School Science Lessons
UNPh24
2018-11-09
Please send comments to: J.Elfick@uq.edu.au

24.0 Change of state
Table of contents

See: Humidity, (Commercial)

13.0 Heat (Primary)

Weather

24.0 Change of state & solutions

24.4.0 Dew point & humidity

24.3.0 Evaporation

24.7.0 Heat engines

24.7.4 Heat loss by human body

24.2.1.0 Latent heat of fusion latent heat of vaporization

24.5.0 Melting Point (M P), Freezing Point (F P)

24.1.0 Phase changes liquid / solid, M P, and F P.

24.2.0 Phase changes, liquid / gas, boiling point, B P.

4.36 Newton's law of cooling

24.0 Change of state and solutions
24.0.0 Change of state and solutions
24.1.07 Critical point, critical temperature
24.1.06 Osmotic pressure equation, Morse equation
24.1.08 Van't Hoff factor, i, for dilute solutions
Experiments
24.1.0 Phase changes, liquid / solid, melting point, m.p.
24.2.0 Phase changes, liquid / gas, boiling point, b.p.

24.4.0 Dew point and humidity
See: Humidity, (Commercial)
24.4.3 Condensation nuclei, supersaturation
24.4.2 Dew point measurement
Experiments
37.29 Cloud in a bottle

24.3.0 Evaporation
See: Evaporating dish (Commercial)
24.3.0 Evaporation
24.3.01 Cooling by evaporation, sweat glands
24.3.5 Cryophorous
24.3.02 Sublimation
24.3.9 Water bag, Coolgardie safe
Experiments
24.3.8 Cooling by evaporation, ether or ethyl chloride
24.3.7 Drinking bird heat engine, drinking duck, dippy bird, dunking bird
24.3.3 Evaporate dichloromethane
24.3.4 Evaporate from a blackboard
24.3.10 Evaporate from a microscope slide
24.3.6 Freezing by evaporation
4.38 Liquids in the sun (Primary)
37.7.0 Moisture enters the air, evaporation (Weather)
24.3.2 Rate of evaporation
24.3.2.1 Rate of evaporation at different air speeds
37.23 Rate of evaporation and moving air
37.24 Rate of evaporation and moisture
24.3.1 Water "lost" by evaporation

24.7.0 Heat engines, steam turbines, refrigerator
24.7.2 Refrigerator, ice chest, portable ice boxes, espresso coffee machine
Experiments
24.7.0 Steam
24.7.1 Steam turbines
24.7.3 Hero's engine, eolipile

24.2.1.0 Latent heat of fusion and latent heat of vaporization
See: Calorimeters (Commercial)
24.1.4 Latent heat, change of state, melting and boiling
Experiments
4.12 Atomizers
5.17 Body temperature (Clinical thermometer)
9.45.7 Body temperature, diurnal variation (See: 2.3 Sweat glands)
24.10.0 Heat energy changes solid to liquid
4.11 Heat energy to change liquid to vapour
4.10 Heat energy to change solid to liquid
24.1.5 Latent heat of fusion of ice to water
24.2.1 Latent heat of steam
24.2.3 Latent heat of vaporization of water

24.5.0 Melting Point (M P), Freezing Point (F P)
See: Melting, (Commercial)
24.1.04 Freezing point depression and boiling point elevation
19.3.06 Freezing, thawing and cold storage of meat
12.11.3.4 Heat substances, sublimation, decrepitation
7.4.3.3 Ice melts, de-icers
7.4.0 Melting point of solids
Experiments
24.3.6 Freezing by evaporation
24.1.05 Freezing point depression of carbonated water, cola
16.1.9 Freezing point determination, (milk testing)
6.24 Freezing water expands (Soils)
4.48 Freezing water expands (ice cubes)
5.5.5 Higher melting point alloys
3.4 Impurities affect the melting point of a substance
24.1.2 Lift an ice cube with salt
5.5.4 Lower melting point alloys
8.1.13 Melt candle wax
32.3.3.2 Melt, Conductivity of melted substances
4.39 Melt different solids (Primary)
24.1.3 Melting point and pressure
7.4.1 Melting point and cooling curve of stearic acid
7.4.2.1 Melting point of 1, 4-dichlorobenzene, C6H4Cl22, toilet deodorant
7.4.3.0 Melting point of ice and freezing point of water, antifreeze
3.2 Melting point of naphthalene
3.3 Melting point of naphthalene with a capillary tube
7.4.2.0 Melting points of substances, candle wax, hexadecan-1-ol (cetyl alcohol)
3.4 Melting points of substances affected by impurities
24.1.14 Molal freezing point constant of cyclohexane solvent
24.1.15 Molar mass of solute from depression of freezing point
3.18 Separate by melting points, tin from tin and carbon mixture
7.4.3.1 Temperature at which ice melts
7.4.3.2 Temperature at which ice and salt mixture freezes
3.63 Tests for melting point of lead, tin, and lead-tin alloys

24.1.0 Phase changes liquid / solid, M P, and F P.
24.1.0 Phase changes liquid / solid, melting point and freezing point
24.1.6 Bitumen foaming and bitumen decay
24.1.13 Hot water freezes faster than cold water, the Mpemba effect
24.1.18 Metglas
24.1.01 Nucleation
24.1.03 Raoult's law
12.3.1 Vapour pressure, Saturation vapour pressure, SVP
12.3.2 Vapour pressure, Saturation vapour pressure over water, Psvp
24.1.19 Wood's metal
Experiments
24.1.17 Crystal growth on the overhead projector
24.1.8 Heat of crystallization of sodium acetate
24.1.9 Heat of crystallization of sodium sulfate, heat of fusion
24.1.7 Heat of crystallization of sodium thiosulfate
24.1.11 Heat of solution of ammonium nitrate crystals
24.1.10 Heat of solution of sulfuric acid
4.41 Ice experiments (Primary)
24.1.2 Lift ice cube with salt
24.1.14 Molal freezing point constant of cyclohexane solvent
24.1.15 Molar mass of solute from depression of freezing point
3.43 Solid, liquid and gas in a plastic drink bottle (Primary)
7.4.3.2 Temperature at which ice and salt mixture freezes
7.4.3.1 Temperature at which ice melts
24.1.02 Vapour pressure of water and non-volatile solvent
24.1.16 Water crystals in soap film

24.2.0 Phase changes, liquid / gas, boiling point, B P.
24.2.1.2 Boil a kettle with a spout
24.2.10 Boil by cooling
3.6 Boiling point of inflammable liquids, ethanol, acetone
3.5.1 Boiling point of sodium chloride solution
3.5.0 Boiling point of water, bar
24.2.8 Broken bottle, ice bomb (Dangerous experiment not suitable for schools!)
4.40 Heat and cool water (Primary)
24.2.2 Heat of condensation of water
24.2.6 Ignite paper with a jet of steam
24.2.9 Liquefying dry ice
22.7.6.2 Peltier effect, supercooled water, Experiments
24.2.4 Pressure and boiling point of water
22.7.6.1 Seebeck effect (thermoelectric effect)
24.2.5 Temperature of steam above water boiling
3.7 Volatility of different liquids

4.10 Heat energy change solid to liquid
Heat energy change solid to liquid, melting point, latent heat of fusion, naphthalene, ethanamide
| See diagram 4.10: Liquid naphthalene solidifies
Put crushed naphthalene or ethanamide (acetamide) in a test-tube in a container of water.
Heat gently until all the substance has melted.
Remove the test-tube from the container and fix a thermometer with its bulb in the melted substance.
Stir the substance with a thermometer while the substance cools and record the temperature every 30 seconds for 6 minutes.
Plot a graph of temperature against time.
At first the temperature drops while the substance remains liquid.
Then the temperature remains the same while the substance changes from liquid to solid.
When all the substance is solid, the temperature starts to drop again.
The melting point, m.p., is the temperature when a solid changes to a liquid.
The specific latent heat of fusion of a substance, L, is the quantity of heat required to change one kilogram of the substance from solid
to liquid without change in temperature.
The unit is joule / kg, J kg-1.
The specific latent heat of fusion of ice = 3.34 X 105 joule / kg, 334 kJ kg-1.

4.11 Heat energy to change liquid to vapour
Heat energy to change liquid to vapour, boiling point, latent heat of vaporization
See diagram 4.11: Heat required to vaporize a liquid
1. Weigh a container, add 50 mL water and weigh again.
Heat the container and water.
Put a thermometer in the water and record the rise in temperature every 30 seconds.
Plot a temperature against time graph.
Draw the line of best fit and calculate the average temperature increase per minute.
Assume that all the heat goes into the liquid and the heat absorbed by the flask is small.
Calculate the heat absorbed by the liquid per minute by multiplying the mass of the liquid by its specific heat and by the temperature
increase per minute.

2. Weigh a container, add 50 mL water and weigh again.
Heat the container and water and allow to boil for 10 minutes.
Leave to cool then weigh the container and water.
Calculate the mass of water lost by evaporation.
This will be the heat of vaporization of the liquid.

3. Put a known mass of water in a boiling flask and a known mass of water in a container.
Record the temperature of the water.
Heat the boiling flask and pass all the steam into the water in the container so that all the steam condenses to water.
When most of the water in the flask has evaporated, stop heating and record the temperature of the water in the container.
Leave the apparatus to cool to room temperature, weigh the water remaining in the flask and the water in the container.
The condensing steam loses latent heat of fusion when it condenses and loses heat when its temperature
(100oC) falls to the temperature of the water in the container.
The specific latent heat of vaporization of water is 2.26 MJ kg-1.

4.12 Atomizers
See:
Atomizer cap (Serrata)
Atomizer is a term used in industry for a heating element that can vaporize a liquid in the mouthpiece of a device so that the vapour it
can be inhaled.
Atomizers are sold separately, so need separate cartridges or drip tips containing the inhalant which must be replaced every few weeks.
Atomizers are used for perfumes and for electronic cigarettes, E-cigarettes.

