School Science Lessons
24. Change of state, cooling by evaporation, dew point and humidity, freezing point, heat engines, latent heat, sublimation
2012-01-28 SP
Please send comments to: J.Elfick@uq.edu.au
Table of contents
24.0.0 Change of state and solutions
24.3.0 Cooling by evaporation
24.4.0 Dew point and humidity
24.5.0 Freezing, freezing point
24.7.0 Heat engines, steam turbines, refrigerator
24.7.3 Heat loss by the human body
24.2.1.0 Latent heat of fusion and vaporization
24.3.0.1 Sublimation
24.0.0
Change of state and solutions
24.1.07 Critical point, critical
temperature
24.1.06 Osmotic pressure equation, Morse equation
24.1.0 Phase changes, liquid / solid, melting
point
24.2.0 Phase changes, liquid / gas, boiling
point
24.1.05 Van't Hoff factor for dilute solutions
24.5.0 Freezing, freezing
point
24.3.6 Freezing by evaporation
24.1.0 Phase changes liquid / solid, melting point and freezing
point
24.1.04 Freezing point depression and boiling
point elevation
24.1.3 Melting point and pressure
7.4.3.0 Melting point of ice and
freezing point of water
24.1.14 Molal freezing point constant of cyclohexane
solvent
24.1.15 Molar mass of solute from depression
of freezing point
24.3.0 Cooling by evaporation
24.3.0 Cooling by evaporation
24.3.8 Cooling by evaporation, ether or ethyl
chloride
24.3.5 Cryophorous
24.3.7 Drinking bird, 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)
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
24.3.9 Water bag, Coolgardie safe
24.3.1 Water "lost" by evaporation
24.4.0 Dew point and humidity
37.15 Atmospheric moisture
37.29 Cloud in a bottle
37.38.2 Cold front
24.4.3 Condensation nuclei, supersaturation
37.48 Dew point
37.5.3 Dew, dew point
37.8.3 Dew point hygrometer
24.4.2 Dew point measurement
37.6 Hair hygrometer
37.38.4 Hurricanes
4.41 Ice experiments (Primary)
37.25 Moisture condenses on cool surfaces
37.20 Moisture from breathing
37.18 Moisture from plants
37.17 Moisture from soil
37.27 Rain cycle
5.22 Rain gauge (Primary)
37.8.2 Sling psychrometer
6.17 Relative humidity (Primary)
37.8.4 Relative humidity table, depression
of the wet bulb
37.5 Rain gauge, precipitation gauge
5.22 Rain gauge (Primary)
37.30 Snowflakes
37.21 Surface area affects evaporation
37.22 Temperature affects the rate
of evaporation
37.38.3 Tornadoes
37.38.1 Warm front
37.26 Water cycle
37.16 Water "lost" by evaporation
37.8.1 Wet and dry bulb hygrometer
(wet and dry bulb thermometers)
24.7.0 Heat engines, steam
turbines, refrigerator
24.7.2 Refrigerator, ice chest, portable ice
boxes, expresso coffee machine
24.7.1 Steam turbines
24.2.1.0 Latent heat of
fusion and vaporization
4.11 Heat energy to change liquid
to vapour, boiling point, latent heat of vaporization
4.10 Heat energy to change solid
to liquid, melting point, latent heat of fusion
24.1.4 Latent heat, change of state, melting
and boiling
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
24.1.0 Phase changes liquid / solid,
melting point and freezing point
24.1.0 Phase changes liquid / solid, melting point and freezing
point
24.1.6 Bitumen foaming and bitumen decay
24.1.17 Crystal growth on the overhead projector
24.1.04 Freezing point depression and boiling
point elevation
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
24.1.13 Hot water freezes faster than cold water,
the Mpemba effect
4.41 Ice experiments (Primary)
24.1.3 Melting point and pressure
24.1.2 Lift ice cube with salt
4.39 Melt different solids (Primary)
7.4.1 Melting point and cooling curve
of stearic acid
7.4.2.1 Melting point of 1,4-dichlorobenzene,
C6H4Cl2
7.4.3.0 Melting point of ice and
freezing point of water
7.4.0 Melting point, m.p. of solids
3.2 Melting point of naphthalene
3.3 Melting point of naphthalene using
a capillary tube
7.4.2.0 Melting point of substances
3.4 Melting point of substances changed
by impurities
24.1.14 Molal freezing point constant of cyclohexane
solvent
24.1.15 Molar mass of solute from depression
of freezing point
24.1.18 Metglas
24.1.01 Nucleation
24.1.03 Raoult's law
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 pure water and solution
of a non-volatile solvent, colligative properties
24.1.16 Water crystals in soap film
24.1.19 Wood's metal
24.2.0 Phase changes, liquid / gas, boiling point
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
7.5.0
Boiling point, b.p. of liquids
3.5.1 Boiling point of sodium chloride
solution
3.5 Boiling point of water
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
24.2.4 Pressure and boiling point of water
3.8 Pressure of the atmosphere affects
the boiling point
24.2.7 Supercooled water
24.2.5 Temperature of steam above water boiling
3.7 Volatility of different liquids
24.0.0 Change of state
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, melting point and freezing point
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
pure water and solution of a non-volatile solvent, colligative properties
See diagram 24.1.02: Vapour Pressure | See also: Table of saturated vapour pressure over water,
Psvp
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
X 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 X kf X 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 X kb X 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 Van't Hoff factor 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.
