Physics / Solidification Cooling Curve
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EXPERIMENT 26 · PHYSICS LABORATORY

Solidification Cooling Curve theory

Understand solidification cooling curve: the equations, assumptions, supporting concepts and measurement method.

THE WORKING RELATIONSHIP
Q_latent = R_loss · Δt = m · L_f ⇔ Y = L_f · X ⇔ L_f = S

Keep the physical quantities, graph variables and units consistent.

01First-Order Phase Transitions and Latent Heat of Fusion

In classical thermodynamics, phase transitions are classified according to the Ehrenfest taxonomy by the lowest derivative of the Gibbs free energy G that exhibits a discontinuity at the transition point. For a first-order phase change such as solid-liquid melting or freezing, the chemical potential μ and Gibbs free energy are continuous across the phase boundary, but the first partial derivatives—entropy S = −(∂G/∂T)_P and volume V = (∂G/∂P)_T—undergo finite step discontinuities. The discontinuity in entropy ΔS_trans = S_liquid − S_solid corresponds directly to the absorption or liberation of a finite quantity of thermal energy at constant temperature: the latent heat of transition Q_latent = T_m ΔS_trans. Per unit mass, this is the specific latent heat of fusion L_f = Q_latent / m. During solidification, disordered molecular bonds in the molten state collapse into an ordered crystalline lattice, shedding intermolecular potential energy into thermal agitation. Because this latent enthalpy release exactly compensates for external heat loss, the coexisting two-phase system possesses zero thermodynamic degrees of freedom under constant pressure (Gibbs phase rule: F = C − P + 1 = 1 − 2 + 1 = 0), holding the temperature rigorously constant at T_m until phase transformation is complete.

L_f = \frac{Q_{\text{latent}}}{m} = T_m \Delta s_{\text{trans}} \quad \text{and} \quad F = C - P + 1 = 0
Why does the temperature of a pure substance remain constant during solidification even while losing heat to the surroundings?

Solidification is a first-order phase transition where latent heat of fusion is liberated as molecules transition from disordered liquid to an ordered crystalline lattice. This internal enthalpy release exactly balances the heat lost to the surroundings, keeping the temperature constant until the entire liquid mass has crystallized.

02The Cooling Curve Method: Energy Balance and Rate of Heat Loss

The determination of latent heat by the method of cooling relies upon establishing an energy conservation balance for a body cooling inside a constant-temperature enclosure. Consider a boiling tube containing mass m of a substance possessing liquid specific heat c_L, solid specific heat c_S, and latent heat of fusion L_f, held in an enclosure maintained at ambient temperature T_0. When the contents are liquid at temperature T > T_m, sensible heat is lost at rate dQ/dt = (m c_L + W_tube) (dT/dt) = −R_loss(T). During the solidification arrest at T = T_m, sensible temperature change ceases (dT/dt = 0). The rate of latent heat liberation dm_solid/dt · L_f must balance the external heat loss rate R_loss(T_m). Integrating over the complete duration of the plateau Δt yields the total latent heat: Q_latent = ∫₀^{Δt} R_loss(T_m) dt = R_loss(T_m) · Δt = m · L_f. If the heat loss rate R_loss at the melting point is independently calibrated (or determined from the sensible cooling slopes immediately preceding and following the arrest), the latent heat of fusion evaluates simply as L_f = (R_loss · Δt) / m. Plotting Q_latent = R_loss · Δt on the vertical axis against mass m on the horizontal axis yields a linear plot passing through the origin with slope equal to L_f.

Q_{\text{latent}} = R_{\text{loss}}(T_m) \cdot \Delta t = m \cdot L_f \iff \Delta t = \left(\frac{L_f}{R_{\text{loss}}}\right) m
How does the duration of the solidification plateau relate to the mass of the substance and the rate of heat loss?

The plateau duration Δt is directly proportional to the sample mass m and inversely proportional to the ambient heat loss rate R_loss at the melting point: Δt = (m · L_f) / R_loss. A larger mass contains more latent enthalpy and requires a longer time to solidify at a given rate of heat dissipation.

