specific latent heat of vaporisation of water experiment

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Specific latent heat of vaporization

The specific latent heat of vaporization (enthalpy of vaporization) is the amount of heat required to vaporize a liquid substance!

Process of vaporization

If a liquid is heated more and more, then at some point the boiling point is reached. At this point, the state of matter changes and the liquid finally begins to vaporize (also referred to as boiling). During vaporization, no further increase in temperature is observed for pure substances, despite the continued supply of heat energy. During vaporization, the energy obviously no longer benefits the increase of the kinetic energy of the molecules, which would otherwise mean an increase in temperature (see also article Temperature and particle motion ).

During vaporization, the transferred energy leads to an increase in the internal energy in terms of changed binding energies between the molecules in the liquid and gaseous state. The intermolecular bonds in the liquid state are being broken by the added heat energy, thus allowing the transition to the gaseous state. In the gaseous state, the molecules are only relatively weakly bonded to each other due to the lower binding forces.

During vaporization, energy must be added to break the intermolecular bonds. In the case of pure substances, the temperature of the liquid remains constant until the process of vaporization is completed!

More detailed information on this can also be found in the article Why does the temperature remain constant during a change of state (phase transition)?

The fact that heat has to be supplied permanently to drive the breaking of intermolecular bonds is evident, for example, in the boiling of water. If water is boiled in a pot, the water only vaporizes as long as the hotplate remains switched on. However, if the heat supply is interrupted, the water also stops boiling.

The question arises as to how much heat must be added in order to completely vaporize a certain amount of a liquid. The heat required for this is also referred to as the heat of vaporization or enthalpy of vaporization . This heat of vaporization does not include the amount of heat required to heat the liquid to the boiling point. The heat of vaporization therefore only includes the heat energy to be added during vaporization if the liquid has already been heated to boiling temperature.

The heat of vaporization (enthalpy of vaporization) is the heat energy to be added to a liquid at its boiling point in order to completely vaporize a certain amount of the substance!

For the difference between the terms heat and enthalpy , see the article Difference between latent heat of vaporization and enthalpy of vaporization .

Since the heat of vaporization added during vaporization is not directly noticeable in an increase in temperature, but can nevertheless be found in the vaporized substance in the form of internal energy , the heat of vaporization is also referred to as latent heat . The term “latent” comes from Latin and means “to be hidden” or “not to appear directly”.

Experimental determination of the heat of vaporization

Experimental setup.

Using water as an example, the heat of vaporization required to vaporize a certain amount of water is to be determined experimentally in the following. For this purpose, water is first heated to boiling temperature with an immersion heater. Then, the vaporization of the water mass over time is observed using a balance on which the experimental setup is placed.

The added heat of vaporization can be determined via the electrical power of the immersion heater, which is completely converted into heat. The heat energy Q v (= heat of vaporization) added at a power P is obtained by the operating time t of the immersion heater according to the following formula:

\begin{align} \label{q} Q_\text{v} = P \cdot t \\[5px] \end{align}

Observation

First, the water is heated to boiling temperature with the immersion heater. When the water begins to vaporize, the experiment can be started at any time. To do this, the balance is reset to zero and the time measurement is started. The water gradually vaporizes and the vaporized mass is displayed on the balance. At regular time intervals, the displayed mass of the balance is recorded. At each time t, the added heat of vaporization Q v up to that point can be determined using formula (\ref{q}). In this way one obtains a statement which amount of heat led to the vaporization of which mass m v .

If one plots the vaporized mass as a function of the added heat, then a proportional relationship becomes apparent. This means that in order to vaporize twice the amount of water, twice the amount of heat must be added. The evaluation of the experimental data shows that about 120 kJ of heat are necessary to vaporize 48 g of water. With a supplied heat energy of about 240 KJ, twice the water mass of 96 g has then finally vaporized.

