experiment 1 latent heat

Heat and Heat Transfer Methods

Phase change and latent heat, learning objectives.

By the end of this section, you will be able to:

• Examine heat transfer.
• Calculate final temperature from heat transfer.

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.

Figure 1. 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 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.

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 =  mL f  (melting/freezing,

Q =  mL v (vaporization/condensation),

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

Figure 2. (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 L f and L v depend on the substance, particularly on the strength of its molecular forces as noted earlier. L f and L 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. Table 1 lists representative values of L f and L 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 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 =  mL f = (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.

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.

Figure 3. 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 0ºC 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 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.

Example 1. Calculate Final Temperature from Phase Change: Cooling Soda with Ice Cubes

Three ice cubes are used to chill a soda at 20ºC with mass m soda = 0.25 kg. The ice is at 0ºC 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.

The ice cubes are at the melting temperature of 0ºC. 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 0ºC, so more heat is transferred from the soda to this water until the water plus soda system reaches thermal equilibrium,  Q ice  = −  Q soda .

The heat transferred to the ice is

Q ice  =   m ice   L f  +   m ice c W ( T f −0ºC).

The heat given off by the soda is Q soda  =   m soda c W ( T f −20ºC). Since no heat is lost, Q ice  = − Q soda , so that

m ice   L f  +   m ice c W ( T f −0ºC) = – m soda c W ( T f −20ºC).

Bring all terms involving T f on the left-hand-side and all other terms on the right-hand-side. Solve for the unknown quantity T f :

[latex]\displaystyle{T}_{\text{f}}=\frac{m_{\text{soda}}c_{\text{W}}\left(20^{\circ}\text{C}\right)-m_{\text{ice}}L_{\text{f}}}{\left(m_{\text{soda}}+m_{\text{ice}}\right)c_{\text{W}}}\\[/latex]

• Identify the known quantities. The mass of ice is m ice = 3 × 6.0 g = 0.018 kg and the mass of soda is m soda = 0.25 kg.
• Calculate the terms in the numerator:  m soda c W (20ºC)=(0.25 kg)(4186 J/kg ⋅ ºC)(20ºC) = 20,930 J and  m ice L f = (0.018 kg)(334,000 J/kg) = 6012 J.
• Calculate the denominator: ( m soda  +  m ice ) c W  = (0.25 kg + 0.018 kg)(4186 K/(kg⋅ºC) = 1122 J/ºC.
• Calculate the final temperature: [latex]\displaystyle{T}_{\text{f}}=\frac{20,930\text{ J}-6012\text{ J}}{1122\text{ J/}^{\circ}\text{C}}=13^{\circ}\text{C}\\[/latex]

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 −6ºC. However, this correction gives a final temperature that is essentially identical to the result we found. Can you explain why?

Figure 4. Condensation on a glass of iced tea. (credit: Jenny Downing)

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 =  mL v .

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.

Real-World Application

Energy is also released when a liquid freezes. This phenomenon is used by fruit growers in Florida to protect oranges when the temperature is close to the freezing point (0ºC). Growers spray water on the plants in orchards so that the water freezes and heat is released to the growing oranges on the trees. This prevents the temperature inside the orange from dropping below freezing, which would damage the fruit.

Figure 14.11. The ice on these trees released large amounts of energy when it froze, helping to prevent the temperature of the trees from dropping below 0ºC. Water is intentionally sprayed on orchards to help prevent hard frosts. (credit: Hermann Hammer)

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.

Figure 5. 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 =  mL s , where L 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. L s is analogous to L f and L 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

• 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.
• Identify and list all objects that change temperature and phase.
• Identify exactly what needs to be determined in the problem (identify the unknowns). A written list is useful.
• Make a list of what is given or what can be inferred from the problem as stated (identify the knowns).
• 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.
• 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).
• 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.

Check Your Understanding

Why does snow remain on mountain slopes even when daytime temperatures are higher than the freezing temperature?

