Phase Transitions — Heating Curve and Enthalpy Changes
Theoretical Background
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Heat required for temperature change: \(q = m c \Delta T\)
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Heat absorbed in a phase transition at constant pressure: \(q = n \Delta H_{\text{trans}}\)
- During an equilibrium phase transition of a pure substance at fixed pressure, the temperature remains constant while heat changes the phase and reorganizes intermolecular interactions.
- Under these conditions, a heating curve combines sloped segments within a single phase and plateaus during phase changes.
Exercise
Calculate the total heat required to bring 50.0 g of ice from –10.0 °C to liquid water at 25.0 °C.
Given data:
- $c_{\text{ice}} = 2.09 \, J\,g^{-1}\,K^{-1}$
- $c_{\text{water}} = 4.18 \, J\,g^{-1}\,K^{-1}$
- $\Delta H_{\text{fus}} = 6.01 \, kJ\,mol^{-1}$
- Molar mass of $H_2O = 18.0 \, g\,mol^{-1}$
Step-by-Step Solution
Step 1. Heating ice from –10 °C to 0 °C
\(q_1 = m c_{\text{ice}} \Delta T = (50.0)(2.09)(10.0) = 1045 \, J\)
Step 2. Melting ice at 0 °C
Moles of water:
\(n = \frac{50.0}{18.0} = 2.78 \, mol\)
Step 3. Heating liquid water from 0 °C to 25 °C
\(q_3 = m c_{\text{water}} \Delta T = (50.0)(4.18)(25.0) = 5225 \, J\)
Step 4. Total heat
\(q_{\text{tot}} = q_1 + q_2 + q_3\)
Convert to consistent units (kJ):
- $q_1 = 1.05 \, kJ$
- $q_2 = 16.7 \, kJ$
- $q_3 = 5.23 \, kJ$
Answer:
\(q_{\text{tot}} = 23.0 \, kJ\)
Notes
- Over the temperature interval considered, the heating curve has three regions: heating the solid, the melting plateau, and heating the liquid.
- The largest energy contribution comes from the phase transition (fusion), which requires far more heat than simply raising the temperature.
- This illustrates the difference between:
- specific heat (energy per unit mass per degree, linked to temperature changes),
- latent heat (energy associated with structural reorganization of matter).
- At constant pressure, the supplied heat equals the enthalpy change, so a heating curve represents how enthalpy is added within and between phases.