Physics

So far we have discussed temperature change due to heat transfer. No temperature change occurs from heat transfer if ice melts and becomes liquid water (i.e., during a phase change). For example, consider water dripping from icicles melting on a roof warmed by the Sun. Conversely, water freezes in an ice tray cooled by lower-temperature surroundings.

Energy is required to melt a solid because the cohesive bonds between the molecules in the solid must be broken apart such that, in the liquid, the molecules can move around at comparable kinetic energies; thus, there is no rise in temperature. Similarly, energy is needed to vaporize a liquid, because molecules in a liquid interact with each other via attractive forces. There is no temperature change until a phase change is complete. The temperature of a cup of soda initially at 0ºC stays at 0ºC until all the ice has melted. Conversely, energy is released during freezing and condensation, usually in the form of thermal energy. Work is done by cohesive forces when molecules are brought together. The corresponding energy must be given off (dissipated) to allow them to stay together Figure 2.

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The energy involved in a phase change depends on two major factors: the number and strength of bonds or force pairs. The number of bonds is proportional to the number of molecules and thus to the mass of the sample. The strength of forces depends on the type of molecules. The heat Q required to change the phase of a sample of mass m is given by

Q = mLf (melting/freezing,

Q = mLv (vaporization/condensation),

where the latent heat of fusion, Lf, and latent heat of vaporization, Lv, are material constants that are determined experimentally. See (Table 1).

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Latent heat is measured in units of J/kg. Both Lf and Lv depend on the substance, particularly on the strength of its molecular forces as noted earlier. Lf and Lv are collectively called latent heat coefficients. They are latent, or hidden, because in phase changes, energy enters or leaves a system without causing a temperature change in the system; so, in effect, the energy is hidden. Table 1 lists representative values of Lf and Lv, together with melting and boiling points.

The table shows that significant amounts of energy are involved in phase changes. Let us look, for example, at how much energy is needed to melt a kilogram of ice at 0ºC to produce a kilogram of water at 0°C. Using the equation for a change in temperature and the value for water from Table 1, we find that Q = mLf = (1.0 kg)(334 kJ/kg) = 334 kJ is the energy to melt a kilogram of ice. This is a lot of energy as it represents the same amount of energy needed to raise the temperature of 1 kg of liquid water from 0ºC to 79.8ºC. Even more energy is required to vaporize water; it would take 2256 kJ to change 1 kg of liquid water at the normal boiling point (100ºC at atmospheric pressure) to steam (water vapor). This example shows that the energy for a phase change is enormous compared to energy associated with temperature changes without a phase change.

Table 1. Heats of Fusion and Vaporization[1] Lf Lv Substance Melting point (ºC) kJ/kg kcal/kg Boiling point (°C) kJ/kg kcal/kg Helium −269.7 5.23 1.25 −268.9 20.9 4.99 Hydrogen −259.3 58.6 14.0 −252.9 452 108 Nitrogen −210.0 25.5 6.09 −195.8 201 48.0 Oxygen −218.8 13.8 3.30 −183.0 213 50.9 Ethanol −114 104 24.9 78.3 854 204 Ammonia −75 108 −33.4 1370 327 Mercury −38.9 11.8 2.82 357 272 65.0 Water 0.00 334 79.8 100.0 2256[2] 539[3] Sulfur 119 38.1 9.10 444.6 326 77.9 Lead 327 24.5 5.85 1750 871 208 Antimony 631 165 39.4 1440 561 134 Aluminum 660 380 90 2450 11400 2720 Silver 961 88.3 21.1 2193 2336 558 Gold 1063 64.5 15.4 2660 1578 377 Copper 1083 134 32.0 2595 5069 1211 Uranium 1133 84 20 3900 1900 454 Tungsten 3410 184 44 5900 4810 1150

Phase changes can have a tremendous stabilizing effect even on temperatures that are not near the melting and boiling points, because evaporation and condensation (conversion of a gas into a liquid state) occur even at temperatures below the boiling point. Take, for example, the fact that air temperatures in humid climates rarely go above 35.0ºC, which is because most heat transfer goes into evaporating water into the air. Similarly, temperatures in humid weather rarely fall below the dew point because enormous heat is released when water vapor condenses.

We examine the effects of phase change more precisely by considering adding heat into a sample of ice at −20ºC (Figure 3). The temperature of the ice rises linearly, absorbing heat at a constant rate of 0.50 cal/g⋅ºC until it reaches 0ºC. Once at this temperature, the ice begins to melt until all the ice has melted, absorbing 79.8 cal/g of heat. The temperature remains constant at 0ºC during this phase change. Once all the ice has melted, the temperature of the liquid water rises, absorbing heat at a new constant rate of 1.00 cal/g⋅ºC. At 100ºC, the water begins to boil and the temperature again remains constant while the water absorbs 539 cal/g of heat during this phase change. When all the liquid has become steam vapor, the temperature rises again, absorbing heat at a rate of 0.482 cal/g⋅ºC.