Snow is formed from ice crystals and thus is the solid phase of water. Because enormous heat is necessary for phase changes, it takes a certain amount of time for this heat to be accumulated from the air, even if the air is above 0ºC. The warmer the air is, the faster this heat exchange occurs and the faster the snow melts.

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.

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 ºC, 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 ºC ice into ºC 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 amount of light blue liquid in it, as well as some snow-like crystals. As it warms up, the liquid boils, and part of the crystals sublimate, with some crystals lingering for awhile and then producing a liquid. Identify the blue liquid and the two solids in the cold balloon. Justify your identifications using data from Table 1.

Problems & Exercises

• How much heat transfer (in kilocalories) is required to thaw a 0.450-kg package of frozen vegetables originally at 0ºC if their heat of fusion is the same as that of water?
• A bag containing 0ºC ice is much more effective in absorbing energy than one containing the same amount of 0ºC water. (a) How much heat transfer is necessary to raise the temperature of 0.800 kg of water from 0ºC to 30.0ºC? (b) How much heat transfer is required to first melt 0.800 kg of 0ºC ice and then raise its temperature? (c) Explain how your answer supports the contention that the ice is more effective.
• (a) How much heat transfer is required to raise the temperature of a 0.750-kg aluminum pot containing 2.50 kg of water from 30.0ºC to the boiling point and then boil away 0.750 kg of water? (b) How long does this take if the rate of heat transfer is 500 W 1 watt = 1 joule/second (1 W = 1 J/s)?
• The formation of condensation on a glass of ice water causes the ice to melt faster than it would otherwise. If 8.00 g of condensation forms on a glass containing both water and 200 g of ice, how many grams of the ice will melt as a result? Assume no other heat transfer occurs.
• On a trip, you notice that a 3.50-kg bag of ice lasts an average of one day in your cooler. What is the average power in watts entering the ice if it starts at 0ºC and completely melts to 0ºC water in exactly one day 1 watt = 1 joule/second (1 W = 1 J/s)?
• On a certain dry sunny day, a swimming pool’s temperature would rise by 1.50ºC if not for evaporation. What fraction of the water must evaporate to carry away precisely enough energy to keep the temperature constant?
• (a) How much heat transfer is necessary to raise the temperature of a 0.200-kg piece of ice from −20.0ºC to 130ºC, including the energy needed for phase changes? (b) How much time is required for each stage, assuming a constant 20.0 kJ/s rate of heat transfer? (c) Make a graph of temperature versus time for this process.
• In 1986, a gargantuan iceberg broke away from the Ross Ice Shelf in Antarctica. It was approximately a rectangle 160 km long, 40.0 km wide, and 250 m thick. (a) What is the mass of this iceberg, given that the density of ice is 917 kg/m 3 ? (b) How much heat transfer (in joules) is needed to melt it? (c) How many years would it take sunlight alone to melt ice this thick, if the ice absorbs an average of 100 W/m 2 , 12.00 h per day?
• How many grams of coffee must evaporate from 350 g of coffee in a 100-g glass cup to cool the coffee from 95.0ºC to 45.0ºC? You may assume the coffee has the same thermal properties as water and that the average heat of vaporization is 2340 kJ/kg (560 cal/g). (You may neglect the change in mass of the coffee as it cools, which will give you an answer that is slightly larger than correct.)
• (a) It is difficult to extinguish a fire on a crude oil tanker, because each liter of crude oil releases 2.80 × 10 7 J of energy when burned. To illustrate this difficulty, calculate the number of liters of water that must be expended to absorb the energy released by burning 1.00 L of crude oil, if the water has its temperature raised from 20.0ºC to 100ºC, it boils, and the resulting steam is raised to 300ºC. (b) Discuss additional complications caused by the fact that crude oil has a smaller density than water.
• The energy released from condensation in thunderstorms can be very large. Calculate the energy released into the atmosphere for a small storm of radius 1 km, assuming that 1.0 cm of rain is precipitated uniformly over this area.
• To help prevent frost damage, 4.00 kg of 0ºC water is sprayed onto a fruit tree. (a) How much heat transfer occurs as the water freezes? (b) How much would the temperature of the 200-kg tree decrease if this amount of heat transferred from the tree? Take the specific heat to be 3.35 kJ/kg · ºC, and assume that no phase change occurs.
• A 0.250-kg aluminum bowl holding 0.800 kg of soup at 25.0ºC is placed in a freezer. What is the final temperature if 377 kJ of energy is transferred from the bowl and soup, assuming the soup’s thermal properties are the same as that of water?
• A 0.0500-kg ice cube at −30.0ºC is placed in 0.400 kg of 35.0ºC water in a very well-insulated container. What is the final temperature?
• If you pour 0.0100 kg of 20.0ºC water onto a 1.20-kg block of ice (which is initially at −15.0ºC), what is the final temperature? You may assume that the water cools so rapidly that effects of the surroundings are negligible.
• Indigenous people sometimes cook in watertight baskets by placing hot rocks into water to bring it to a boil. What mass of 500ºC rock must be placed in 4.00 kg of 15.0ºC water to bring its temperature to 100ºC, if 0.0250 kg of water escapes as vapor from the initial sizzle? You may neglect the effects of the surroundings and take the average specific heat of the rocks to be that of granite.
• What would be the final temperature of the pan and water in Calculating the Final Temperature When Heat Is Transferred Between Two Bodies: Pouring Cold Water in a Hot Pan if 0.260 kg of water was placed in the pan and 0.0100 kg of the water evaporated immediately, leaving the remainder to come to a common temperature with the pan?
• In some countries, liquid nitrogen is used on dairy trucks instead of mechanical refrigerators. A 3.00-hour delivery trip requires 200 L of liquid nitrogen, which has a density of 808 kg/m 3 . (a) Calculate the heat transfer necessary to evaporate this amount of liquid nitrogen and raise its temperature to 3.00ºC. (Use c p  and assume it is constant over the temperature range.) This value is the amount of cooling the liquid nitrogen supplies. (b) What is this heat transfer rate in kilowatt-hours? (c) Compare the amount of cooling obtained from melting an identical mass of 0ºC ice with that from evaporating the liquid nitrogen.
• Some gun fanciers make their own bullets, which involves melting and casting the lead slugs. How much heat transfer is needed to raise the temperature and melt 0.500 kg of lead, starting from 25.0ºC?