24.0.0 Change of state and solutions
See: Ice models (Commercial)
State of matter, solid (s), liquid (l), gas (g), aqueous solution (dissolved in water) (aq)
Change of State, melting point and freezing point, boiling point and air pressure, evaporation and condensation, cooling by evaporation
PVT Surfaces (for an ideal gas, the pressure, volume, temperature 3 dimensional plot on a graph, 3-dimensional models)
Change of state
The three states of matter are solid, liquid and gas.
When a substance changes from one state of matter to another, it has undergone a change of state (change of phase).
Change of state always occurs with a change of heat energy that flows into or out of the material but no change of temperature occurs.
During change of liquid to solid, freezing, heat leaves the liquid as it freezes.
During change of liquid to gas, vaporization (boiling and evaporation), heat goes into the liquid as it vaporizes.
During change of gas to liquid, condensation, heat leaves the gas as it condenses.
During change of solid to gas, sublimation, heat goes into the solid as it sublimates.
The heat energy exchanges during a change of state are changes in potential energy, not kinetic energy.
The heat energy that comes into ice during a phase change breaks the bonds between the molecules of water.
The molecules are now at a higher potential energy state, but not, on the average, moving any faster, so their average kinetic energy
remains the same, and the temperature remains the same.

24.1.0 Phase changes, liquid / solid, M P and F P.
Change of state, melting point and freezing point
The melting point is the temperature at which a substance changes from solid form to liquid form.
A pure substance under standard conditions of pressure has a definite melting point.
If you supply heat to a solid at its melting point, the temperature does not change until the melting process is complete.
The melting point of ice is 0oC.
Boiling is rapid conversion of a liquid into vapour that takes place when the liquid reaches a certain temperature.
It involves the formation of vapour bubbles within the body of a liquid, whereas evaporation occurs only at the surface.
Boiling point, for any given liquid, is the temperature at which the application of heat raises the temperature of the liquid no further,
but converts it to vapour.
The boiling point of water under normal pressure is 100oC.
The lower the pressure, the lower the boiling point and vice versa.
Freezing is the change of state from liquid to solid that occurs at the freezing point of a substance.
For a given substance, freezing occurs at a definite temperature, called its freezing point, which is invariable under similar conditions of
pressure.
The temperature remains at this point until all the liquid is frozen.

24.1.01 Nucleation
The phenomenon of nucleation refers to the start of many types of new processes, including crystallization, bubble formation in a
saturate liquid, liquid droplets in a saturated vapour, fracture.
The places where nucleation occurs are called nucleation sites, e.g. suspended particles, dust, cracks in the inner surface of a container,
iodine crystals used for cloud seeding, spaces in solid materials.
Such nucleation is called heterogeneous nucleation.
However, when nucleation occurs without nucleation sites it is called homogeneous nucleation, e.g. in supercooled liquids.

24.1.02 Vapour pressure of water and non-volatile solvent
| See diagram 24.1.02: Vapour Pressure
| See diagram 24.1.03: Vapour pressure of a liquid
Put a beaker of deionized water under a strong container, e.g. a bell jar and evacuate the air with a strong pump.
Measure the pressure in the container with an aneroid barometer.
At 25oC, the air pressure falls from atmospheric pressure to the vapour pressure of pure water, 23.8 mm Hg (3.169 kPa) as water
molecules evaporate from the water surface in the beaker.

Repeat the experiment with a 1 molal solution in water of ethylene glycol (ethane-1, 2-diol, antifreeze).
This lowering of vapour pressure is a colligative property, i.e. a property that depends on the relative number of particles and not their
chemical properties.
Colligative properties include relative lowering of vapour pressure, elevation of boiling point, depression of freezing point and osmotic
pressure.
The colligative properties of a dilute aqueous solution of a non-ionized solute, e.g. glucose, urea are used to measure their relative
molecular masses.
The colligative properties of ionized solutes are used to measure the percentage ionization.

24.1.03 Raoult's law
The vapour pressure of a solvent above a solution, P1 = Xsolvent × P1o, where Xsolvent = the mole fraction of the solvent in the
solution and P1o = the vapour pressure of the pure solvent.
Xsolvent = number of moles of solvent / total number of moles.
For example, if 0.1 moles of sugar are dissolved in 10 moles of water, Xsolvent = 10 / 10.1 = 0.99. Raoult's law is colligative
because the solute added must not react with the solvent so the solution should behave as an ideal solution.
In general, Raoult's law only applies to very dilute solutions, and is based on the principle that when a solute is added to a solvent there
is a decrease in the relative number of solvent molecules at the surface of the liquid to evaporate.

24.1.04 Freezing point depression and boiling point elevation
See diagram 24.1.04: Freezing point depression and boiling point elevation | See diagram 24.1.04.1: Elevation of boiling point
The presence of a solute raises the boiling point and lowers the freezing point.
This phase diagram shows that when a non-volatile solute is dissolved in a solvent the solvent vapour pressure is lowered.
The vapour pressure curve for the solution is lower than the vapour pressure curve for pure water but not parallel to it.
The vapour pressure curve of the solution intersects the sublimation pressure curve at a lower temperature than the vapour pressure
curve.
The freezing points curves for the solution and water begin at the triple points, the intersections of the sublimation pressure curve and
the vapour pressure curve, where solid, liquid and gas phases can simultaneously exist.
So the freezing point curve for the solution is at a lower temperature than freezing point curve for water.
Freezing point depression can be used to calculate the molar mass of a substance.
The freezing point depression is proportional to the sum of the concentrations of all the different dissolved particles so it can be used to
measure biological concentrations, e.g. urine.
A solution has no boiling point because the solvent tends to evaporate more than the solute, so the concentration of the solution increases.
However, a constant boiling mixture (azeotrope) is an exception to Raoult's law where the composition of the vapour is the same as the
composition of the boiling liquid, e.g. a mixture of 4% water and 96% ethanol.
So a mixture of water and alcohol can never be converted to absolute alcohol by boiling.
Freezing point depression, dTf = i × kf × m, where dTf = difference between freezing point of solvent and freezing point of solution, as
a positive value, and kf = molal freezing point depression constant, e.g. water = 1.86oC kg / mole.
Boiling point elevation, dTb = i × kb × m, where dTb = difference between boiling point of solvent and boiling point of solution, as a
positive value, and kb = molal boiling point elevation constant, e.g. water = 0.52 oC kg / mole
m = molality (molal concentration) mol kg-1.
(Molality = moles of solute / kg of solvent.)
Each solvent has a specific value for kb and kf, e.g. for benzene, kb = 2.53oC kg / mole and kf = 5.10oC kg / mole.
The colligative properties of a solution include depression of freezing point, elevation of boiling point and osmosis.
The effects of alcohols on the temperature range decrease with the increase of molecular weight because the size of colligative
properties depends on the number of particles.
So if equal weights of methanol (32.04 g / mole) and 1-butanol (74.12 g / mole) methanol would have more particles.
While boiling point elevation can be explained by decrease of relative number of solvent molecules to evaporate, freezing point
depression cannot be explained by saying that solute particles "get in the way of the solvent particles", but rather by the difference in the
chemical potential of the solvent particles.
Depression of freezing point is used when salting ice for making ice cream at home and salting roads in cold climate winters.
Elevation of boiling point is used when adding ethylene glycol ("antifreeze") to the water of car radiators when "winterizing" the car.

24.1.05 Freezing point depression of carbonated water, cola
See: Aprons rubberised, (Commercial)
Carbonic acid, soda water, H2CO3: 3.34.3
Standard atmospheric pressure = 101.325 kPa.
Pressure of Coca Cola Classic = 380 kPa.
A leading cola company suggests it should be consumed at 2o-3oC, for optimal taste and best digestion.

1. A young boy in China was badly injured when he opened a can of semi-frozen soft drink that exploded in his face.
Any attempt to "speed-cool" a soda can may result in a busted aluminium drink can and a stick coat on the inside of the freezer.
The boy would not have been hurt if the can had been filled with water alone.
Once the carbon dioxide gas has been forced out of the water as the water crystallizes, the carbon dioxide accumulates in the remaining
space in the drink can and its pressure increases rapidly.
The expansion of water as it freezes is important to soda detonation not because it puts pressure on the can but because it puts pressure
on an increasingly cramped reservoir of gaseous carbon dioxide.

2. Soda pop drink cans may explode in the freezer, but only when ice begins to form.
See: Aprons rubberised, (Commercial)
The freezing point of a pure solvent will be lowered by the addition of a solute.
The temperature at which a liquid freezes decreases as the concentration of dissolved materials increases.
Dissolved gases, e.g. carbon dioxide, can act as a solute and its concentration is dependent on pressure.
Carbon dioxide dissolves as follows:
CO2 + H2O --> H+ + HCO3-
The dissolved HCO3- in solution changes the freezing point.
Do the following experiment away from students behind a safety screen.
Wear chemical splash goggles, chemical-resistant gloves, and a chemical-resistant apron.
Remove the label from the plastic soda water bottle and cool it in a refrigerator overnight.
Prepare an ice salt mixture with rock salt sprinkled over crushed ice in a beaker.
Put the cooled bottle of soda water in the beaker and cover it with layers of ice and rock salt.
Insert a thermometer into the ice salt mixture so that it touches the plastic bottle.
Allow the temperature to drop to -8 C and stay there for about 10 minutes.
Do not allow the temperature to drop to below -10 C, because the plastic bottle may explode.
Carefully remove the plastic bottle from the ice water mixture, but do not shake it.
The soda water is still liquid.
The liquid soda water turns to instant ice when the bottle is opened.