Especially with regard to the comparability of the heats of vaporization of different liquids, it therefore makes sense to always relate the heats of vaporization Q v to a standardized amount of mass to be vaporized (e.g. 1 kilogram or 1 gram). This constant ratio between the heat of vaporization and the mass m v to be vaporized is called specific heat of vaporization or specific enthalpy of vaporization q v :

\begin{align} &\boxed{q_\text{v} = \frac{Q_\text{v}}{m_\text{v}}}~~~[q_\text{v}]=\frac{\text{J}}{\text{kg}}~~~~~\text{specific heat of vaporization} \\[5px] \end{align}

From the experiment, a specific heat of vaporization of around 2500 kJ/kg is finally obtained for water. This means that 2500 kJ of heat is required to vaporize 1 kilogram of water. However, with the experimentally determined heat of vaporization using the described experimental setup, it must be noted that the heat emitted by the immersion heater does not completely benefit the vaporization of the water. Some of the heat is also used to heat the vessel and is thus transferred to the surroundings. Therefore, a lower amount of heat is used for the vaporization of the water than calculated with formula (\ref{q}). The literature value for the specific heat of vaporization of water is therefore somewhat lower with 2257 kJ/kg.

Specific heat of vaporization is the heat of vaporization to be added per unit mass of a liquid to be vaporized!

The specific heat of vaporization q v describes the relationship between the mass to be vaporized m v and the heat of vaporization to be added for this purpose Q v :

\begin{align} &\boxed{Q_\text{v} = q_\text{v} \cdot m_\text{v}} ~~~\text{heat of vaporization} \\[5px] \end{align}

In the case of water, the heat of vaporization to be added is more than five times as great as the amount of heat that would have had to be used to heat the water from 0 °C to 100 °C. This relatively large heat of vaporization is one of the reasons why a fire is extinguished excellently with water.

Specific heat of vaporization of selected substances

If the experiment described above is carried out with a water-alcohol mixture or with other liquids instead of pure water (e.g. molten metals that are vaporized), it can be seen that these substances vaporize at different rates. Consequently, more or less heat energy is required to vaporize a given mass of the substance. The specific heat of vaporization is therefore dependent on the substance.

The greater the specific heat of vaporization of a substance, the more heat is required to vaporize a given mass. Substances with large specific heats of vaporization therefore do not vaporize as quickly. The table below shows the specific heats of vaporization of selected liquids.

It should be noted that the specific heats of vaporization are indirectly influenced by the ambient pressure, since this changes the boiling temperatures (see also the article Why does water boil faster at high altitudes? )! However, since most liquids are vaporized at an ambient pressure of 1 bar, the specific heat of vaporization usually refers to the boiling temperature at 1 bar.

Note that the temperature of not all liquids remains constant during vaporization! For example, petroleum as a mixture of different substances does not vaporization at a boiling point, but within a boiling range. In the case of petroleum, this boiling range is between 180 °C and 330 °C. The heat added during the phase transition is therefore used both to raise the temperature and to drive the vaporization. It is therefore not possible to determine exactly what proportion of the heat added is used for the temperature increase or for the vaporization. Consequently, no (specific) heat of vaporization can be assigned to such a mixture of substances. In general, such a boiling range occurs with mixtures of substances, whereas pure substances usually have a boiling point.

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Measurement of the specific latent heat of vaporisation of water

When steam, at 100 degrees celsius, is bubbled into water, it first condenses to water and then its temperature decreases as it loses heat to the surrounding water. If this occurs in an insulated container then the heat lost by the steam is equal to the heat gained by the water and the container. The specific latent heat of vaporisation of water (l) can then be calculated using the formula: m s l + m s c w (fall in temp of condensed steam) = m c c c (rise in temp of calorimeter) + m w c w (rise in temp of water) where m s , m c and m w are the masses of steam, calorimeter and water respectively and c w and c c are the specific heat capacities of water and of the material of the calorimeter. Note: c copper = 390 J / kg / K, c aluminium = 910 J / kg / K, c polystyrene = 0.