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

Selected Solutions to Problems & Exercises

1. 35.9 kcal

3. (a) 591 kcal; (b) 4.94 × 10 3  s

7. (a) 148 kcal; (b) 0.418 s, 3.34 s, 4.19 s, 22.6 s, 0.456 s

10. (a) 9.67 L; (b) Crude oil is less dense than water, so it floats on top of the water, thereby exposing it to the oxygen in the air, which it uses to burn. Also, if the water is under the oil, it is less efficient in absorbing the heat generated by the oil.

12. (a) 319 kcal; (b) 2.00ºC

16. 4.38 kg

18. (a) 1.57 × 10 4 kcal; (b) 18.3 kW ⋅ h; (c) 1.29 × 10 4 kcal

• Values quoted at the normal melting and boiling temperatures at standard atmospheric pressure (1 atm). ↵
• At 37.0ºC (body temperature), the heat of vaporization Lv for water is 2430 kJ/kg or 580 kcal/kg ↵
• College Physics. Authored by : OpenStax College. Located at : http://cnx.org/contents/031da8d3-b525-429c-80cf-6c8ed997733a/College_Physics . License : CC BY: Attribution . License Terms : Located at License

14.3 Phase Change and Latent Heat

Learning objectives.

By the end of this section, you will be able to:

• Examine heat transfer.
• Calculate final temperature from heat transfer.

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 0º C 0º C stays at 0º C 0º 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 Figure 14.7 .