3. The concentration of carbon dioxide in soft drinks remains constant only as long as the drink can or bottle is kept sealed.
When the containers are opened the carbon dioxide fizzes out of the bottle, so the concentration of solutes in the water decreases and
the freezing point rises so the soft drink will freeze more quickly.
However, another explanation for the fizzing is that the soft drink is a supercooled solution ready for freezing.
The carbon dioxide bubbles that form when the container is opened provide sites for the nucleation of the ice crystals.
Tapping the chilled bottle without opening it may allow some carbon dioxide bubbles to form followed by freezing within the
unopened container.

24.1.06 Osmotic pressure equation, Morse equation
The osmotic pressure of a dilute solution, Π (capital pi) = iMrT, where i = Van't Hoff factor, M = molarity, R 00821 (gas constant),
T = absolute temperature.

24.1.07 Critical point, critical temperature
See diagram 24.1.05: Critical point of carbon dioxide
19.2.9.3 Caffeine, extraction with supercritical carbon dioxide, critical point
Critical temperature is the temperature above which a gas cannot be liquefied by pressure.
For water, the upper limit of the vaporization curve is at 374oC and 218 atm, called the critical point.
Above the critical temperature, a gas cannot exist in the liquid state no matter what pressure is applied.
Helium has the lowest critical temperature, -268oC.
The critical point for carbon dioxide is 31oC and 73 atm.
At the critical point the liquid and gas phases have the same density.
At the critical point the physical properties of the liquid and gaseous phases are the same.
At the critical point (critical state) the liquid state of the matter ceases to exist.
When a liquid is heated, its density decreases and the pressure and density of its vapour increases until the critical temperature is
reached where the two densities are equal and the phase boundary between liquid and gas disappears.
The heat of vaporization becomes zero.
The critical point of water is at 647K (374C) and 22.064 MPa (3200 PSIA or 218 atmospheres, atm.).

Critical point of ether
See diagram 24.1.07: Critical point of ether
The apparatus has a glass tube containing ether surrounded by a heating element connected to a 12 Volt power supply.
The meniscus is made visible by focussing with the apparatus placed on a projector.

24.1.08 Van't Hoff factor, i, for dilute solutions
i = Van't Hoff factor.
(If not ionized particles, i =1, e.g. sucrose.
If particles of two ions, i = 2, e.g. NaCl.
If particles of 3 ions, i = 3, e.g. MgCl2.)
If particles associate in solution, i <1. e.g. ethanoic acid in benzene.
If particles dissociate in solution, > 1, e.g. NaCl --> Na+ + Cl-.
If particles neither associate nor dissociate, i =1, e.g. glucose in water.
Van't Hoff factor is the ratio = concentration of particles of dissolved substance / concentration of substance based on its mass,
i.e. the number of particles in solution after dissociation / the number of formula units dissolved in the solution.

24.1.2 Lift ice cube with salt
See: Ice models (Commercial)
See diagram 24.2.7: Lift ice cube with salt
1. Ice melts and freezes again by the action of salt.
Put an ice cube in a full cup of cold water so that it floats.
Let a thread lie on the ice cube and sprinkle salt on it, see the diagram.
The aim of sprinkling salt is to wet the thread by melting the ice near the thread.
Wait until the melted ice near the thread condenses again.
Lift one end of the thread to check if water on it has frozen completely.
If it is, lift two ends of the thread to hold the ice cube from the water.
Salt lowers the freezing temperature of water so some ice at the top of the ice cube melts to form salty water that is absorbed by the
dry string.
The melting water washes aside the salt so the concentration of salt water near the string drops and the freezing point raises again
allowing the water at the top of the ice cube to freeze again and trap the string.

2. Float an ice cube in water.
Put a matchstick on the ice.
Sprinkle a very small amount of salt on to the ice at each side of the matchstick.
After one minute, lift the ice cube by lifting the matchstick.
Salt water has a lower melting point or freezing point than pure water.
The salt sprinkled on the ice dissolves in the ice lowering its freezing point.
The ice around the matchstick melts and the matchstick floats in it.
Some of the water may fall off the ice cube but also the salt keeps dissolving in the ice so lowering the concentration of salt around the
matchstick and raising the freezing point back towards 0oC, when it freezes again trapping the matchstick in the ice.
The layer of water between the matchstick and the ice cube is now frozen so the matchstick sticks to the ice.

24.1.3 Melting point and pressure
See diagram 24.1.3: Melting by pressure | See diagram 24.1.3a: Regelation
Increase of pressure increases the freezing point of substances but increase of pressure on substances that expand on freezing lowers
the freezing point.
So increased pressure on ice turns it to liquid but when increased pressure stops the ice freezes again without change in temperature,
regelation.

1. Regelation
Cut through a block of ice with a copper wire loop that has a heavy mass hanging from each end.
Copper wire cuts through faster than iron or cotton thread.
At the start of the experiment the copper wire should be the same temperature as the ice, otherwise it is just transferring heat and
melting the ice.
When the copper wire exerts pressure on the ice below it the ice melts because its melting point is now lower.
The ice under pressure is still at 0oC, which is above its melting point.
So the wire sinks down through the melt water.
However, above the wire the pressure decreases to near atmospheric pressure.
The melt water still at 0oC refreezes because the melting point is again 0oC.
The refreezing water above the wire loses latent heat of fusion that is conducted down to the copper wire and to some extent
increases the melting process.
The copper wire keeps passing down through the whole block of ice, cutting it on two and joining the cut regions together again.

2. When you apply pressure to ice, you lower the freezing point of it.
Freeze a rectangular ice cube in a freezer.
Put it on a wooden board.
Tie 500 g weights to each end of a thin nylon thread.
Spread the thread across the upper surface of the ice and hang the weights at each side.
Observe the ice melting below the thread as it cuts slowly into the ice.
If the temperature is low enough, as the ice freezes the ice above the thread can freeze again, regelation.
Then you can lift the ice cube by nylon thread.
Substances that expand on freezing show a lowering melting point under pressure.

3. Tie the ends of a 50 cm length of iron wire or nylon cord to two 10 cm lengths of cut broomstick to act as handles.
Lean over a block of ice, grip the handles and use them to push the wire or cord down onto it.
Move the wire from side to side and with a downward motion, as if you are sawing on the ice.
Eventually you can move the wired or cord down through the whole block of ice but it remains as one block of ice.

4. Push your finger down on the middle of an ice cube.
Some ice at the upper surface melts because of the increased temperature and pressure.
Sprinkle salt on the ice cube and press it again.
The ice cube melts slower because salt lowers the melting point of ice.

5. Squeeze crushed ice in a thick walled cylinder to form a solid block.
In a snowball fight, throw loose snow only.
If you squeeze the snow in your gloved hand you create a rocky piece of ice that is dangerous to throw.

6. The ice under a skaters ice skates quickly melts under the skate then refreezes again when the skater has passed.
However, some people claim that pressure from ice skates is not enough to melt ice, except when the temperature is a fraction of a
degree below 0oC.
In this view, skates move easily over ice because a very thin layer of water already on the ice lowers the friction and melting from
frictional heating give more water lubricant.

24.1.4 Latent heat, change of state, melting and boiling
1. A change in the physical state (solid, liquid, or gas) of a material is called a change of state, e.g. melting, boiling, evaporation,
solidification and condensation.

2. Latent heat of fusion: The amount of heat per unit mass that has to be removed to freeze a substance is a constant for any given
substance, and is called the latent heat of fusion.
When a pure substance changes from solid to liquid the temperature does not rise but remains at the melting point until all the solid has
melted.
The heat required to melt one kilogram of the substance is called the latent heat of fusion,
Water = 3.33 × 105 joule / kg.
The exceptionally high latent heat of fusion of water, 333 kj kg-1, prevents water temperature from changing quickly at temperatures
near 0oC above or below, because of the extra energy needed to freeze or melt water.

3. Latent heat of vaporization: The heat required to evaporate one kilogram of the substance is called the latent heat of vaporization,
e.g. Water = 22.6 × 105 joule / kg.
When a pure substance changes from liquid to vapour the temperature does not rise but remains at the boiling point until all the liquid
has boiled away.
A pure substances absorbs heat energy to change from solid to liquid at the constant temperature of the melting point.
The heat required to melt 1 kilogram of the substance is called the latent heat of fusion (lf).
The lf of water = 3.33 × 105 joule / kg.
A pure substances absorbs heat energy to change from liquid to vapour at the constant temperature of the boiling point.
The heat required to evaporate 1 kilogram of the substance is called the latent heat of vaporization (lv).
The lv of water = 22.6 × 105 joule / kg.
The latent heat of evaporation of water is the highest of all substances, 2260 kj kg-1, which reduces water loss and heat loss to the
atmosphere.

4. The assembled state of molecules or atoms in matter is called phase.
The transformation of the matter from a state to another is called phase change.
The amount of heat absorbed or emitted by an object of unit mass in the process of phase change is called latent heat in phase change.
The heat of vaporization, the heat of solution and the heat of sublimation are all latent heat in phase change.
Under the same pressure the same kind of matters are transformed from liquid state to steam state, the heat of vaporization needed is
equal to that from steam state back to liquid state.
As the pressure and temperature are in the condition of matter that is in such a state that the solid, liquid and steam, three phases exist
together, there is a relation of value: the heat of sublimation = the heat of solution + the heat of vaporization.

24.1.5 Latent heat of fusion of ice to water
See: Ice models (Commercial)
1. Weigh a 500 mL beaker.
Half fill with crushed ice and weigh again.
Record the temperature of the ice.
Heat the beaker and ice for a minute and again note its temperature.
Repeat until all the ice has been melted for several minutes.
Record your observations as: State of matter, Time in minutes, Temperature in "C.
Draw a graph of your results by plotting temperature against time and note the shape of the graph.
When there is no ice but only ice water, weigh the container and contents.
There should be no noticeable change in mass because few particles from the original ice or water have escaped.