03Supercooling, Nucleation Kinetics, and Metastable States

A ubiquitous phenomenon encountered in cooling curves of liquids is supercooling (undercooling), wherein the liquid temperature dips significantly below the thermodynamic melting point T_m without crystallizing. Classical nucleation theory explains that the formation of a microscopic crystalline embryo of radius r inside a homogeneous liquid requires creating a solid-liquid interface with interfacial surface energy γ. The total change in Gibbs free energy to form an embryo is ΔG(r) = 4π r² γ − (4/3)π r³ Δg_v, where Δg_v is the volumetric thermodynamic driving force, which is proportional to the undercooling degree ΔT = T_m − T. For small undercooling, the positive interfacial energy term dominates, creating a free energy barrier ΔG* at a critical nucleus radius r* = 2γ / Δg_v. Embryos smaller than r* spontaneously redissolve. Consequently, a clean, undisturbed liquid remains trapped in a metastable state. Once a critical nucleus forms spontaneously or is introduced via artificial seeding, crystal growth proceeds rapidly. The rapid liberation of latent heat of fusion heats the surrounding liquid back up to the equilibrium transition temperature T_m, producing a characteristic sharp dip followed by an immediate temperature recovery to the horizontal plateau.

\Delta G^* = \frac{16\pi \gamma^3}{3 (\Delta g_v)^2} \quad \text{and} \quad r^* = \frac{2\gamma}{\Delta g_v} \propto \frac{1}{T_m - T}
What causes supercooling, and why does the temperature rise back to the melting point once crystallization begins?

Supercooling occurs because forming initial microscopic crystal nuclei requires overcoming an interfacial energy barrier. In the absence of nuclei, the liquid remains metastable below T_m. Once nucleation is triggered, rapid crystal growth liberates latent heat faster than it can escape to the surroundings, warming the mixture back to the equilibrium melting point T_m.

04Contrasting Crystalline Solids and Amorphous Glass Formers

The cooling curve provides a decisive diagnostic test to differentiate true crystalline materials from amorphous or vitreous substances. In a crystalline material like salol (phenyl salicylate), naphthalene, or water, the molecules possess long-range periodic translational order. The solid-liquid transition occurs at a sharply defined thermodynamic temperature T_m, manifesting on the cooling curve as an abrupt change in slope from negative to zero, forming an extended horizontal plateau. In contrast, amorphous substances like paraffin wax, glass, or pitch lack long-range periodic order. As molten paraffin wax cools, its long, tangled alkane chains experience increasing van der Waals attractions and steric hindrance, leading to a continuous, progressive increase in dynamic viscosity. Rather than a sharp first-order phase arrest, the cooling curve of paraffin displays a continuous, smoothly downward-sloping sigmoidal trajectory with only a slight inflection marking the glass transition or gradual softening range. Comparing the cooling curves of salol and paraffin wax vividly demonstrates the distinction between true latent heat release at a singular melting point and continuous vitrification without latent enthalpy discontinuity.

\text{Crystalline: } \left(\frac{dT}{dt}\right)_{T_m} = 0 \quad \text{vs.} \quad \text{Amorphous: } \left(\frac{dT}{dt}\right) < 0 \text{ everywhere}
How does the cooling curve of a crystalline substance differ fundamentally from that of an amorphous substance like wax?

A crystalline substance exhibits a distinct horizontal arrest plateau (dT/dt = 0) at its exact melting point due to the liberation of latent heat. An amorphous substance has no fixed melting point or latent heat; its cooling curve slopes continuously downward through a broad softening and glass transition zone.

05Calibration of Heat Loss Rate R_loss from Pre- and Post-Arrest Slopes

To evaluate the latent heat of fusion quantitatively without relying on an external electrical calibration heater, the heat loss rate R_loss at the melting point can be extracted directly from the cooling curve geometry. Just prior to the onset of crystallization, the liquid of mass m and specific heat c_L cools with instantaneous slope (dT/dt)_L at T = T_m. The rate of heat dissipation at this instant is R_loss = (m c_L + W_tube) |dT/dt|_L. Similarly, immediately following complete solidification, the solid substance of specific heat c_S cools with slope (dT/dt)_S, giving R_loss = (m c_S + W_tube) |dT/dt|_S. Averaging these two independent estimates accounts for minor differences between liquid and solid specific heat capacities: R_loss(T_m) ≈ ½ [ (m c_L + W_tube) |dT/dt|_L + (m c_S + W_tube) |dT/dt|_S ]. For standardized boiling tubes of salol in our apparatus configuration, the total rate of radiative and convective heat loss at 41.5 °C into ambient air at 20.0 °C is R_loss = 270.0 J min⁻¹ (4.50 W). This calibration allows the measured plateau duration Δt (min) to be converted directly into total latent enthalpy Q_latent = R_loss · Δt.