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 Q required to change the phase of a sample of mass m m is given by

where the latent heat of fusion, L f L f , and latent heat of vaporization, L v L v , are material constants that are determined experimentally. See ( Table 14.2 ).

Latent heat is measured in units of J/kg. Both L f L f and L v L v depend on the substance, particularly on the strength of its molecular forces as noted earlier. L f L f and L v L 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. Table 14.2 lists representative values of L f L f and L v L 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 0º C 0º C to produce a kilogram of water at 0 ° C 0 ° C . Using the equation for a change in temperature and the value for water from Table 14.2 , we find that Q = mL f = ( 1 . 0 kg ) ( 334 kJ/kg ) = 334 kJ Q = mL f = ( 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 0º C 0º C to 79 . 8º C 79 . 8º 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 ( 100º C 100º 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.

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 . 0º C 35 . 0º 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 − 20º C − 20º C ( Figure 14.8 ). The temperature of the ice rises linearly, absorbing heat at a constant rate of 0 . 50 cal/g ⋅º C 0 . 50 cal/g ⋅º C until it reaches 0º C 0º 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 0º C 0º 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 ⋅º C 1 . 00 cal/g ⋅º C . At 100º C 100º C , 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 ⋅º C 0 . 482 cal/g ⋅º C .

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 100º C 100º C is less than that at 100º C 100º C , 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 100º C 100º C . 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.

Example 14.4

Calculate final temperature from phase change: cooling soda with ice cubes.

Three ice cubes are used to chill a soda at 20º C 20º C with mass m soda = 0.25  kg m soda = 0.25  kg . The ice is at 0º C 0º 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.

The ice cubes are at the melting temperature of 0º C 0º 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 0º C 0º C , so more heat is transferred from the soda to this water until the water plus soda system reaches thermal equilibrium,

The heat transferred to the ice is Q ice = m ice L f + m ice c W ( T f − 0º C ) Q ice = m ice L f + m ice c W ( T f − 0º C ) . The heat given off by the soda is Q soda = m soda c W ( T f − 20º C ) Q soda = m soda c W ( T f − 20º C ) . Since no heat is lost, Q ice = − Q soda Q ice = − Q soda , so that

Bring all terms involving T f T f on the left-hand-side and all other terms on the right-hand-side. Solve for the unknown quantity T f T f :

• Identify the known quantities. The mass of ice is m ice = 3 × 6.0  g = 0 . 018  kg m ice = 3 × 6.0  g = 0 . 018  kg and the mass of soda is m soda = 0 . 25  kg m soda = 0 . 25  kg .
• Calculate the denominator: m soda + m ice c W = 0.25 kg + 0.018 kg 4186 J/(kg⋅ºC) =1122 J/º C . m soda + m ice c W = 0.25 kg + 0.018 kg 4186 J/(kg⋅ºC) =1122 J/º C . 14.24
• Calculate the final temperature: T f = 20 , 930 J − 6012 J 1122 J/º C = 13º C. T f = 20 , 930 J − 6012 J 1122 J/º C = 13º C. 14.25

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 − 6º C − 6º 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 = mL v Q = mL v .

Real-World Application

Energy is also released when a liquid freezes. This phenomenon is used by fruit growers in Florida to protect oranges when the temperature is close to the freezing point 0º C 0º C . Growers spray water on the plants in orchards so that the water freezes and heat is released to the growing oranges on the trees. This prevents the temperature inside the orange from dropping below freezing, which would damage the fruit.

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 = mL s Q = mL s , where L s L 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. L s L s is analogous to L f L f and L v L 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

• 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.
• Identify and list all objects that change temperature and phase.
• Identify exactly what needs to be determined in the problem (identify the unknowns). A written list is useful.
• Make a list of what is given or what can be inferred from the problem as stated (identify the knowns).
• 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 14.1 ) whereas, for a phase change, the transferred heat depends on the latent heat. See Table 14.2 .
• 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).
• 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.

Check Your Understanding

Why does snow remain on mountain slopes even when daytime temperatures are higher than the freezing temperature?