2. Continue as in the last experiment heating more strongly and note the temperature every minute until the water boils.
Again there would be no noticeable change in mass until a considerable number of molecules have been lost as steam.
Again record your results and make a graph of temperature against time.
Note the rise in temperature as heat is supplied.

3. The heat of solution of ice can be measured by measuring how much ice can be melted by a piece of hot solid covered on the ice.
Measure the mass of a piece of metal, its specific heat has known.
Hang it by a thread to let it go into water and heat water until boiling.
Then put it into a funnel as quickly as possible, cover pieces of ice on it.
Collect the melted water to a measuring cylinder through the funnel.
Record the volume and calculate the mass of melted water.
Note the funnel should be big enough, and the pieces of ice are more enough, in other words, there should be left a few pieces of ice,
which cannot be melted.
Calculate the amount of heat emitted by metal in the process of the temperature dropping to 0oC, the ratio of it to the mass of water
melted is the heat of solution of ice.
4. Plot the cooling curve of ethanamide, acetamide.

24.1.6 Bitumen foaming and bitumen decay
The behaviour of bitumen during foaming is primarily due to physical processes although chemistry does also play a role.
When a cold water droplet at ambient temperature makes contact with the bitumen at 170oC to 180o̊C, the following chain of events
occurs:
1. The bitumen exchanges energy with the surface of the water droplet heating the droplet to a temperature of 100̊C and cooling the
bitumen.
2. The transferred energy of the bitumen exceeds the latent heat of steam resulting in explosive expansion and the generation of steam.
3. Steam bubbles are forced into the continuous phase of bitumen under pressure.
4. With emission from the bitumen spray nozzle the encapsulated steam expands until a thin film of slightly cooler bitumen holds the
bubble intact through its surface tension.
5. During expansion, the surface tension of the bitumen film counteracts the ever diminishing steam pressure until a state of equilibrium
is reached.
Due to the low thermal conductivity of bitumen and water, the bubble can remain stable for a few seconds.
6. So many bitumen bubbles form to together produce foamed bitumen.
7. As the colloidal mass cools to ambient temperature, the steam in the bubbles condenses causing bubbles to collapse and the
bitumen foam to "decay".

24.1.7 Heat of crystallization of sodium thiosulfate
1. Put 2 cm of sodium thiosulfate crystals in a test-tube, put the test-tube in a beaker of water and slowly heat the beaker with a
Bunsen burner.
When the crystals have melted, take out the test-tube which should be still at room temperature, dry it with a cloth and, while holding
it firmly in your hand, drop in a small crystal of sodium thiosulfate.
Crystals form in all directions from the small crystal and the liquid becomes completely solid with crystals.
You can feel the rise in temperature in the test-tube because the formation of crystals is an exothermic process.

2. Heat a half-filled test-tube of sodium thiosulfate crystals (sodium thiosulfate pentahydrate, Na2S2O3.5H2O) in a beaker of hot water.
Gently heat the beaker so that the sodium thiosulfate dissolves in its own water of crystallization.
So heat gently otherwise you will drive off all the water of crystallization to form anhydrous sodium hyposulfite (hypo).
When the crystals have all melted, cover the test-tube and leave to cool in an insulated beaker containing tap water at the height of the
melted crystals in the test-tube.
Put one thermometer in the melted crystals and fix another thermometer in the tap water in the beaker.
If the temperature of the solution is below the melting point you now have a supercooled solution that will form crystals if a crystal of
sodium thiosulfate or even dust is added to the solution, or if the solution is stirred.
Stir the melted crystals with the thermometer very carefully so that you do not break the thermometer when crystals form.
Record the temperatures in the two thermometers as the sodium thiosulfate solidifies to form crystals again.
Let all the crystals form around the thermometer and later remove the thermometer by dissolving the crystals in running tap water.
The temperature of the sodium thiosulfate remains constant as the crystals form and only then does the temperature fall because the
room temperature is lower than the melting point of sodium thiosulfate, 48.3o.
However, the temperature of the tap water in the beaker rises as the sodium thiosulfate cools and continues to rise as the sodium
thiosulfate crystals form because the reaction is exothermic.

24.1.8 Heat of crystallization of sodium acetate
Heat of crystallization is defined as the increase in enthalpy when 1 mole of a substance is transformed into its crystalline state at
constant pressure.
1. Prepare a supersaturated solution of sodium acetate and drop in a crystal to trigger crystallization.
A thermocouple will show the change in temperature.
Observe the change in temperature when a flask of supercooled sodium acetate crystallizes.

2. Gently heat sodium acetate trihydrate crystals to 54oC so that they dissolve in their own water of crystallization.
Keep heating the solution up to 100oC, then leave to cool.
This supersaturated solution can be cooled to room temperature without forming crystals so it is used in a sodium acetate self-heating
pack.
These heat packs are used for medical purposes to provide heat externally to parts of the body.
The heat pack sold in pharmacies consists of a sodium acetate solution and a copper disc sealed in a plastic envelope.
The patient can bend the disc by squeezing it through the heat pack to cause friction between the disc and the sodium acetate to
change the liquid solution to become solid in an exothermic reaction.
The reaction generates heat in seconds up to 54oC.
The crystals in the pack can be returned to the liquid state by immersing the pack in boiling water for about 10 minutes so the sodium
acetate heat pack is very handy because the change is reversible and the heat pack can be used again and again.
The latent heat of fusion of sodium acetate is about 264289 kJ / kg.

24.1.9 Heat of crystallization of sodium sulfate, heat of fusion
Sodium sulfate decahydrate crystals, Na2SO4.10H2O, (Glauber's salt), melt at a conveniently low temperature of 32oC, but they do
not store as much heat as sodium acetate crystals.
The heat of fusion of sodium sulfate decahydrate is 25.53 kJ / mol.
The term heat of fusion is used to describe the energy required to change an amount of a substance from solid to liquid without
changing its temperature.
The heat of fusion is also called standard enthalpy of fusion or specific melting heat, to describe the change in heat energy required for
a substance to change its state from solid to liquid or vice versa at the melting point of the substance.
The specific heat of fusion refers to heat of fusion referenced to a unit of mass.
The molar heat of fusion refers to the heat of fusion referenced to the enthalpy change per amount of substance in moles.
Substances with high heat of fusion are called PCMs (phase change materials) or LHS units (latent heat storage units).

24.1.10 Heat of solution of sulfuric acid
Observe heating when sulfuric acid is added to water, do NOT water added to sulfuric acid!

24.1.11 Heat of solution of ammonium nitrate crystals
Add water to ammonium nitrate and observe the change in temperature.
Instant cold packs for treating sports injuries formerly contained water and ammonium nitrate that could be mixed by squeezing the
pack to cause the endothermic reaction and resulting drop in temperature.

24.1.13 Hot water freezes faster than cold water, the Mpemba effect
Hot water, e.g. 90oC does appear to freeze faster than the same amount of cold water, e.g. 18oC.
This phenomenon was observer by Aristotle 4th century BC and is named after Erasto Mpemba, a schoolboy in Tanzania, who drew
attention to it.
The initially hot water has less of the apparently frozen ice solid because it contains trapped liquid water.
The initially cold water freezes at a lower temperature to a solid ice with less included liquid water.
The initial lower temperature causes intensive nucleation and a faster crystal growth rate.
At a freezing temperature of -6o C, the initially hot water apparently freezes first, but eventually the initially cold water completely
freezes before the initially hot water.
Initially cold water will have the maximum concentration of such 20-face clusterings that do not easily allow the rearrangement of
water molecules for formation of hexagonal ice crystals.
Initially, hot water has lost its ordered clustering, so if the cooling time is short, clustering will not occur fully before freezing.

24.1.14 Molal freezing point constant of cyclohexane solvent
Find the freezing point of cyclohexane.
Fit a test-tube with a stopper, thermometer and stirrer.
Weigh the test-tube and stopper, add 10 mL of cyclohexane, C6H12, and weigh again.
Put the test-tube in a water and ice mixture and record the temperature that crystals form.
Add 0.5 g of p-dichlorobenzene, C6H4Cl2, then warm and stir the mixture to aid dissolving.
Put the test-tube into a water, ice and salt mixture and record the temperature when the first crystals appear.
Calculate the molal freezing point constant for cyclohexane, the change in freezing point caused by 1 mol of p-dichlorobenzene per
kilogram of cyclohexane.
The freezing point depression, dtf = i × Kf × m, where freezing point depression =
(freezing point of pure solvent - freezing point of solution), m = molality of particles in the solution
(molality = moles of solute / weight of solvent, kg), and Kf = the molal freezing point depression constant for the solvent =
20.2oC kg / m for cyclohexane.
Repeat the experiment with cyclohexanol, C6H11OH, as the solvent.

24.1.15 Molar mass of solute from depression of freezing point, molar mass of alcohols
Measure the freezing point of known mass, kg, of a solvent, t1, with known freezing point constant, Kf,
Add known mass of a solute, kg, with known Van't Hoff factor, e.g. i =1 if solute is molecular
Measure the freezing point of the solution (solute in solvent) t2, freezing point depression, dtf = iKfm, so molality = (t1 -t2) / Kf
and molality of the solution = moles of solute / kg of solvent.
Molecular weight = mass of solute / moles of solute
1. Add 10 g of a known alcohol to 100 g of water.
Put the mixture in a container of dry ice from an ice cream seller.
Be careful! Dry ice is frozen carbon dioxide.
Do not touch it or lick it because it will cause severe burns.
Record the freezing point of the mixture.
Repeat the experiment with other known alcohols.
Draw a graph to show the relationship between molecular weight of alcohol and depression of freezing point, dtf = Kf m (for alcohols
the Van't Hoff factor, i = 1)
2. Record the freezing point of 100 g of water.
Record the freezing point of a mixture of 30 g of an unknown alcohol in 100 g of water, m = dtf / Kf, where m = moles of solute / kg
solvent, Kf water = 1.86 oC kg / m.
Molecular weight of unknown alcohol solute = weight of unknown alcohol / m, moles of solute.