• To select different calorimeters 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 Steam" to allow steam from the steam generator to pass to the water in the calorimeter.
• After a temperature rise of 10 to 15 degrees, the steam pipe is removed (happens automatically here) so heating is stopped. Record temperature.
• Press "Get Total Mass" to find the mass of calorimeter + water + steam. Record.
• Press "Reset".
• Repeat the experiment using a variety of calorimeters and masses of water.
• Ensure that only steam (not water) enters the water in the calorimeter. Use a "steam trap" (it actually traps water) if available.
• 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 heat supplied.
• 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 cooled water (about 5 deg. below room temp.) at the start of the experiment so that, overall, heat is neither lost nor gained from the surroundings.

Specific latent heat of vaporisation measurement

The specific latent heat of vaporization of a liquid may be measured by a modification of the method of Ramsey and Marshall (1896). The apparatus is shown in Figure 1.

The double-walled glass vessel is fitted to a condenser and mounted vertically. The inner section contains the liquid in which is a heater made of platinum wire. When the liquid boils the vapour passes through small holes into the outer vessel and then down into the condenser. Here it condenses, runs down and is collected in the beaker. It is essential that evaporation is rapid, for then the vapour in the outer vessel acts as a heat shield and eliminates heat losses from the inner vessel. When a steady state has been reached - that is, when liquid drips into the beaker at a constant rate - a clean beaker is placed under the condenser and the mass of liquid m condensing, and hence being evaporated, in a measured time t can be found. The specific latent heat of vaporisation L of the liquid can then be found from the equation

where VI is the power supplied to the coil. If a joule-meter is available the energy input E may be measured directly; then Energy input (E) = mL The large specific latent heat of vaporization of water explains why it is much more painful to be scalded by steam at 100 o C than by an equal mass of liquid water at 100 o C. The steam first condenses before it cools to your body temperature and in doing so releases roughly ten times as much heat energy as it does in the cooling phase.

Measurement of the specific latent heat of fusion of ice

The simplest method for measuring this quantity is the method of mixtures. Ice is dropped into water a few degrees above room temperature, and the resulting fall in temperature is recorded after all the ice has melted. Since the water falls from a few degrees above the temperature of the surroundings to a few degrees below the heat losses may be ignored - the mixture is assumed to gain as much heat as it losses and a cooling correction need not be applied. (See also: Latent heat of fusion experiment

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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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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AP®︎/College Biology

Course: ap®︎/college biology   >   unit 1.

• Hydrogen bonding in water
• Hydrogen bonds in water
• Capillary action and why we see a meniscus
• Surface tension
• Cohesion and adhesion of water
• Water as a solvent

Specific heat, heat of vaporization, and density of water

• Importance of water for life
• Lesson summary: Water and life
• Structure of water and hydrogen bonding

Introduction

Water: solid, liquid, and gas, density of ice and water, heat capacity of water, heat of vaporization of water, attribution:, additional references:, want to join the conversation.

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The Latent Heat of Vaporization Experiment Report

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Introduction

Conceptual framework, risks and risk control, experiment procedures and requirements, results obtained and analysis, uncertainties, explanations.

During my academic life, I have always observed that companies that use water as a heating medium prefer using steam to boiling water. This tendency has always amazed me since the boiling water and steam have the same temperature which is 100 degree Celsius. I have been arguing that if water and steam have the same temperature, none of them should be preferred to the other on basis of the heating capability. Additionally, I noted that after sweating, I feel cold for a moment although the ambient temperatures are very high. Based on these mysteries of nature I decided to investigate what happens during vaporization of water in order to explain the phenomenon. The figure below is a graph representing the heating curve of pure water where we observe that water maintains 100 degrees Celsius for a moment before turning to steam.

There are concepts and theories that explain why steam is a better heating medium than boiling water although they might have the same temperature. Scientists argue that before substances change from liquids to gas, they absorb heat without showing increase in the temperature. This heat is hidden in a manner that it cannot be identified by increase in temperature implying that they are not even sensed by a thermometer. The hidden heat is referred to as the latent heat of vaporization since it is invisible to apparatus that detect changes in temperature. When the heat is measured per unit mass, which is mostly considered as one kilogram, the measure is referred to as the specific latent heat of vaporization. Various substances have different amounts of latent heat of vaporization since they have differing complexions. The table 1 shows the various substances along with their corresponding specific latent heat of vaporization.