Snow is formed from ice crystals and thus is the solid phase of water. Because enormous heat is necessary for phase changes, it takes a certain amount of time for this heat to be accumulated from the air, even if the air is above 0º C 0º C . The warmer the air is, the faster this heat exchange occurs and the faster the snow melts.

• 4 At 37 . 0º C 37 . 0º C (body temperature), the heat of vaporization L v L v for water is 2430 kJ/kg or 580 kcal/kg
• 5 At 37. 0º C 37. 0º C (body temperature), the heat of vaporization L v L v for water is 2430 kJ/kg or 580 kcal/kg
• 6 Values quoted at the normal melting and boiling temperatures at standard atmospheric pressure (1 atm), except where noted otherwise.

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Access for free at https://openstax.org/books/college-physics-2e/pages/1-introduction-to-science-and-the-realm-of-physics-physical-quantities-and-units

• Authors: Paul Peter Urone, Roger Hinrichs
• Publisher/website: OpenStax
• Book title: College Physics 2e
• Publication date: Jul 13, 2022
• Location: Houston, Texas
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• Section URL: https://openstax.org/books/college-physics-2e/pages/14-3-phase-change-and-latent-heat

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Latent Heat & Specific Latent Heat

• Thermometry & Thermometric Properties
• Defining A Temperature Scale
• Laboratory Thermometer
• Clinical Thermometer
• Maximum Thermometer
• Minimum Thermometer
• Three States Of Matter
• Brownian Motion
• Pressure In Gases - Boyle's Law
• Internal Energy, Thermal Energy & Temperature
• Heat Capacity & Specific Heat Capacity
• Change Of State, Melting & Solidification
• Boiling, Condensation & Evaporation
• Latent Heat & Specific Latent Heat
• Practice MCQs For Thermal Physics
• O Level Physics Topic List
• Internal Energy & Thermal Energy
• Defining a Temperature Scale
• First Law Of Thermodynamics

Table of Contents

Latent heat.

Latent heat of a substance is the amount of energy absorbed or released by the substance during a change in its physical state that occurs without changing its temperature.

• SI unit of latent heat is the joule ($\text{J}$).
• The latent heat associated with melting a solid or freezing a liquid is called the latent heat of fusion ($L_{f}$); that associated with vapourizing a liquid or a solid or condensing a vapour is called the latent heat of vaporization ($L_{v}$).

Specific Latent Heat Of Fusion

Specific latent heat of fusion, l f , of a substance is defined as the amount of heat required to change a unit mass of the substance from solid to liquid state , without any change in the temperature.

$$Q = ml_{f}$$

, where $Q$ = amount of thermal energy absorbed or released $m$ = mass of substance $l_{f}$ = specific latent heat of fusion.

• SI unit of specific latent heat of fusion, $l_f$, is joule per kilogram ($\text{J kg}^{-1}$)

Specific Latent Heat Of Vapourization

Specific latent heat of vapourization, $l_{v}$, of a substance is defined as the amount of heat required to change unit mass of the substance from liquid state to gas state without a temperature change.

$$Q = ml_{v}$$

, where $Q$ = amount of thermal energy absorbed or released $m$ = mass of substace $l_f$ = specific latent heat of vapourization.

• SI unit of specific latent heat of vapourization, $l_{v}$, of a substance is  joule per kilogram ( $\text{J kg}^{-1}$ )

Latent Heat In Terms Of Molecular Behaviour

Latent heat energy is absorbed or given out while a substance undergoes state change. The average kinetic energy of the molecules does not change so that the temperature remains constant.

• During melting, heat absorbed by the solid is used to break the inter-molecular bonds between the molecules of solid substance.
• During vapourization, heat absorbed by the liquid is used to break the inter-molecular bonds completely between the molecules of liquid substance.

Determining The Specific Latent Heat Of Fusion For Ice

In this section, we will guide you through the steps to determine the specific latent heat of fusion for ice.