R_{\text{loss}}(T_m) = (m c_L + W_{\text{tube}}) \left|\frac{dT}{dt}\right|_L \approx 270.0 \text{ J min}^{-1}
How can the rate of heat loss R_loss at the freezing point be determined from the cooling curve itself?

By measuring the cooling rate (dT/dt) immediately before the plateau begins (in the liquid phase) and immediately after it ends (in the solid phase). Multiplying these slopes by the respective heat capacities of the tube and contents gives the instantaneous heat loss rate at T_m.

06Chemical Properties, Purity, and Physical Constants of Salol

Salol, the common chemical name for phenyl salicylate (C₁₃H₁₀O₃, molecular weight 214.22 g mol⁻¹), is an ester synthesized by the condensation of salicylic acid with phenol. It serves as an exemplary teaching substance for phase change calorimetry due to several advantageous physical properties: it is non-toxic, chemically stable, non-hygroscopic, and possesses a convenient melting point of T_m = 41.5 °C (314.65 K), which is safely above ambient room temperature (20 °C) yet low enough to handle without hazardous heating equipment. Its accepted literature value for the latent heat of fusion is L_f = 79.50 kJ kg⁻¹ (79,500 J kg⁻¹ = 17.03 kJ mol⁻¹). Salol has a liquid density of approximately 1,250 kg m⁻³ and a specific heat capacity of c_L ≈ 1,420 J kg⁻¹ K⁻¹ in the liquid phase and c_S ≈ 1,250 J kg⁻¹ K⁻¹ in the solid crystalline phase. Its pronounced propensity to supercool makes it ideal for demonstrating nucleation seeding and the sharp recovery to the equilibrium phase arrest plateau.

\text{Salol: } \text{C}_{13}\text{H}_{10}\text{O}_3, \; M = 214.22 \text{ g/mol}, \; T_m = 41.50 \text{ }^\circ\text{C}, \; L_f = 79.50 \text{ kJ/kg}
Why is phenyl salicylate (salol) particularly well-suited for university laboratory demonstrations of cooling curves and latent heat?

Salol has a convenient melting point (41.5 °C) easily accessible with a warm water bath, is non-toxic and chemically stable, possesses a well-documented latent heat of fusion (79.5 kJ/kg), and exhibits clean supercooling and a flat, distinct crystallization plateau.

07Systematic Uncertainties, Incomplete Crystallization, and Radial Gradients

A rigorous analysis of cooling curve calorimetry must account for non-ideal transport mechanisms that perturb the idealized isothermal arrest plateau. In a cylindrical boiling tube, heat escapes radially through the cylindrical glass perimeter and axially through the open tube mouth. Because solid salol is a poor thermal conductor (thermal conductivity k ≈ 0.20 W m⁻¹ K⁻¹), a radial temperature gradient develops across the tube: a solid crust forms against the inner glass wall while the center remains molten. The thermometer bulb, positioned along the central axis, may record a slightly rounded termination of the plateau as the final crystallizing front contracts toward the stem. Furthermore, if the sample is not stirred gently, localized supercooling can cause dendritic crystal growth that traps pockets of liquid, skewing the apparent plateau duration. To minimize these errors, the boiling tube should have a modest diameter (≤ 25 mm), the outer air jacket must remain draft-free to preserve radial symmetry, and sample charges should span a two-fold mass range to ensure robust ordinary least-squares slope extraction.

\frac{\partial T}{\partial t} = \frac{k}{\rho c} \left(\frac{\partial^2 T}{\partial r^2} + \frac{1}{r}\frac{\partial T}{\partial r}\right) \quad \text{(radial thermal conduction)}
Why does the end of the solidification plateau often round off gradually rather than ending with a sharp corner in real experiments?

Because salol is a poor thermal conductor, solidification proceeds from the colder outer tube walls inward. The central liquid core around the thermometer freezes last, and sensible cooling begins at the perimeter before the core is completely solid, producing a rounded transition into the post-arrest cooling curve.

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Connect the equations to a measurement.

The simulation evaluates the solidification plateau duration using the exact thermodynamic conservation relation Δt = (m · L_f) / R_loss, where m is the sample mass in kilograms, L_f = 79,500 J kg⁻¹ (79.50 kJ kg⁻¹) is the reference latent heat of fusion of salol, and R_loss = 270.0 J min⁻¹ (4.50 W) is the calibrated heat loss rate. Plateau durations are rounded to 0.01 min to model standard laboratory stopclock precision, and ordinary least-squares linear regression of Q_latent = R_loss · Δt against mass m provides the reported latent heat of fusion.