24.1.16 Water crystals in soap film
A ring with a soap film is cooled in a chamber surrounded by dry ice on the overhead projector.
Water crystals form water crystals in soap film.

24.1.17 Crystal growth on the overhead projector
Melt together tartaric acid and benzoic acid and observe the crystal growth on cooling between crossed Polaroid on the overhead
projector.

24.1.18 Metglas
Metglas is a metal that has been quenched from liquid to solid very quickly so not allowing enough time for crystallization of the metal.
So the atoms are randomly arranged to give the metal three times its normal electrical resistivity and high magnetic permeability.

24.1.19 Wood's metal
This fusible alloy contains one part of cadmium, two parts of tin, four parts of lead, seven parts of bismuth and melts between 66oC
and 71oC.

24.2.1 Latent heat of steam
When liquid becomes steam, its energy increases and needs more energy provided by outside.
The boiling is the phenomenon of violent vaporization, during boiling the temperature of the liquid remains constant.
Pour a certain amount of liquid into a beaker.
Heat the beaker and stir the liquid with a thermometer.
Record the increased temperature of the liquid every 10 seconds.
Take time as horizontal axis, increased temperature as vertical axis, mark the values in a time / temperature graph, connect recording
points to form a smooth line.
The points in the graph shows the average increased temperature of the liquid in every unit time.
Calculate the amount of heat absorbed by the liquid every minute.
The amount of heat absorbed by the beaker is far less than absorbed by the  liquid absorbed, so it is omitted in the calculation.
Heat the liquid continuously until it boils.
Although heating the liquid, its temperature no longer increases while producing steam.
After boiling for 10 minutes, stop heating and let the liquid cool to room temperature.
Measure the weight of the liquid again.
Calculate the loss mass of the liquid that is also the mass to become steam.
The amount of heat needed for unit mass to become steam is called the latent heat of the steam of the liquid.

24.2.1.2 Boil a kettle with a spout
Heat water in a tea kettle with a spout.
Look carefully just outside the spout of a boiling tea kettle.
You see nothing because water vapour is colourless.
However, a short distance away from the spout you see condensed vapour with drops big enough to be seen.

24.2.2 Heat of condensation of water
See diagram 24.3.3: Heat of condensation of water
If an object, specific heat s, mass m, temperature (t1), placed in steam.
Collect all liquid produced by condensation.
The amount of heat emitted by condensation of steam are used to increase the temperature of the object.
Measure the mass of the liquid m1 and the temperature t2 increasingly, from: ms (t2- t1) = λ × m So λ = m (t2- t1) / mL.
Use a test-tube diameter of 4 cm, height 20 cm, with a rubber stopper.
Punch a hole at the side of the centre of the stopper.
Insert a thermometer.
Use another small test-tube with length is less than 10 cm.
Use a copper wire with diameter approx. 1 mm.
Wrap it around the upper part of the small test-tube to allow the test-tube to be inserted freely.
Stretch it to the lower part of the tube.
Wrap one circle again.
Bend it upward to make a small hook.
The frame should allow the small test-tube be at the centre of the large test-tube and not touch the bottom.
Hang a 4 cm iron nail with a thread.
The other end of the thread is taped to the centre of the bottom of stopper with white tape.
The length of the thread should make the nail higher than mouth of the small test-tube after putting it into the large test-tube.
The thermometer neither touches the wall of the large test-tube nor touches the nail and small test-tube.
Fix the whole apparatus to a stand.
Add suitable amount of water into the large test-tube.
Dry the small test-tube, then measure its mass m1.
Put small test-tube into the copper wire frame, then put it into the large test-tube.
Measure the mass of nail m, then tie it to the thread and tape it to the rubber stopper.
Heat the water in the large test-tube, drive the air in the tube completely.
Absorb the possible little water in the small test-tube by blotting paper.
Insert the thermometer into rubber stopper, record the temperature in the room t1.
Then hang the stopper right above the large test-tube.
After the nail is stable, cover the mouth of the tube slowly downward.
Check if the nail is just opposite to the small test-tube.
Observe that water drops into the small test-tube from the nail.
Record the temperature t2 as water does not drop completely.
Take out the small test-tube, absorb water outside, then measure the mass, m2.
Calculate λ = ms(t2 - t1) / (m2 -m1).

24.2.3 Latent heat of vaporization of water
| See diagram 24.2.3: Temperature / energy (time) heating curve for water
| See diagram 24.3.4: Beaker, test-tube, thermometer and alcohol burner.
Use a beaker, a test-tube, a thermometer and an alcohol burner.
Pour water into the beaker and test-tube, 150 mL in beaker and 50 mL in a test-tube.
Put them on a supporter, adjust the height so the test-tube is near the bottom of the beaker.
Measure the temperature of water in the beaker, then in the test-tube with a thermometer held by a clip.
Heat the water in the beaker over the alcohol burner.
Record the temperatures in the beaker and test-continuously.
After the water boils, is the temperature in the beaker the same as the temperature in the test-tube?
If water in test-tube boils?
Keep heating until water in beaker has completely been dry.
Observe if water in test-tube can boil finally?
The temperatures in the beaker and test-tube read from thermometer in the whole process of boiling may neither be 100oC,
This is due to thermometer itself exists error and the atmospheric pressure is not just standard pressure.
However, the temperatures inside and outside the tube are always equal to each other.
It is they are in a state of heat equilibrium, the conduction of heat has been stopped that the water in conduct tube cannot use the heat
of vaporization to boil.

24.2.4 Pressure and boiling point of water
See diagram 3.8: Decreasing the pressure on boiling water
See diagram 24.239: Pressure affects boiling point of water: A Steam, B Boiling tube, C Water
Boiling depends on bubbles of water vapour forming within the liquid being heated.
A liquid boils when its saturation vapour pressure = external pressure.
The pressure inside a bubble near the surface of a liquid = pressure of the gas in the bubble, Pa + saturation vapour pressure of the
liquid, Pv.
If P = pressure acting on the surface of the liquid, i.e. atmospheric pressure, and 2S / r = pressure in the bubble caused by surface
tension, P + 2S / r = Pa + Pv.
When temperature increases, Pv increases and Pa decreases because P is constant, so the bubble expands.
At the temperature when P = Pv and Pa approximately = 0, 2S / r is negligible, the volume of the bubble expands greatly and the
water boils.
The liquid boils when the saturation pressure = the external pressure, so increase of external pressure raises the boiling point and vice
versa.
The boiling point, b.p., of water at standard atmospheric pressure, 760 mmHg, 101 325 Pa, is 100oC.
Boiling point changes about 1oC for each 28 mmHg change in pressure.
At sea level water boils at 100oC.
Height above sea level and boiling point: 600 m (98oC), 1500 m (95oC), 2000 m (93oC), 3000 m (90oC).
It is reported that at 3658 m on a mountain in the Himalayas, the boiling point of water was 88oC, but in a home pressure cooker the
boiling point reaches 120oC, so the food cooks faster.
Motor vehicle radiators have a pressure cap to keep water in a liquid state at temperatures over 100oC.
Water boils faster with a lid completely or partially on a container, not because of increased pressure, but because steam can condense
on the bottom surface of the lid instead of escaping and taking away some heat.

Experiments
See diagram 3.8: Decreasing the pressure on boiling water
1. Put water in a sidearm test-tube or in a round bottom flask with a one-hole stopper.
Push a thermometer through a hole in the stopper so that the bulb of thermometer reaches, but does not touch, the bottom of the
test-tube or flask.
Add boiling chips to prevent bumping.
Boil the water and read the temperature.
Stop heating.
Connect a water pump to the sidearm or to the second hole of the 2-hole stopper.
When the water stops boiling, turn on the water pump to reduce the pressure.
Read the temperature, heat to boiling and read the temperature again.

2. Go to a high mountain and put a raw egg in an open cooking pot containing cold water.
Heat the water in the pot and note the boiling point.
The water boils and evaporates at a low temperature so you cannot cook the egg!
People who climb Mount Everest complain that they cannot get a good cup of tea, because the water boils at too low a temperature!
Put a potato in a pressure cooker containing cold water.
Be sure that the valve on the lid is in the open position.
Heat the water until it boils and steam comes through the valve.
Close the valve.
The potato cooks very quickly in the high temperature and pressure.

3. This method does not allow you to measure the pressure or the temperature inside the flask, but it is the safest method.
Boil water in half full round bottom flask.
Stop heating and insert a tight one-hole stopper fitted with a thermometer.
Support the inverted flask on a ring stand, thermometer pointing down.
Pour cold water on the bottom of the flask.
As the steam condenses, the pressure on the water lowers and the water boils again at a lower temperature.
Pour more cold water on the bottom of the flask.
The water boils again at a still lower temperature.

4. Fit a 3-hole stopper with a thermometer, an open manometer, and an outlet tube for steam.
Put plenty of grease around the stopper then insert it firmly but not tightly into a flask that is half full of water.
Heat the flask slowly.
Note the bubbles of air and later the larger bubbles of steam that form at the bottom of the flask, rise to the top, and condense on
leaving the outlet tube and striking the air.
Observe the temperature rise to 100oC and then remain steady.
Carefully apply a screw clamp and partly close the outlet valve.
Be careful! Do not close completely or an explosion can result!
With the outlet partly closed the steam cannot escape as quickly so the manometer show an increase of pressure inside the flask.
The temperature of the water rises and the bubbling stops because the temperature that water boils at which water boils depends on
the pressure.
Motor car radiators have a pressure cap to keep water in a liquid state at temperatures over 100oC.
Remove the Bunsen burner so that the water begins to cool.
Connect an aspirator to the outlet tube to reduce the pressure in the flask.
With reduced pressure the water begins to boil again at a lower temperature.