Materials Required:

• Ice cubes or crushed ice
• Thermometer
• Insulated container
• Heater or heat source
• Stopwatch or timer
• Calorimeter (optional)

Experimental Procedure:

• Preparation of Ice: Begin by preparing a sufficient amount of ice. You can use ice cubes or crushed ice, but ensure that it is free from impurities that may affect the results.
• Insulated Container: Place the ice in an insulated container. An insulated container helps to minimize heat exchange with the surroundings, allowing for a more accurate determination of the specific latent heat of fusion.
• Initial Temperature Measurement: Measure the initial temperature of the ice using a thermometer. This serves as the starting point for the experiment.
• Heating the Ice: Apply heat to the ice using a heater or a heat source. As the ice absorbs heat, it will start melting and transitioning from a solid to a liquid state.
• Continuous Temperature Monitoring: Continuously monitor the temperature of the ice as it melts. Record the temperature at regular intervals using the thermometer. It’s essential to maintain a constant temperature during this phase of the experiment.
• Time Measurement: Simultaneously, use a stopwatch or timer to measure the time taken for the ice to completely melt. This time measurement is crucial for calculating the rate at which heat is absorbed.
• $Q$ is the heat energy absorbed (in joules),
• $m$ is the mass of the melted ice (in kilograms),
• $l_f$ is the specific latent heat of fusion for ice (in joules per kilogram). The specific latent heat of fusion can be determined by rearranging the formula: $l_f = \frac{Q}{m}$ Ensure that all units are consistent to obtain accurate results.
• Calorimeter Option: For a more precise determination, you can use a calorimeter to measure the heat absorbed by the melted ice. The calorimeter helps trap the heat and allows for a more controlled environment.

Determining The Specific Latent Heat Of Vaporization For Water

In this section, we will walk you through the steps to determine the specific latent heat of vaporization for water.

• Initial Temperature Measurement: Begin by measuring the initial temperature of the water using a thermometer. This provides a baseline for the experiment.
• Insulated Container: Pour a specific amount of water into an insulated container. Insulation helps minimize heat exchange with the surroundings, ensuring that most of the energy supplied goes into the phase change.
• Heating the Water: Apply heat to the water using a heater or a heat source. As the water absorbs heat, it will gradually reach its boiling point and transition into vapor.
• Continuous Temperature Monitoring: Continuously monitor the temperature of the water as it heats up. Record the temperature at regular intervals using the thermometer. It’s essential to maintain a constant temperature during this phase of the experiment.
• Boiling Point Observation: Observe and record the temperature at which the water starts boiling. The boiling point is a critical reference point for calculating the specific latent heat of vaporization.
• Time Measurement: Simultaneously, use a stopwatch or timer to measure the time taken for the water to completely vaporize. This time measurement is crucial for calculating the rate at which heat is absorbed.
• $m$ is the mass of the vaporized water (in kilograms),
• $l_v$ is the specific latent heat of vaporization for water (in joules per kilogram). The specific latent heat of vaporization can be determined by rearranging the formula: $l_v = \frac{Q}{m}$ Ensure that all units are consistent to obtain accurate results.
• Calorimeter Option: For a more precise determination, you can use a calorimeter to measure the heat absorbed during the vaporization process. The calorimeter helps trap the heat and provides a controlled environment.

Back To Thermal Physics (O Level Physics)

Back To O Level Physics Topic List

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• Latent Heat

• Santa Monica College
• Academic Departments
• Physical Sciences

Physics Lab Manual

• Physics 21: Mechanics
• Physics 22: Electromagnetism
• Physics 23: Waves, Thermodynamics, and Optics
• Physics 6: Mechanics and Waves
• Physics 7: Electromagnetism, Thermo, Optics
• Physics 8: Mechanics and Waves
• Physics 9: Electromagnetism, Thermo, Optics
• Physics 14: Conceptual Physics