Table 1: It shows the different substances along with their specific latent of vaporization.

From the table above, it is evident that water has a specific latent heat of vaporization of 2260kJ/Kg implying that the any mass of water absorbs heat that is proportional to this ratio (Lefrois, 1979). When calculating the amount that a mass of substance need to vaporize, we multiply the specific latent heat of vaporization with the mass.

Heat absorbed= Specific latent heat of vaporization x Mass of the vaporizing substance

In this case, Q is the amount of heat absorbed, m represents the mass of substance which will evaporate, and H v is the specific latent heat of vaporization of the substance. For example, when 0.3 kilograms of alcohol is evaporating it absorbs 256.5 kJ of heat. This is obtained by multiplying 855 kJ/Kg by 0.3 Kilograms.

Q= 855 kJ/Kg x 0.3 Kgs = 256.5 Kilojoules

There are various risks that are conjoined with this experiment due to the technicality of the experiments. This list below shows the risks and how they will be controlled.

• Since the experiment involves heating water the steam might burn inflicting injuries on the body. Additionally, the hot apparatus might cause burns on the skin leading to pain and wound. In order to control this risk, holders will be used ensuring that the apparatus are not held with bare hands.
• Secondly, the apparatus might break due to when heating especially if they experience temperature imbalances. In this case, hot apparatus that are made of glass will not be washed using cold water before they cold down. Thermometer will be kept in their bags when they are not in use.
• Lastly, spillage of water on the floor might make the floor become wet and lead to slipping that can cause adverse bodily injuries. Lags will be used to wipe the floor and benches ensuring there is not water on the floors.

This experiment needs an electrical heater, a boiling trough, one stop watch, and a balance for measuring mass.

• Set up the tools that are described above as shown in figure 2 in order to set for the experiment.

• Measure the original mass of the boiling trough, water and the heater. Record this mass as M1.
• Identify the heater’s watt and record it as W.
• Immerse the heater in the water and put on the heater
• Start the watch when the water starts boiling and not before it boils.
• Record the mass of the trough and water after every three minutes. This should be recorded on the M2 column.
• The time should be recorded besides the time column

• Calculate the mass of evaporated liquid by subtracting M2 from M1 and record this on the corresponding column.

During calculation, it should be considered that the heat releases during by the heater is equal to the heat that is absorbed by the water and trough.

(W heater ) x Time = m x H v.

Therefore, the specific latent heat of vaporization can be obtained by making the subject of the above formula and compute the result.

H v = ((W heater ) x Time)/m

Taking the first data set of the results that were obtained we can calculate the latent heat of vaporization.

P= 126 x 180s = 22680

Hv = 22680/ 10= 2268 j/Kg

We can find the average by adding all the value of latent heat and dividing by three. (2268+2061+1890)/3= 2073 J/Kg

This value could be compared with the theoretical value which is 2260 J/Kg. However, there is a significant error that is reflected on the value. This error is attributed to the prevailing temperature conditions that could have caused loss of heat to the surrounding environment. Additionally, draught might have affected the results since it additionally cause the loss of heat (Hunter, 1997). Lastly, the error might have been caused due to some impurities which could be found in the water implying that the water was not pure during experiment. This implies that the amount of heat required for heating and vaporization might have been lower than the common. However, the valued can be relied on since it is on the range of error that is associated with the latent heat of vaporization in this experiment.

• The power of the heater was not uncertain since the reading was not obtained directly from the heater
• Uncertainty of beam balance was +/- 1gram
• Uncertainty of time was +/- 0.1 s
• The uncertainty of the latent heat of vaporization is obtained by adding up all the results and finding the average. This could be

From the above experience, it is evident that water absorbs a hidden amount of energy so that it can vaporize and change its state of matter from liquid to steam. However, this heat is not visible since thermometer does not show difference temperature. This scenario explains why the steam is a better heating medium that the boiling water. In this case, steam absorbs latent heat of vaporization enabling it to have more heat than the boiling water. According to my experiment, 1 gram of steam could have additional 2273 joules of heat more than boiling water which is at the same temperature. This implies that it produce more heat than the boiling water, in the same light, when the steam cools down it releases this heat which is absorbed by the object which is being heated by the medium. On the other hand, boiling water would only have the heat that enabled it to reach 100 degree Celsius.