5. Food cooks quickly in a pressure cooker, because the pressure of the steam above the water in the cooker can be twice the
atmospheric pressure with the water boiling at about 120oC, so food can be cooked more quickly.
Put the cover on the pot before heating the pot because the pot will expand on heating and then you cannot put the top on the pot.

6. Half fill a round bottom flask with water and insert a two-hole stopper fitted with a thermometer and a glass outlet tube.
Support the flask in a ring stand.
The bulb of the thermometer should be in the water.
Do not heat the flask.
Use rubber tubing to connect an exhaust pump to the glass outlet.
Record the temperature.
Start pumping air, and water vapour, out of the flask.
Students can see the air bubbles rise first and can then watch the water boil at room temperature.

7. Fit a 3-hole stopper with a thermometer, an open manometer, and an outlet tube for steam.
Put plenty of grease around the stopper then insert it firmly but not tightly into a flask that is half full of water.
Heat the flask slowly.
Bubbles of air, and later larger bubbles of steam, form at the bottom of the flask then rise to the top.
The steam condenses on leaving the outlet tube and striking the air.
Note the temperature rise to 100oC and then remains steady.
Carefully apply a screw clamp and partly close the outlet valve.
Be careful! Do not close the flask completely or an explosion can result.
With the outlet partly closed, the steam cannot escape as quickly so the manometer shows an increase of pressure inside the flask.
The temperature of the water rises and the bubbling stops because the temperature that water boils at which water boils depends on
the pressure.
Remove the Bunsen burner so that the water begins to cool.
Connect an aspirator to the outlet tube to reduce the pressure in the flask.
With reduced pressure the water begins to boil again at a lower temperature.

8. Use a thermometer which measuring range is over 100oC, a pressure cooker and a aluminium cooker with a lid on which there is
hole.
Add water in both cookers.
Cover the pressure cooker, take off the valve used to reduce the pressure.
Cover the aluminium cooker, tighten the lid by several clips.
Heat the two cookers over a roaring fire until water in two cookers boils violent.
Wear a pair of gloves, hold the top of the thermometer, put the measuring bulb of thermometer into the steam outlet on pressure
cooker first, then into the hole on the lid of aluminium cooker.
The thermometer stays in steam until the liquid column in thermometer does not rise up.
Record from the thermometer in two cases.
The readings you have read may have difference from boiling point of water in the cooker, but the difference between two readings
shows the difference of temperatures caused by pressure in the cooker.
Pour flask with half full of water, cover a stopper which is inserted with a thermometer.
Put it on a frame, heat it over an alcohol burner until the water boils, record the temperature at the moment.
Then remove the flask from the heat source by using a test-tube clip.
Place the flask over an empty container.
Wait for a while, as the readings in thermometer begins to drop (do not open the lid), pour cold water on the flask, observe that the
water in the flask boils again.
Record the temperature again.
As the steam condenses while it becomes cold, the pressure in the flask reduces, so water boils again.
Pour a flask with half full of water, insert a stopper with two holes.
One hole is for inserting thermometer, the other is for inserting a tube used for conducting out.
Insert a thermometer into water in the flask.
Insert a short glass tube, the end downward must be far from the surface of water, connect the upper end of it a pump apparatus with
a rubber tube.
Fix the flask stable with a pad, then begin to pump air out.
Observe that with the pressure in the flask decreases, first you observe some air bulbs rise up from water, then water begins to boil.
Record the temperature of water in the flask.
If temperature of water is very high, pump only by using a large syringe.
This time, to consider the second pumping, add a rubber tube screw clamping apparatus on rubber tube as shown in the diagram.
After pump screw the clamping tightly, then take out the syringe from rubber tube.
Push out air inside the syringe (i.e. push the inner cylinder to the bottom of the outer one), connect the rubber tube again.
Unscrew the clamping, pump secondly.
If temperature of water is merely that in the room, or to avoid the trouble of pumping secondly, the pump apparatus is better to be the
pump or air pump.

24.2.5 Temperature of steam above water boiling
See diagram 24.4.2: Temperature of steam above water boiling
Do this experiment behind a safety screen.
This experiment can be very dangerous because the glass flask could explode! Heat the flask with the pinch clip open.
When boiling commences at 100oC, close the clip just long enough to see the temperature start to rise, then open the clip and turn off
the gas.
Temperature of boiling water or rather the temperature of the steam above water boiling under pressure greater than in the normal
atmosphere.
The steam is not allowed to escape and the thermometer reading rises.
In a pressure cooker, the pressure of the steam above the boiling water can be twice normal atmospheric pressure and the boiling
point can be 120oC.
The high pressure cooks the food more quickly.

24.2.6 Ignite paper with a jet of steam
See diagram 24.4.3: Ignite paper with superheated steam
The steam from the boiling can passes through the copper pipe with the Bunsen burners under it and becomes very hot.

24.2.8 Broken bottle, ice bomb
(Dangerous experiment not suitable for schools!)
1. Completely fill a small bottle with water, wrap it in two old tea towels and put it in the freezer.
After a few days remove the broken bottle.

2. Completely fill a cast iron bomb with water, put in a dry ice / acetone mix at -77oC, then put this in a strong box.
Be careful! When the water in the bomb freezes, the cast iron bomb explodes showing that the volume of ice is greater than the
volume of an equal mass of liquid water and that freezing water produces great pressure.

24.2.9 Liquefying dry ice
Press down on a piston on dry ice in a clear tube until at 5 atmospheres liquidation occurs.
Put some carbon dioxide in a small transparent syringe and squeeze to liquefy.

24.2.10 Boil by cooling
See diagram 24.2.10: Boil by cooling
Use a flat bottom flask (Erlenmeyer flask) or a flask with a bottom curving up in the centre (Cincinnati flask).
Heat water in the flask to boiling, fix a stopper and invert the flask.
Pile crushed ice on the bottom of the flask.
The water in the neck of the flask begins to boil as the air in the flask cools and contracts.

24.3.01 Cooling by evaporation, sweat glands
During evaporation the faster particles escape from the surface of the liquid so the net kinetic energy of all the particles in the liquid
is lowered.
So evaporation has a cooling effect.
We feel cooler when evaporation of liquids from our sweat glands occur but when the humidity is high we feel uncomfortable because
evaporation does not occur easily.
Evaporation is the process in which a liquid turns to a vapour without its temperature reaching boiling point.
At any time, a proportion of its molecules will be fast enough (have enough kinetic energy) to escape from the attractive intermolecular
forces at the liquid surface and into the atmosphere.
As the mean kinetic energy of the molecules of the liquid rises, the number of the molecules of the liquid possessing enough energy to
escape rises.
So the rate of evaporation rises with increased temperature.
Condensation is the conversion of a vapour to a liquid as it loses heat.
This is frequently achieved by letting the vapour come into contact with a cold surface.

24.3.02 Sublimation
Sublimation is the change of state from solid to vapour, or from vapour to solid, which occurs without passing through an intermediate
liquid state.
Under normal temperature and pressure, only ice, dry ice, camphor, sulfur, phosphor may sublimate.
In the formation of white frost, for example, water vapour in the atmosphere passes directly from the vapour state to the solid crystals
of ice that make up frost.

24.3.0 Evaporation
See: Evaporating dish (Commercial)
Evaporation occurs when water particles have enough kinetic energy (latent heat) to change from the liquid phase to the vapour phase.
The speed of evaporation depends on atmospheric pressure, relative humidity, speed of air flow, level of solar radiation (especially
infrared radiation), water purity, water surface area, water temperature.
Ignoring solar radiation, evaporation rate, E = (30.6 +32.1U) (Pw -PA) / δH
where, E = kg m2 per hour
U = wind speed, metres per second
PW = saturation vapour pressure for that specific water temperature, (mm Hg)
PA = saturation vapour pressure (SVP), at the air dew point temperature, (mm Hg)
δH = latent heat of water for that specific water temperature (KJ per kg).

24.3.1 Water "lost" by evaporation
See diagram 24.3.1: Evaporation from a wet towel
Wet a towel then screw it until no water drops.
Hand it on a hanger and make it flat.
Use a stick and a piece of rope.
Tie the rope to the middle of the stick.
Hang the stick up with the rope.
Prepare a weight and two pieces of short rope.
Tie the weight and the towel to separate two sides of the stick.
Adjust their positions to make the stick balance.
Leave them aside without change in position for one hour.
The weight and the towel no longer balance, the weight is heavier
Where did the water in the towel go?
Add else weight on the hanger to make the stick balance again.
Take the added weight on a beam to measure its mass.
The mass is just that of lost water.
Repeat the experiment but place an electric radiator near the towel.
Repeat the experiment again but let the towel pile together when hang it on the hanger.
The evaporation speed is related to the temperature and the evaporation area.
The higher the temperature, faster the evaporation speed, the larger the evaporation area, faster the evaporation speed.
On the contrary the evaporation speed is slower.

24.3.2 Rate of evaporation
See diagram 24.2.2
Fill a large flat dish with half full of water.
Use another container, the diameter of which is smaller than that of dish, height of which is higher than that of dish, fill the same
amount of water in it.
Mark the level of water, arrange them side by side in a place where the temperature and case of air flowing are the same.
The next day, observe in which container the level of water is lower.
If the level of water varies a little, you may wait for another day.