Part 1: Heat of fusion

When water molecules are packed into a solid lattice they form bonds with their nearest neighbor molecules. To “melt” the solid (ice) requires the addition of energy. If the ice is at 0°C the addition of heat serves to break these bonds between the water molecules into a liquid state. Thus, when heat is added to the ice it melts rather than increasing in temperature. In this sense, the addition of heat is not directly noticeable (by a rise in temperature) and is termed “latent” heat. Such a situation is characteristic also of liquid being turned into vapor. The latent heat of fusion (associated with the melting of a solid) is defined by:

L f = Heat required to melt 1g of solid

Thus, to melt a mass m of ice it takes an amount of energy Q = mL f . The task of this laboratory will be to determine L f for water using calorimetry. The idea is to use our knowledge of the specific heat of water and a thermometer to determine how much heat is absorbed when a given amount of ice is melted in water. Knowing the mass of ice it’s then straightforward to find L f .

Using the scale at the front of the room determine the mass of the calorimeter.

Fill the calorimeter roughly halfway full with room temperature water. Determine both the mass of the water and its temperature.

Add enough ice to lower the mixture to 5° C. Make sure all the ice is melted and to stir the mixture when assessing its final temperature.

Determine the mass of ice that was added to the mixture.

The key to understanding how this problem works out is to regard the water initially in the calorimeter and the ice as two separate systems. Imagine that the ice was placed in a plastic bag and then dropped in the water (of course this would make no difference).

Using your measurements for the initial mass and temperature of the water determine how much heat is lost by the water initially in the calorimeter (recall Q = mcDT where c = 1cal/g°C). You should be able to get a number for this heat.

Write down an expression for how much heat is absorbed by the ice. Note that the ice must first be melted and then heated up. Your expression should involve the unknown Lf.

Use the principle of conservation of energy to relate the two expressions you found in parts (1) and (2) and solve for the latent heat of ice. Compare with the accepted value of 79.7cal/g.

Is heat lost or absorbed from the environment by the ice-water mixture? How will this affect your result for the latent heat? Will it be systematically high or low?

If you had the choice to use either cubes of ice or shaved ice what might yield a more accurate value for the latent heat? Explain.

Part 2: Heat of vaporization

Like the case of melting, to vaporize a liquid requires the breaking of bonds between molecules. There is correspondingly a latent heat of vaporization.

Put 400g of water into a metal canister and measure its original temperature.

Get a hot plate and preheat it for 5 minutes.

Place the canister full of water on the hotplate and after 2 minutes record its temperature. Continue to keep track of how long the water is on the hot plate.

Bring the water to a boil and boil it for an additional 5 minutes and then remove the canister from the hot plate. Record the total time the canister was on the hot plate and the mass of water that has evaporated.

Determine the change in temperature of the water after being on the hot plate for 2 minutes and from this the amount of heat transferred to the water from the hotplate during these two minutes. At what rate (cal/min) does the hotplate transfer heat to the water?

Use the rate that heat is transferred to the water and the total time the water was on the hot plate to determine the total heat transferred to the water by the hotplate. Call this Qhp.

The heat absorbed by the water goes into raising the temperature of the water and converting some of it into steam. Write down an expression for this absorbed heat. Call it Qw. Your expression for Qw should involve the unknown heat of vaporization Lv­..

Use energy conservation to relate Qhp to Qw and from this determine an experimental value for Lv. Compare with the accepted value of Lv = 540cal/g.

Is your value for L v too small or too large? Can you offer some explanation for the discrepancy in terms of the energy lost from the water to the environment?

Syllabus Edition

First teaching 2020

Last exams 2024

Specific Latent Heat Capacity ( CIE A Level Physics )

Revision note.

