Giáo trình

College Physics

Science and Technology

Phase Change and Latent Heat

Tác giả: OpenStaxCollege

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.

Heat from the air transfers to the ice causing it to melt. (credit: Mike Brand)

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 C stays at C 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 [link].

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 size 12{Q} {} required to change the phase of a sample of mass m size 12{m} {} is given by

Q=mLf (melting/freezing), size 12{Q= ital "mL" rSub { size 8{f} } } {}
Q=mLv (vaporization/condensation), size 12{Q= ital "mL" rSub { size 8{v} } } {}

where the latent heat of fusion, Lf size 12{L rSub { size 8{f} } } {}, and latent heat of vaporization, Lv size 12{L rSub { size 8{v} } } {}, are material constants that are determined experimentally. See ([link]).

(a) Energy is required to partially overcome the attractive forces between molecules in a solid to form a liquid. That same energy must be removed for freezing to take place. (b) Molecules are separated by large distances when going from liquid to vapor, requiring significant energy to overcome molecular attraction. The same energy must be removed for condensation to take place. There is no temperature change until a phase change is complete.

Latent heat is measured in units of J/kg. Both Lf size 12{L rSub { size 8{f} } } {} and Lv size 12{L rSub { size 8{v} } } {} depend on the substance, particularly on the strength of its molecular forces as noted earlier. Lf size 12{L rSub { size 8{f} } } {} and Lv size 12{L rSub { size 8{v} } } {} 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. [link] lists representative values of Lf size 12{L rSub { size 8{f} } } {} and Lv size 12{L rSub { size 8{v} } } {}, 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 C to produce a kilogram of water at 0°C. Using the equation for a change in temperature and the value for water from [link], we find that Q=mLf=(1.0kg)(334kJ/kg)=334kJ 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 C to 79.C. 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 size 12{"100"°C} {} 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.

Heats of Fusion and Vaporization
Values quoted at the normal melting and boiling temperatures at standard atmospheric pressure (1 atm).
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
At 37.C (body temperature), the heat of vaporization Lv size 12{L rSub { size 8{v} } } {} for water is 2430 kJ/kg or 580 kcal/kg
539
At 37.C (body temperature), the heat of vaporization Lv for water is 2430 kJ/kg or 580 kcal/kg
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.C, 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 ([link]). The temperature of the ice rises linearly, absorbing heat at a constant rate of 0.50 cal/g⋅ºC until it reaches C. 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 C 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.

A graph of temperature versus energy added. The system is constructed so that no vapor evaporates while ice warms to become liquid water, and so that, when vaporization occurs, the vapor remains in of the system. The long stretches of constant temperature values at C and 100ºC reflect the large latent heat of melting and vaporization, respectively.

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 size 12{"100"°C} {} 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.

Calculate Final Temperature from Phase Change: Cooling Soda with Ice Cubes

Three ice cubes are used to chill a soda at 20ºC size 12{"20"°C} {} with mass msoda=0.25 kg. The ice is at C and each ice cube has a mass of 6.0 g. Assume that the soda is kept in a foam container so that heat loss can be ignored. Assume the soda has the same heat capacity as water. Find the final temperature when all ice has melted.

Strategy

The ice cubes are at the melting temperature of C. Heat is transferred from the soda to the ice for melting. Melting of ice occurs in two steps: first the phase change occurs and solid (ice) transforms into liquid water at the melting temperature, then the temperature of this water rises. Melting yields water at C, so more heat is transferred from the soda to this water until the water plus soda system reaches thermal equilibrium,

Qice=Qsoda. size 12{Q rSub { size 8{"ice"} } = - Q rSub { size 8{"soda"} } } {}

The heat transferred to the ice is Qice=miceLf+micecW(TfC). The heat given off by the soda is Qsoda=msodacW(Tf20ºC). Since no heat is lost, Qice=Qsoda, so that

miceLf+micecWTfC=-msodacWTf20ºC.

Bring all terms involving Tf size 12{T rSub { size 8{f} } } {} on the left-hand-side and all other terms on the right-hand-side. Solve for the unknown quantity Tf size 12{T rSub { size 8{f} } } {}:

Tf=msodacW20ºCmiceLf(msoda+mice)cW.

