Physics · Ch 7 — Thermal Properties of Matter
Latent Heat
Latent Heat
Whenever a substance changes state, heat is either absorbed or released, WITHOUT any accompanying change in temperature -- exactly the flat plateaus seen in the heating curve of §7.8 (Fig. 7.9). The latent heat of a substance is defined as the quantity of heat required to change the state of UNIT MASS of the substance, at constant temperature. For a mass changing state completely:
where , the latent heat, is characteristic of the substance (and of which change of state is involved), with SI unit ; its value is usually quoted at one standard atmosphere.
Two named latent heats matter most: the latent heat of FUSION is the heat needed to convert unit mass of a substance from solid to liquid at its melting point, without any temperature change; the latent heat of VAPOURIZATION is the heat needed to convert unit mass from liquid to vapour at its boiling point, again without any temperature change. Plotting temperature against total heat added for water (Fig. 7.11) makes both plateaus visible directly on a heat axis rather than a time axis, and shows clearly that the boiling plateau is far WIDER (needs far more heat) than the melting plateau. …
What this figure shows. A temperature (vertical axis, °C) versus heat energy added (horizontal axis, joules) graph for a fixed mass of water starting as ice below 0 °C. The curve rises (ice warming) up to 0 °C, then becomes perfectly FLAT/horizontal over a wide range of added heat while ice melts to water at constant 0 °C (this flat segment is noticeably SHORTER in horizontal extent than the boiling plateau, since Lf is much smaller than Lv), then rises again (liquid water warming from 0 °C to 100 °C) with a different, gentler slope than the ice-warming segment, then becomes flat again, this time over a MUCH WIDER range of added heat, while water boils to steam at constant 100 °C, and finally rises again as the resulting steam is heated above 100 °C. The caption notes the differing slopes of the phase lines show that ice, water and steam have different specific heats, and the very …
Substance | Melting point (°C) | Lf (x10^5 J/kg) | Boiling point (°C) | Lv (x10^5 J/kg)
Gold | 1063 | 0.645 | 2660 | 15.8
Lead | 328 | 0.25 | 1744 | 8.67
Water | 0 | 3.33 | 100 | 22.6
Ethyl alcohol | -114 | 1.0 | 78 | 8.5
Mercury | -39 | 0.12 | 357 | 2.7 …
Worked out. 0.1 kg of ice at 0 °C is mixed with 0.32 kg of water at 35 °C, settling at a final temperature of 7.8 °C; the example equates the heat lost by the warm water (36434.944 J) to the sum of the heat needed to melt the ice (m(ice) x Lf) and the heat needed to then warm the melted ice-water from 0 °C to 7.8 °C (3265.08 J), solving for Lf = 3.317x10^5 J/kg, closely matching the standard value in Table 7.6. …