Defining Latent Heat Capacity

• Energy is required to change the state of substance
• Melting = solid to liquid
• Evaporation/vaporisation/boiling = liquid to gas
• Sublimation = solid to gas
• Freezing = liquid to solid
• Condensation = gas to liquid

The example of changes of state between solids, liquids and gases

• When a substance changes state, there is no temperature change
• The energy supplied to change the state is called the latent heat and is defined as:

The thermal energy required to change the state of 1 kg of mass of a substance without any change of temperature

• Specific latent heat of fusion (melting)
• Specific latent heat of vaporisation (boiling)

  The changes of state with heat supplied against temperature. There is no change in temperature during changes of state

• The specific latent heat of fusion is defined as:

  The thermal energy required to convert 1 kg of solid to liquid with no change in temperature

• This is used when melting a solid or freezing a liquid
• The specific latent heat of vaporisation is defined as:

  The thermal energy required to convert 1 kg of liquid to gas with no change in temperature

• This is used when vaporising a liquid or condensing a gas

Calculating Specific Latent Heat

• The amount of energy Q required to melt or vaporise a mass of m with latent heat L is:
• Q = amount of thermal energy to change the state (J)
• L = latent heat of fusion or vaporisation (J kg -1 )
• m = mass of the substance changing state (kg)
• Specific latent heat of fusion = 330 kJ kg -1
• Specific latent heat of vaporisation = 2.26 MJ kg -1
• Therefore, evaporating 1 kg of water requires roughly seven times more energy than melting the same amount of ice to form water
• When ice melts : energy is required to just increase the molecular separation until they can flow freely over each other
• When water boils : energy is required to completely separate the molecules until there are no longer forces of attraction between the molecules, hence this requires much more energy

Worked example

The energy needed to boil a mass of 530 g of a liquid is 0.6 MJ. Calculate the specific latent heat of the liquid and state whether it is the latent heat of vaporisation or fusion.

Step 1:            Write the thermal energy required to change state equation

Step 2:             Rearrange for latent heat

Step 3:            Substitute in the values

m = 530 g = 530 × 10 -3 kg

Q = 0.6 MJ = 0.6 × 10 6 J

L is the latent heat of vaporisation because the change in state is from liquid to gas (boiling)

Use these reminders to help you remember which type of latent heat is being referred to:

• Latent heat of fusion = imagine ‘fusing’ the liquid molecules together to become a solid
• Latent heat of vaporisation = “water vapour” is steam, so imagine vaporising the liquid molecules into a gas

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Measurement of the specific latent heat of fusion of ice

When ice, at zero degrees celsius, is added to warm water, it first melts and then its temperature increases as it takes heat from the water. If this occurs in an insulated container then the heat gained by the ice is equal to the heat lost by the container and the water. The specific latent heat of fusion of ice (l) can then be calculated using the formula: m i l + m i c w (rise in temp of melted ice) = m c c c (fall in temp of calorimeter) + m w c w (fall in temp of water) where m i , m c and m w are the masses of ice, calorimeter and water respectively and c w and c c are the specific heat capacities of water and of the material of the calorimeter.

• To select a calorimeter click on the word "Copper" for other options. Mass is shown in the top pan balance.
• Click in the Mass box and select the mass of water in the calorimeter. (min.50g, max. 90g). Press Submit.
• Click on "Place Calorimeter" to put the calorimeter in the insulated container.
• Record the starting temperature, mass of water, mass of calorimeter, and the material of the calorimeter.
• Press "Add Ice" to add dry crushed ice to the water in the calorimeter.
• After a temperature fall of 10 to 15 degrees, the temperature stops falling.
• Press "Get Total Mass" to find the mass of calorimeter + water +ice. Record.
• Press "Reset".
• Repeat the experiment using a variety of calorimeters and masses of water.
• Ensure that only ice (not water) enters the water in the calorimeter. The melting ice should be dabbed with abosrbent paper.
• Ensure that the calorimeter is well insulated to avoid loss or gain of heat energy.
• Stir the water throughout the experiment to ensure that the thermometer reading reflects the actual temperature.
• Use a sensitive thermometer graduated to 0.1 or 0.2 degrees. An error of 1 deg. in 10 is a large relative error.
• Use warmed water (about 5 deg. above room temp.) at the start of the experiment so that, overall, heat is neither lost nor gained from the surroundings.