24.3.2.1 Rate of evaporation at different air speeds
Use two identical flat rigid dishes <300 mm depth, control dish and experiment dish, and water from the tap.
Fix a millimetre scale vertically in both dishes.
Put identical volumes of water in both dishes and record the depths of water.
Record the surface area of the water in the dishes.
Fix a three speed fan to blow wind evenly over the water surface of the experiment dish.
Record the water temperature, atmospheric pressure, wet bulb and dry bulb temperatures.

24.3.3 Evaporate dichloromethane
Float a small test-tube of dichloromethane in a larger beaker of water.
Inset a glass tube into the dichloromethane and blow down the tube.
The volatile dichloromethane evaporates at room temperature, loses latent heat of vaporization and some touching the small test-tube
freezes.

24.3.4 Evaporate from a blackboard
1. Use a moist sponge or cloth wet two areas of equal size some distance apart on a blackboard surface.
Then fan one area with a piece of paper board, leave another to evaporate without fanning.
Observe which area evaporates first.

2. Fasten a piece of cloth over a wooden frame that is about 30 cm2 and 3 cm thick.
Wet the cloth, but do not drop water.
Next make two wet areas on blackboard surface.
Cover one wet area with a frame with wet cloth, leave the other area open.
After a few moments, remove the frame and observe which of the two areas evaporates faster.

24.3.5 Cryophorous
A cryophorus shows the freezing of water by its own evaporation.
It has two glass bulbs, connected by a glass and contains only water and water vapour, and no air.
The water is in one of the bulbs is cooled below 0oC in a ice / salt mixture or an alcohol / dry ice mixture, the equilibrium shifts to
produce more water vapour.
This cools the water at the top of the tube and it quickly freezes.

24.3.6 Freezing by evaporation
Freeze water in a watch glass over a dish of sulfuric acid in a bell jar.

24.3.7 Drinking bird heat engine, drinking duck, dippy bird, dunking bird
See: Thermodynamics, (Drinking bird, Dippy bird), Commercial)
"Drinking Bird", crude heat engine, temperature difference causes cyclical motion (toy productl) .

See diagram 24.3.7: Drinking bird
1. The drinking bird is a small heat engine that converts a temperature difference into cyclical motion.
It is not a toy, and may be adjusted to work properly by moving the sleeve up or down the connecting tube.
However, it contains toxic and hazardous liquids, so only the teacher, not the students, should make any adjustments to it.
Some toy product versions of drinking bird do not work well below a room temperature of 25oC!

2. The drinking bird consists of two glass bulbs for head and body connected by a glass connecting tube that dips down into a volatile
liquid in the body, usually methylene chloride (dichloromethane, CH2Cl2, "methelene chloride").
Methylene chloride is an organic solvent, in paint strippers, and has a low boiling point of 39.6oC.
Previously ether, Freon and other volatile liquids were used.
A sleeve attached to a pivot, fulcrum, clasps the glass tube.
Each end of the pivot is inserted into holes at the top of the legs.
The head bulb contains vapour and is covered with absorbent material, e.g. felt, to form a head covering and a beak.

3. Put a container full of cold water in front of the drinking duck such that the height of the sides of the container are just lower than the
height of the pivot.
Hold your warm fingers around the body bulb to increase the vapour pressure and force liquid up the connecting tube into the head bulb.
The centre of gravity lowers so the head tips forward.
The beak and head becomes immersed in the water which the absorbent material absorbs.
The lower end of the connecting tube rises above the level of liquid in the body.
Bubbles of vapour pass up through the connecting tube to equalize the vapour pressure in the head bulb and body bulb.
Liquid is displaced down the connecting tube to re-enter the body.
The centre of gravity rises and the head of the drinking bird rises again.
The body bulb + liquid is heavier than the head bulb.

4. Cooling by evaporation.
As the water in the absorbent material evaporates, vapour in the head loses latent heat of vaporization, becomes cooler, and condenses.
The vapour pressure in the head bulb decreases relative to the vapour pressure in the body bulb.
Liquid rises up the connecting tube and enters the head bulb.
The head becomes heavier than the body.
The drinking duck tips down and the absorbent material in the beak absorbs water from the container.
Water starts to evaporate from the wet absorbent material and the cycle starts again.
A similar apparatus called a "hand boiler" or "libido detector" is held in the palm of the hand.
In the last century, if a lady was to shake hands with a man for the first time, she would first cool the palm of her hand by holding an
egg-shaped piece of cool marble.

24.3.8 Cooling by evaporation, ether or ethyl chloride
Use ether or ethyl chloride is used to freeze water in a small dish or cool a thermometer.

24.3.9 Water bag, Coolgardie safe
People living in the bush keep water in a canvass water bag hung under a tree in the shade or attached to the bumper car of a vehicle.
Water seeps through the canvass and evaporates from its outside surface to keep the water cool.
Water evaporates from the bag, latent heat is required to change the water from liquid to vapour state, this heat is taken from the
water in the bag, and therefore its temperature falls, a steady state being reached when the quantity of heat taken from the water to
evaporate part of it is equal to the heat received from the surroundings on account of the temperature difference between the water in
the bag and the surroundings.
The more rapidly the water evaporates, the greater the temperature difference between the water and the surroundings.
So the bag should be placed in a current of air in the shade.
On a hot, muggy day the relative humidity is high, therefore evaporation is slow, and the water will be only slightly cooler than the
surroundings.
On a hot, dry day the relative humidity is low, evaporation is rapid, and the water will be considerably cooler than the surroundings.

24.3.10 Evaporate from a microscope slide
1. Put a drop of a volatile liquid, e.g. propanone, on a microscope slide.
Observe the drop and note the time for the drop to evaporate.
Note the comparative time for evaporation under different conditions, e.g. drop spread out, drop on warmed microscope slide, drop
cooled by fanning, drop warmed by your blowing on it.
Propanone (CH3COCH3, acetone) is highly flammable so use a teat pipette as in Universal indicator bottles.
Before doing this experiment make sure that there are no flames or lighted Bunsen burners in the laboratory.

2. Put one drop of acetone on a microscope slide then record the time for it to evaporate under the following conditions:
Still air at room temperature, with a fan turned on, while blowing across the microscope slide to create a current of warm air, with the
microscope slide first warmed with the palm of your hand or held over hot water.
List the relative speeds of evaporation in rank order.
Repeat the experiment with one drop of acetone on a microscope slide, spread thinly with a tooth pick or matchstick.
Compare the results of experiments with the drop and the drop spread thinly.

24.4.1 Sling psychrometer to find relative humidity.
Two thermometers one with a wet wick are mounted on a device swung around the head.

24.4.2 Dew point measurement
Evaporating alcohol cools a shiny surface until dew forms.

24.4.3 Condensation nuclei, supersaturation
Superasaturation in the atmosphere refers to a vapour that has a higher pressure or partial pressure than the vapour pressure of that
compound.
When perfectly clean air is slightly cooled below its dew point temperature the air becomes supersaturated but no droplets form
because saturation vapour pressure is greater over a curved surface than over a plane surface.
The smaller the radius of curvature the greater the effect such that if a small drop does form it immediately evaporate again.
In supersaturated air water droplets may precipitate upon being disturbed as in a cloud chamber or by increasing the rate of
supersaturation to a four-fold value and more so that precipitation occurs, first on negative ions, later on positive ions and later on the
water molecules themselves.
However, particles of dust provide condensation nuclei on which condensation may occur because the radius of the dust particles are
large enough to make the curved surface effect negligible.
1. Hold an extinguished match in the steam from a tea kettle.
2. Place moistened cotton in a bell jar and evacuate it until fog forms.

24.7.0 Steam
When liquid water is heated it forms a vapour called steam, the invisible gaseous form of water formed by boiling.
On cooling, this vapour forms liquid water.
Boil water in a kettle and observe the apparent space just at the opening of the spout.
Hold the prongs of a cold metal fork in the space and observe the condensation of water vapour.
At a short distance away from the spout, observe a white cloud of water vapour condensed in the air when the water vapour hits
particles in the air.

24.7.1 Steam turbines
See diagram 24.4.1: Simple steam turbine
Attach a sharpened pencil with eraser end to the clamp which attached to the ring stand.
Cover a test-tube downward on top of the pencil.
Then cut a circular piece of cardboard about 6 cm in diameter, punch a hole exactly the same to test-tube in diameter to fit it tightly
around the test-tube.
It is better to tape it to the test-tube.
Use 10 tooth sticks, glue them around the edge of the cardboard.
Note to let each of them half out of the cardboard.
Cut 10 piece of aluminium foil each takes a shape of square.
Fold each one in half to get a shape of rectangular, tape each on the tooth stick which should be in centre of the folded surface and
their inclined direction should be the same.
Pour water into an Erlenmeyer flask, cover a rubber stopper with a bent glass tube tightly.
Put the flask on the tripod and heat it over an alcohol burner.
At the same time, adjust the position of the tube clip to make aluminium foil blades are just opposite to the open end of the bent glass
tube and at the same level.
As the water in flask boils and the steam has produced, observe what happens.

24.7.2 Refrigerator, ice chest, portable ice box, espresso coffee machine
See diagram 24.7.2: Refrigerator
1. Refrigerator
In a refrigerator, food loses heat by conduction, convection and radiation to the air, shelves and inner cabinet.
Volatile refrigerant fluids in a coil around the freezer evaporate and so take in latent heat to cause cooling.
The vapour from this evaporation is removed by an electric pump to the black heat exchanger coil with cooling fins behind the
refrigerator which radiates heat to the surrounding air.
When the electric pump removes vapour from the coil around the freezer it also reduces the pressure in the coil and lowers the boiling
point and induces more evaporation.
In the heat exchanger coil the vapour compresses, loses latent heat of vaporization, and is returned as a liquid to the coil around the
freezer.
The electric pump can be adjusted to switch on and off to control the temperature of the refrigerator.
Industrial refrigerators use ammonia and Freon-22 refrigerants and other gases, e.g. carbon dioxide.
Domestic refrigerators use less toxic refrigerants, e.g. Freon 11.
2. Ice chest, portable ice box
Ice chests and portable ice boxes, e.g. "Esky" cooler, are insulated boxes containing an ice block or ice in small cubes.
Heat is lost from food and drink by conduction if touching the ice, convection currents and radiation into the ice.