24.1.06 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
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 (374°C) and 22.064
MPa (3200 PSIA or 218 atmospheres, atm.).
24.1.2 Lift an ice cube with salt
See diagram 24.2.7
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 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.
3. 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.
4. 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.
5. 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 X 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 X 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 x 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 x 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 which 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
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 54°C so that they dissolve
in their own water of crystallization. Keep heating the solution up to 100°C,
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 commercial heat pack 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 264–289
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 32°C
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 X Kf X
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 a mobile 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 without
crystallization. It has interesting electrical and magnetic properties.
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,
at the moment of boiling the temperature of the liquid remains. Pour a
certain amount of liquid into a beaker. Heat the beaker in a place where
there is not wind flow and note to stir the liquid by 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 experiment recording and graph 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. Use
the mass of the liquid multiply its specific heat, again multiply the increased
temperature in every unit time use ed from the above graph, you can calculate
the amount of heat absorbed by liquid every minute. As the amount of heat
absorbed by beaker is far less than that of liquid absorbed, so the former
is omitted in calculation. Heat the liquid continuously until it boils. After
that although heat it but the temperature of the liquid no longer increases
and produces the steam, i.e. some liquids become gases. After boiling 10
minutes, stop heating and cool the liquid until its temperature becomes back
to that of indoors. Measure the weight of the liquid again. Calculate the
loss mass of the liquid that is also the mass becomes steam. Use the total
amount of heat absorbed in 10 minutes by liquid divides the loss mass of
the liquid, that is the amount of heat needed for liquid every unit mass
to becomes steam, 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 you see steam 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
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) = lambda x m So lambda
= 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 lambda = 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
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 to allow the test-tube is near
the bottom of beaker. Measure the temperature of water in beaker, then
in test-tube by a thermometer which is held by a clip. Heat the water in
beaker over the alcohol burner, measure the temperatures in beaker and
test-tube each continuously. Especially after the water boils, see if the
temperature in beaker is equal to that in 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 24.4.1: Simple steam turbine |
See 3.8: Pressure affects the boiling point
| See diagram 24.239: Pressure affects boiling point
of water
A Steam, B Boiling tube, C Water
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.
Motor car radiators have a pressure cap to keep water in a liquid state
at temperatures over 100oC.
Height above sea level and boiling point: 600 m 98oC, 1500
m 95oC, 2000 m 93oC, 3000 m 90oC.
1. Go to a high mountain and put a raw egg in an open cooking pot containing
cold water. Heat the water in the pot. The water boils and evaporates at
a low temperature so you cannot cook the egg! 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.
2. 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.
3. 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.
4. Food cooks quickly in a pressure cooker because the pressure of the
steam above the water in the cooker can be twice 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.
5. Half fill a round bottom flask with water and insert a two-holes
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.
6. Fit a three-holes 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.
Use a thermometer which measuring range is over 100oC, a
pressure cooker and a aluminium cooker with a lid on which there is a 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, you should 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
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
The steam from the boiling can passes through the copper pipe with the
Bunsen burners under it and becomes very hot.
24.2.7 Supercooled water
Cool a small test-tube of water in a Peltier device or dry ice /
alcohol bath and use a thermocouple to record the temperature. Shake to
freeze and the temperature will rise. A Peltier device is a very small solid
state device that functions as heat pump made of layers of ceramic plates
and bismuth telluride. Apply d.c. current to move heat from one side to the
other so the cold side can cool a small electronic device. They contain no
moving parts and no Freon refrigerant.
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-bottomed 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.
24.3.0 Cooling by evaporation
During evaporation the faster particles escape from the surface of the liquid
so the net kinetic energy of 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.
24.3.0.1 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.
24.3.01 Evaporation
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 = windspeed, 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
Understand evaporation is a ubiquitous phenomenon and study the factors
related to evaporation sheep. 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 poison for one hour. What happened?
Where did the water of 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 0°C
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, drinking duck, dippy bird,
dunking bird
See diagram 24.3.7 | See diagram 24.3.7 photograph
Order online: Drinking Bird,
crude heat engine, temperature difference causes cyclical motion
See pdf: 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 commercial 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"). 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 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.
5. A similar apparatus called a "hand boiler" or "libido detector" is
held in the palm of the hand.
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.
24.7.1 Steam turbines
See diagram 24.4.1
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 boxes,
expresso coffee machine
See diagram 24.7.2
1. 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 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. Expresso 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 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.