Water can evaporate at temperatures below the boiling point. More energy is required than at the boiling point, because the kinetic energy of water molecules at temperatures below 100ºC is less than that at 100ºC, hence less energy is available from random thermal motions. Take, for example, the fact that, at body temperature, perspiration from the skin requires a heat input of 2428 kJ/kg, which is about 10 percent higher than the latent heat of vaporization at 100ºC. This heat comes from the skin, and thus provides an effective cooling mechanism in hot weather. High humidity inhibits evaporation, so that body temperature might rise, leaving unevaporated sweat on your brow.

We have seen that vaporization requires heat transfer to a liquid from the surroundings, so that energy is released by the surroundings. Condensation is the reverse process, increasing the temperature of the surroundings. This increase may seem surprising, since we associate condensation with cold objects—the glass in the figure, for example. However, energy must be removed from the condensing molecules to make a vapor condense. The energy is exactly the same as that required to make the phase change in the other direction, from liquid to vapor, and so it can be calculated from Q = mLv.

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Condensation forms in Figure 4 because the temperature of the nearby air is reduced to below the dew point. The air cannot hold as much water as it did at room temperature, and so water condenses. Energy is released when the water condenses, speeding the melting of the ice in the glass.

Sublimation is the transition from solid to vapor phase. You may have noticed that snow can disappear into thin air without a trace of liquid water, or the disappearance of ice cubes in a freezer. The reverse is also true: Frost can form on very cold windows without going through the liquid stage. A popular effect is the making of “smoke” from dry ice, which is solid carbon dioxide. Sublimation occurs because the equilibrium vapor pressure of solids is not zero. Certain air fresheners use the sublimation of a solid to inject a perfume into the room. Moth balls are a slightly toxic example of a phenol (an organic compound) that sublimates, while some solids, such as osmium tetroxide, are so toxic that they must be kept in sealed containers to prevent human exposure to their sublimation-produced vapors.

All phase transitions involve heat. In the case of direct solid-vapor transitions, the energy required is given by the equation Q = mLs, where Ls is the heat of sublimation, which is the energy required to change 1.00 kg of a substance from the solid phase to the vapor phase. Ls is analogous to Lf and Lv, and its value depends on the substance. Sublimation requires energy input, so that dry ice is an effective coolant, whereas the reverse process (i.e., frosting) releases energy. The amount of energy required for sublimation is of the same order of magnitude as that for other phase transitions.

The material presented in this section and the preceding section allows us to calculate any number of effects related to temperature and phase change. In each case, it is necessary to identify which temperature and phase changes are taking place and then to apply the appropriate equation. Keep in mind that heat transfer and work can cause both temperature and phase changes.

Problem-Solving Strategies for the Effects of Heat Transfer

1. Examine the situation to determine that there is a change in the temperature or phase. Is there heat transfer into or out of the system? When the presence or absence of a phase change is not obvious, you may wish to first solve the problem as if there were no phase changes, and examine the temperature change obtained. If it is sufficient to take you past a boiling or melting point, you should then go back and do the problem in steps—temperature change, phase change, subsequent temperature change, and so on.
2. Identify and list all objects that change temperature and phase.
3. Identify exactly what needs to be determined in the problem (identify the unknowns). A written list is useful.
4. Make a list of what is given or what can be inferred from the problem as stated (identify the knowns).
5. Solve the appropriate equation for the quantity to be determined (the unknown). If there is a temperature change, the transferred heat depends on the specific heat (see Table 1 in Temperature Change and Heat Capacity) whereas, for a phase change, the transferred heat depends on the latent heat. See Table 1.
6. Substitute the knowns along with their units into the appropriate equation and obtain numerical solutions complete with units. You will need to do this in steps if there is more than one stage to the process (such as a temperature change followed by a phase change).
7. Check the answer to see if it is reasonable: Does it make sense? As an example, be certain that the temperature change does not also cause a phase change that you have not taken into account.

Section Summary

• Most substances can exist either in solid, liquid, and gas forms, which are referred to as “phases.”
• Phase changes occur at fixed temperatures for a given substance at a given pressure, and these temperatures are called boiling and freezing (or melting) points.
• During phase changes, heat absorbed or released is given by: Q = mL where L is the latent heat coefficient.

Glossary

heat of sublimation: the energy required to change a substance from the solid phase to the vapor phase

latent heat coefficient: a physical constant equal to the amount of heat transferred for every 1 kg of a substance during the change in phase of the substance

sublimation: the transition from the solid phase to the vapor phase

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