U6.5 · Energy of Phase ChangesPractice · 相变的能量

The temperature stays constant during the phase change, and q = n × ΔH_vap = 0.500 × 38.6 = 19.3 kJ, which goes into overcoming the attractions (including hydrogen bonds) between ethanol molecules.

The temperature is constant, so the average kinetic energy is unchanged. The heat absorbed pulls the molecules apart against their attractions (hydrogen bonds), raising their potential energy.

Vaporizing only pulls molecules away from each other; it doesn’t break covalent bonds within them. O is very electronegative, so water molecules hydrogen-bond, while $\ce{H2S}$ molecules have only weaker dipole and dispersion attractions. $\ce{H2S}$ actually has the larger molar mass (34 vs 18). The higher boiling point and the larger enthalpy of vaporization are both results of hydrogen bonding, not explanations for each other.

Three stages: ① steam 110 → 100 °C: 20.0 × 1.86 × 10 = 372 J; ② condensing at 100 °C: 1.110 × 40700 = 45177 J; ③ water 100 → 80 °C: 20.0 × 4.18 × 20 = 1672 J. Total 47221 J ≈ 47.2 kJ, most of it from condensation.

Each sample absorbs the same heat per second, and ΔT = q ÷ (m·c), so the smaller the specific heat, the faster the rise. Steam has the smallest specific heat (1.86), so it warms fastest; liquid water has the largest and warms slowest, which is why the liquid segment of a heating curve is the least steep.

Read the two plateau lengths: about 12 kJ at 0 °C (melting) and about 81 kJ at 100 °C (boiling), which are 2.00 mol × 6.01 and 2.00 mol × 40.7 kJ/mol. On a plateau the temperature, and so the average kinetic energy, stays the same; all the heat goes into overcoming intermolecular forces. After melting the molecules are still touching, with only some hydrogen bonds broken; vaporizing separates them completely, which takes much more energy. Neither change breaks covalent O–H bonds.