3. Espresso coffee machine
An espresso coffee machines make steam that can be passed through water containing ground coffee or milk.
The steam condenses quickly liberating latent heat to brew coffee quickly and froth the milk for cappuccino or other milk coffees.

24.7.3 Hero's engine, eolipile
See: Hero's Engine (Commercial)
Hero's engine (Hero of Alexandria, 10-70 AD), a radial steam engine, is a sphere or cylinder metal container with oppositely curving
nozzles.
When water in the container is heated, the steam leaves through the nozzles in opposite directions to cause opposite thrust, torque, and
rotate the container.
It demonstrates the power of steam, as jets of steam cause the rotation.

24.7.4 Heat loss by the human body
The human body loses heat in raising the temperature from room temperature to body temperature of the surrounding air and objects,
e.g. a chair, clothes, food and drink entering the body, cold air inhaled after the warm air has been exhaled.
The amount of heat lost is the sum of ms(t2-t1), where m = mass of objects or air, s = specific heat of objects or air, t2-t1 = change in
temperature.
Also, the human body loses heat in evaporating water as perspiration from the skin body and as vapour in the breath.
The amount of heat lost is the sum of mL, where m = mass of water vaporized and L= latent heat.
The human body loses heat by conduction, convection and radiation, and evaporation of moisture.
In cold climates, the loss of heat can be reduced by suitable clothing made of bad conducting material, e.g. wool or fur so that the air
does not contact the skin except at the face, and so reduce convection currents.
Also, evaporation into the atmosphere occurs only from the uncovered portions of the body.
Evaporation from other parts of the body occurs into a closed air spaces that ceases when the spaces becomes saturated.
Loss of heat by radiation is reduced by wearing bad radiators, e.g. white, or shiny surfaces for the outer garment, unless in direct
sunlight.
In hot climates, the gain of heat can be reduced by suitable clothing that absorb little direct radiation as possible, e.g. white clothes.
Clothes should be loose so that the air can move freely round the body, and so increase the loss of heat by evaporation.
If the air temperature is lower than the body temperature, clothes should be good conductors to conduct heat away from the body and
as loose or scanty to facilitate cooling by convection and evaporation.
If the air temperature is higher than the body temperature, cooling depends on evaporation into air with free access to the body and in
circulation to increase evaporation.
On a cold still day, heat radiated from the body raises the temperature of the air in contact with the body, so that the temperature of the
immediate surroundings is higher than the actual air temperature.
On a windy day the air in contact with the body is continually being removed, and so has approximately the true air temperature.
Also, on a windy day evaporation occurs more rapidly than on a similar still day, so that the loss of heat by evaporation is also increased.

24.10.0 Heat energy changes solid to liquid
Heat energy changes solid to liquid, m.p., latent heat of fusion, naphthalene, ethanamide (acetamide)
See diagram 24.10.0: Liquid naphthalene solidifies
Put crushed naphthalene or ethanamide (acetamide) in a test-tube in a container of water.
Heat gently until all the substance has melted.
Remove the test-tube from the container and fix a thermometer with its bulb in the melted substance.
Stir the substance with a thermometer while the substance cools and record the temperature every 30 seconds for 6 minutes.
Plot a graph of temperature against time.
At first the temperature drops while the substance remains liquid.
Then the temperature remains the same while the
Substance changes from liquid to solid.
When all the substance is solid, the temperature starts to drop again.
The melting point, m.p., is the temperature when a solid changes to a liquid.
The specific latent heat of fusion of a substance, L, is the quantity of heat required to change one kilogram of the substance from solid
to liquid without change in temperature.
The unit is joule / kg, J kg-1.
The specific latent heat of fusion of ice = 3.34 × 105 joule / kg, 334 kJ kg-1.

4.36 Newton's law of cooling
Newton's law of cooling states that the rate of loss of heat from a body both by radiation and convection is proportional to the
difference between the temperature of the body and the temperature of the surroundings.
It applies only to small ranges in temperature.
Test whether a hot cup of coffee cools faster than a warm cup of coffee.
Record the room temperature, e.g. 17.5oC.
Use identical coffee cups.
Put the same volume of hot coffee or warm coffee in the coffee cups.
Insert a thermometer and use it to keep stirring gently.
Record the temperature every two minutes for twenty minutes while still stirring.
In the second column, record the temperature of the cooling water every two minutes.
In column D, record the difference between the temperatures every two minutes and the room temperature.
Calculate F / D for each two minute interval.
The mass of coffee in the coffee cup is constant so the rate of heat loss of the coffee is proportional to the fall in temperature.
The rate of fall of temperature is proportional to the mean difference of temperature between the coffee and the surroundings.
So the fall in temperature during time interval / mean difference in temperature between the coffee and surroundings = constant.
As the temperature of the body is higher and the temperature of surroundings is lower, the difference of the two temperatures is
greater, so the rate of heat loss of the body is faster.

Table 4.36
Time in
minutes
Temp.
of coffee
F = fall in temp.
in the last 2 minutes
Mean temp.in last 2 minutes
(to nearest 0.1oC)
D = difference between
water temp.and room temp.
F / D
(constant)
0 44.7oC .
.
. .
2 41.4oC 3.3oC 43.1oC 25.6oC 0.13
4 38.7oC 2.9oC 40.1oC 22.6oC 0.13
6 36.1oC 2.6oC 37.4oC 19.9oC 0.13
8 33.7oC 2.4oC 34.9oC 17.4oC 0.13
37.23 Moving air affects the rate of evaporation
With a moist sponge or cloth, wet two areas of equal size some distance apart on a cool blackboard surface.
Fan one area with a piece of cardboard and leave the other to evaporate without fanning.
The wind difference causes the difference in rate of evaporation.

37.24 Moisture in the air affects the rate of evaporation
Half fill two identical dishes with water.
Suspend a wet cloth over one dish.
Note the different rates of evaporation in the two dishes.

37.29 Cloud in a bottle
See diagram 37.135: Cloud in a bottle
1. Add a small amount of water to a plastic drink bottle.
Light a match, blow it out and drop the smoking match stick into the bottle.
Immediately screw on the lid and start squeezing the bottle.
The cloud in the bottle from the burning match stick smoke disappears when you squeeze the bottle.
Inside the bottle water vapour condenses on the smoke particles that act as nucleation sites.
When you squeeze the bottle the temperature of the air inside it increases and the condensed water vapour on the smoke particles
evaporates.

2. Pour a little water into a jar wider than your fist.
Screw on the lid tightly and leave for half an hour.
Put a stick of white blackboard chalk in a plastic bag and use a hammer to crush it to a fine powder.
Cut the neck off a party balloon.
Open the jar and put in the chalk powder and immediately stretch the balloon over the opening of the jar until the rubber is horizontal
over the opening.
Fix a rubber band around the neck of the jar to keep the balloon stretched tight.
Push the balloon down with your fist and hold it down for one minute.
Raise your fist and take off the balloon to see a cloud formed in the jar.
Your fist compressed the air in the jar to warm it and absorb more water vapour.
When you remove the balloon cover, the air cools and some of the water vapour condenses on the chalk dust to form cloud.

3. Use a big bottle with a one-hole stopper fixed with a 10 cm length of glass tubing.
Put 3 cm of warm water in the bottle and drop chalk dust into the air inside the bottle.
Connect the end of the glass tubing to a bicycle pump with rubber tubing.
Hold the stopper tight in the bottle and pump air in.
When the air becomes compressed inside the bottle, close the rubber tubing with a strong clamp and remove the bicycle pump.
Keep a tight hold on the stopper and open the clamp to let air blow out through the glass tubing.
A cloud forms in the bottle.
When the air comes out of the bottle, the air left in the bottle expands and cools, reducing the temperature in the bottle below the dew
point.
The moisture then condenses and forms a cloud.
Similarly, when warm air rises above the Earth the air pressure is reduced.
The air expands and cools, and clouds form when the cooling goes below the dew point.

4. Make a better cloud by using smoke instead of chalk dust or by dropping a lighted match into the bottle.
The smoke or lighted match provides small particles of the carbon that float in the air and act as condensation cores as steam sticks to
them and condenses to water.

5. Pour several mL water in a bottle, cover the bottle tightly with a rubber stopper on which a hole is punched.
Insert a short glass tube into the stopper, then connect the pump used to pump the bike with the glass tube on the stopper by a piece
of rubber tube and short glass tube.
The mouth in the pump should be removed first.
Pump air to the bottle and let the students observe the phenomenon in the bottle, especially at the moment that the stopper jumps up
suddenly.
If it is not easy to observe clearly, you may throw a lighted piece of match in the bottle, then a little fog appears in the bottle.
Repeat the step, much fog will appear.
When the stopper jumps up suddenly, the volume of the air in the bottle increase suddenly leading to reduce the pressure in the bottle,
and the temperature decreases with it.
This results in the evaporation of water and cool condensation later.
The reason that the burning match can make the process of condensation faster is due to the burning match provided the small
particles of the carbon.
It is these small solid particles floating in the air that acted as a condensation core to cause the water steam sticks to them and
condenses.
The experiment shows why it is easy to produce fog and rain in the areas contaminated.
For example, in those cities of heavy industry there are more fog days than the non-industrial cities.