text{C}$) → liquid → vaporization (25 $\,{}^{\circ}\text{C}$) → vapour (75 $\,{}^{\circ}\text{C}$), and the two totals.

Approach. Each single-phase leg uses $\Delta S=nC_p\ln(T_2/T_1)$ and $\Delta H=nC_p\Delta T$; each phase change (at constant $T$, reversible, 1 atm) uses $\Delta S=\Delta H_{trs}/T_{trs}$ and $\Delta H=\Delta H_{trs}$ directly. Sum all five legs for the two totals.

(a) Solid, $-25\,{}^{\circ}\text{C}\to5.4\,{}^{\circ}\text{C}$ (248.15 K → 278.55 K). $$\Delta S_a=C_{p,s}\ln\frac{278.55}{248.15}=118.4\ln(1.1225)=\boxed{13.7\ \text{J/K}}.$$
(b) Fusion at 5.4°C (278.55 K). $$\Delta S_b=\frac{\Delta H_{fus}}{T_{fus}}=\frac{9900}{278.55}=\boxed{35.5\ \text{J/K}}.$$
(c) Liquid, $5.4\,{}^{\circ}\text{C}\to25\,{}^{\circ}\text{C}$ (278.55 K → 298.15 K). $$\Delta S_c=C_{p,l}\ln\frac{298.15}{278.55}=134.8\ln(1.0704)=\boxed{9.2\ \text{J/K}}.$$
(d) Vaporization at 25°C (298.15 K). $$\Delta S_d=\frac{\Delta H_{vap}}{T_{vap}}=\frac{33{,}900}{298.15}=\boxed{113.7\ \text{J/K}}.$$
(e) Vapour, $25\,{}^{\circ}\text{C}\to75\,{}^{\circ}\text{C}$ (298.15 K → 348.15 K). $$\Delta S_e=C_{p,v}\ln\frac{348.15}{298.15}=82.4\ln(1.1677)=\boxed{12.8\ \text{J/K}}.$$
(f) Total entropy change. $$\Delta S_{total}=13.7+35.5+9.2+113.7+12.8=\boxed{184.9\ \text{J/K}}.$$
(g) Solid enthalpy, $-25\,{}^{\circ}\text{C}\to5.4\,{}^{\circ}\text{C}$. $$\Delta H_g=C_{p,s}\Delta T=118.4(30.4)=\boxed{3599\ \text{J}}=3.60\ \text{kJ}.$$
(h) Liquid enthalpy, $5.4\,{}^{\circ}\text{C}\to25\,{}^{\circ}\text{C}$. $$\Delta H_h=C_{p,l}\Delta T=134.8(19.6)=\boxed{2642\ \text{J}}=2.64\ \text{kJ}.$$
(i) Vapour enthalpy, $25\,{}^{\circ}\text{C}\to75\,{}^{\circ}\text{C}$. $$\Delta H_i=C_{p,v}\Delta T=82.4(50)=\boxed{4120\ \text{J}}=4.12\ \text{kJ}.$$
(j) Total enthalpy change. Summing all five legs including both latent heats: $$\Delta H_{total}=3599+9900+2642+33{,}900+4120=\boxed{54{,}161\ \text{J}}=54.2\ \text{kJ}.$$

Part	ΔS (J/K)	Part	ΔH (kJ)
(a) solid, −25→5.4°C	13.7	(g) solid, −25→5.4°C	3.60
(b) fusion, 5.4°C	35.5	—	9.90 (given)
(c) liquid, 5.4→25°C	9.2	(h) liquid, 5.4→25°C	2.64
(d) vaporization, 25°C	113.7	—	33.90 (given)
(e) vapour, 25→75°C	12.8	(i) vapour, 25→75°C	4.12
(f) Total	184.9	(j) Total	54.2

Concept & Theory

Topic. Entropy and enthalpy of heating a pure substance through solid→liquid→vapour, combining sensible-heat and latent-heat contributions.

Key relations. Sensible: $\Delta S=nC_p\ln(T_2/T_1)$, $\Delta H=nC_p\Delta T$. Reversible phase change at constant $T$, 1 atm: $\Delta S_{trs}=\Delta H_{trs}/T_{trs}$, $\Delta H=\Delta H_{trs}$.

Why this works. $S$ and $H$ are state functions, so the total change along the path is just the sum of the five legs regardless of how the path is subdivided; the phase-change legs use $\Delta S=\Delta H_{trs}/T_{trs}$ because a reversible transition at fixed $T$ and $P$ has $\Delta G=0$, i.e. $\Delta H=T\Delta S$ exactly at $T_{trs}$.

Common pitfall. Using Celsius instead of Kelvin inside $\ln(T_2/T_1)$ (the ratio is only meaningful on an absolute scale), and forgetting to convert the two transition temperatures to Kelvin before dividing $\Delta H_{trs}$ by $T_{trs}$.

Source. Gaskell, Introduction to the Thermodynamics of Materials, Ch. 5 (entropy of heating and of phase transformation).

Question 3 of 7: Entropy and Enthalpy of Heating Through Two Phase Transitions

Question 3: Entropy and Enthalpy of Heating Through Two Phase Transitions (20 marks)