The same explanation can be used to explain why I feel cold when I am sweating during a hot day. In this case, the sweat contains water that is the main component of sweat from the body. On a hot day, high temperature make the water on the skin to evaporate to the atmosphere. As the water evaporates, it absorbs the latent heat of vaporization from the body. This absorption of heat form the body makes the skin to feel cold for a moment. Consequently, the latent heat of vaporization becomes crucial when cooling object. In fact, the cooling systems that use water use the same principle of latent heat of vaporization. In this case, water gets into contact with the hot metals that make the cooling system. Water is heated up to 100 degree Celsius leading where it vaporizes. During vaporization, it absorbs the latent heat of vaporization according to the mass of water that is used for cooling. This heat reduces the amount of heat which was originally in the system resulting to cooler temperatures than the ones which were originally in the machine.

According to this experiment, the latent heat of vaporization is estimated to be 2073J/ Kg which is roughly accurate as compared to the theoretical values that is 2260J/Kg. it shows that water absorbs hidden amount of heat that is not detectable on a thermometer since it does not cause change in temperature. This heat makes steam a better heating medium than boiling water where steaming is one of the best cooking methods. It, also, explains why people feel cold when sweat evaporates from the skin on a hot day. Additionally, the concept explains the ability of water to act as a cooling liquid in machine since it absorbs the latent heat of vaporization causing reduction in temperature (Wilson & Hall, 2005).

Bellium, A, Specific Latent Heat of Vaporization , Home Page .

Hunter, C, 1997, Latent heat, Winnipeg, Man, Nuage Editions.

Lefrois, R, 1979. Active heat exchange system development for latent heat thermal energy storage , Washington, Dept. of Energy, Office of Energy Technology, Division of Energy Storage.

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Wilson, J, & Hall, 2005, Physics laboratory experiments (6th ed.), Boston, Houghton Mifflin.

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IvyPanda . 2022. "The Latent Heat of Vaporization Experiment." April 20, 2022. https://ivypanda.com/essays/the-latent-heat-of-vaporization-experiment/.

1. IvyPanda . "The Latent Heat of Vaporization Experiment." April 20, 2022. https://ivypanda.com/essays/the-latent-heat-of-vaporization-experiment/.

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Comparison of the water vapor budget evolution of developing and non-developing disturbances over the western north pacific.

1. Introduction

2. materials and methods, 2.2. identification of developing and non-developing disturbances, 2.3. water vapor budget equation, 2.4. box difference index, 3. basic characteristics of disturbances, 4. water vapor budget of developing and non-developing disturbances, 4.2. evaporation, 4.3. moisture flux convergence, 4.4. precipitation, 4.5. analysis of bdi, 5. conclusions and discussion, author contributions, data availability statement, acknowledgments, conflicts of interest.

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Sun, Z.; Gao, S.; Jian, M. Comparison of the Water Vapor Budget Evolution of Developing and Non-Developing Disturbances over the Western North Pacific. Remote Sens. 2024 , 16 , 2396. https://doi.org/10.3390/rs16132396

Sun Z, Gao S, Jian M. Comparison of the Water Vapor Budget Evolution of Developing and Non-Developing Disturbances over the Western North Pacific. Remote Sensing . 2024; 16(13):2396. https://doi.org/10.3390/rs16132396

Sun, Zhihong, Si Gao, and Maoqiu Jian. 2024. "Comparison of the Water Vapor Budget Evolution of Developing and Non-Developing Disturbances over the Western North Pacific" Remote Sensing 16, no. 13: 2396. https://doi.org/10.3390/rs16132396

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