Solution

  1. Identify the known quantities. The mass of ice is mice=3×6.0 g=0.018 kg size 12{m rSub { size 8{"ice"} } =3 "." 6`g=0 "." "018"`"kg"} {} and the mass of soda is msoda=0.25 kg size 12{m rSub { size 8{"soda"} } =0 "." "25"`"kg"} {}.
  2. Calculate the terms in the numerator:
    m soda c W 20º C = 0 .25 kg 4186 J/kg ⋅º C 20º C = 20,930 J

    and

    miceLf=0.018 kg334,000 J/kg=6012 J. size 12{ ital "mL" rSub { size 8{f} } = left (0 "." "018"`"kg" right ) left ("333","000"`"J/kg" right )"=5995"`J} {}
  3. Calculate the denominator:
    msoda+micecW=0.25  kg + 0.018 kg4186 K/(kg⋅ºC=1122 J/ºC. size 12{ left (m rSub { size 8{"ice"} } +m rSub { size 8{"soda"} } right )c rSub { size 8{W} } = left (0 "." "25"" kg+0" "." "018 kg" right ) left ("4186 K/" \( "kg" cdot °C right )"=1122 J/"°C} {}
  4. Calculate the final temperature:
    Tf=20,930 J6012 J1122 J/ºC=13ºC.

Discussion

This example illustrates the enormous energies involved during a phase change. The mass of ice is about 7 percent the mass of water but leads to a noticeable change in the temperature of soda. Although we assumed that the ice was at the freezing temperature, this is incorrect: the typical temperature is C. However, this correction gives a final temperature that is essentially identical to the result we found. Can you explain why?

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 size 12{Q= ital "mL" rSub { size 8{f} } } {}.

Condensation forms on this glass of iced tea 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. (credit: Jenny Downing)

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.

Direct transitions between solid and vapor are common, sometimes useful, and even beautiful. (a) Dry ice sublimates directly to carbon dioxide gas. The visible vapor is made of water droplets. (credit: Windell Oskay) (b) Frost forms patterns on a very cold window, an example of a solid formed directly from a vapor. (credit: Liz West)

All phase transitions involve heat. In the case of direct solid-vapor transitions, the energy required is given by the equation Q=mLs size 12{Q= ital "mL" rSub { size 8{s} } } {}, where Ls size 12{L rSub { size 8{s} } } {} 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 size 12{L rSub { size 8{s} } } {} is analogous to Lf size 12{L rSub { size 8{f} } } {} and Lv size 12{L rSub { size 8{v} } } {}, 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 [link]) whereas, for a phase change, the transferred heat depends on the latent heat. See [link].
  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.

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, size 12{Q= ital "mL"} {}

    where L size 12{L} {} is the latent heat coefficient.

Conceptual Questions

Heat transfer can cause temperature and phase changes. What else can cause these changes?

How does the latent heat of fusion of water help slow the decrease of air temperatures, perhaps preventing temperatures from falling significantly below 0ºC size 12{"0°C"} {}, in the vicinity of large bodies of water?

What is the temperature of ice right after it is formed by freezing water?

If you place 0ºC size 12{"0°C"} {} ice into 0ºC size 12{"0°C"} {} water in an insulated container, what will happen? Will some ice melt, will more water freeze, or will neither take place?

What effect does condensation on a glass of ice water have on the rate at which the ice melts? Will the condensation speed up the melting process or slow it down?

In very humid climates where there are numerous bodies of water, such as in Florida, it is unusual for temperatures to rise above about 35ºC(95ºF). In deserts, however, temperatures can rise far above this. Explain how the evaporation of water helps limit high temperatures in humid climates.

In winters, it is often warmer in San Francisco than in nearby Sacramento, 150 km inland. In summers, it is nearly always hotter in Sacramento. Explain how the bodies of water surrounding San Francisco moderate its extreme temperatures.

Putting a lid on a boiling pot greatly reduces the heat transfer necessary to keep it boiling. Explain why.

Freeze-dried foods have been dehydrated in a vacuum. During the process, the food freezes and must be heated to facilitate dehydration. Explain both how the vacuum speeds up dehydration and why the food freezes as a result.

When still air cools by radiating at night, it is unusual for temperatures to fall below the dew point. Explain why.

In a physics classroom demonstration, an instructor inflates a balloon by mouth and then cools it in liquid nitrogen. When cold, the shrunken balloon has a small