WEEK 2 NOTE

LATENT HEAT
OBJECTIVE
Students should be able to
i. Define latent heat
ii. Ii, differentiate between latent heat of fusion and vaporization
iii. Explain how to determine the latent heat of fusion and vaporization
iv. Explain boiling. Melting and evaporation
v. List and explain the factors that affect boiling, melting and evaporation

LATENT HEAT
Heating a substance does not always increase its temperature. If heat is supplied to a block of ice at 0°C, it melts
(change of state) and becomes water at the same temperature (0°C). A thermometer dipped in the melting ice
does not show an increase in temperature despite the fact that heat energy is being supplied to it. The constant
reading of the thermometer does not indicate that energy is not given to the ice while it melts. The heat supplied
to the ice during the process of melting is called latent or hidden heat. Latent heat is used to break the bonds
between molecules and to push them apart. It does not increase the kinetic energy of the molecules therefore the
temperature remains constant.
NOTE:
When a substance is changing state, it absorbs/gives out thermal/heat energy which does not result in
temperature change. This heat absorbed or given out during change of state is known as latent heat.

Specific latent heat of fusion

Specific latent heat of fusion is the heat energy needed to change 1kg of a solid to liquid at a constant
temperature.
Heat absorbed = Mass × Specific latent heat
Q = mlf
Where lf is the specific latent heat of fusion. The unit of specific latent heat of fusion is joule per kilogram (J kg -1) or
joule per gram (J g-1). The specific latent heat fusion of ice is 336000 J kg-1 or 336 J g-1. This means that 336000 J is
needed to melt completely 1 kg of ice.

Latent heat of fusion

Latent heat of fusion is the heat energy needed to change any given mass of a solid to liquid at a constant
temperature.
Worked examples
1. Given that the latent heat of fusion of ice is 336000 J kg-1, the specific heat capacity of ice and water are 2100 J
kg-1 K-1and 4200 J kg-1 K-1 respectively, calculate the heat needed to:
(a) melt 250 g of ice at 0°C;
(b) change 2 kg of ice at 0°C to water at 30°C;
(c) change 1.2 kg of ice at -20°C to water at 0°C.
Solution
(a) Mass of ice 250g = 0.25kg, lf= 336000 J kg-1
Q = mlf= 0.25 × 336000 = 84000 J
(b) Two stages are involved in this change of state. Firstly, the ice melts and becomes water at 0°C, and lastly water
gains heat energy as the temperature increases from 0°C to 30°C.
Q = mlf + mcwΔ θ
Q = 2×336000 + 2×4200×30
Q = 672000 + 252000
Q =924000 J
(c) Q = mcΔ θ + mlf
Q = 1.2×2100×20 + 1.2×336000
Q = 50400 + 403200
Q = 453600 J

2. Ice of 0.1 kg at 0°C is added to 1.0 kg of water at 30°C and stirred until all the ice melts completely. If the
temperature of the mixture is 20°C, calculate the specific latent heat of fusion of the ice.
{Specific heat capacity of water = 4200 J kg-1 K-1}
Solution

3. A copper cup of mass 0.5 kg contains 1.5 kg of water at 30°C. If the cup and its content are placed in a deep
freezer at -10°C, find the total heat lost by the cup and water to reach the same temperature as the deep freezer.
{Specific heat capacity of water = 4200 J kg-1 K-1, specific latent heat of fusion of ice = 336000 J kg-1 and specific
heat capacity of copper = 400 J kg-1 K-1} specific heat capacity of ice = 2100 J Kg-1 K-1
Solution
Copper loses heat in only one stage; from 30°C to -10°C. Heat lost by copper Q1 = mcccΔθc Heat lost by copper =
0.5×400×40 = 8000 J.
Water lost heat in three stages;
(i) Stage one; to cool from 30°C to 0°C
(ii) Stage two; to freeze to ice at 0°C;
(iii) Stage three; ice cools from 0°C to -10°C.
Q2 = mwcw Δ θw + mlf+ miciΔ θi
Q2= 1.5×4200×30+1.5×336000+1.5×2100×10
Q2= 189000+504000+31500
Q2 = 724500 J
Total heat lost Q = Q1 + Q2 = 8000 + 724500 = 732500 J.
Determination of specific latent heat of fusion of ice
Apparatus: Calorimeter, thermometer, blocks of melting ice, beaker of water, blotting paper and Bunsen burner.
Method
(i) Clean the calorimeter, put the stirrer inside, weigh and record the mass of the empty calorimeter and stirrer
(mc);
(ii) Pour water into the calorimeter until it is half filled and weigh the calorimeter a second time to determine the
mass of water (m1);
(iii) Warm the water until the temperature is about 5°C above room temperature, stir and record the temperature
θl;
(iv) Use a blotting paper to dry the melting ice and add to the water in the calorimeter. Stir until the ice melts
completely before adding another piece of ice;
(v) Keep adding ice and stirring until the temperature falls to about 5°C below the room temperature. Measure
and record the temperature θ2.
(vi) The calorimeter is weighed a third time to determine the mass of melted ice (m2).
Calculations
Mass of water mw = (ml – mc)
Mass of ice melted mi = (ml -mc)
Heat gained by ice Q = mlf + micwΔ θi
Heat lost by water Q = mwcwΔθw
Heat lost by calorimeter Q = mcccΔθc
Heat lost by water and calorimeter = Heat gained by the block of ice
mwcwΔ θw + mcccΔ θc =mlf + micwΔθi
Precautions
(i) The ice is dried with blotting paper before it is added to the water in the calorimeter.
(ii) The calorimeter and its content are stirred continuously to ensure even temperature.
(iii) The calorimeter and the water inside are warmed slightly above the room temperature and cooled below the
room temperature by the same degree to compensate for exchange of heat with the surroundings.

Melting and freezing points of a substance

The melting point of a substance is the constant temperature at which it changes state from solid to liquid.
Ice melts at a constant temperature (0°C). The heat supplied during the change of state is not registered by a
thermometer and is called latent heat of fusion. This is the flat portion (PS) of the heating curve.

The flat portion of the heating curve also represents the melting point of the substance. When heat is removed
from water, the temperature steadily falls until it freezes at 0°C.
The constant temperature at which a substance changes state from liquid to solid is the freezing point of the
substance.
Factors affecting melting point of a substance
The melting point of a pure substance can be changed in two ways: by adding impurities and changing the
pressure on it.
(i) Impurities alters the melting point of a substance: Addition of salt and liquids with low boiling point to ice will
lower its melting point. Instead of ice melting at 0°C, it will melt at a temperature lower than 0°C. The degree of
lowering of the melting point depends on the concentration of the impurity added. Adding salt to water in the
ratio of 1:10 will lower the freezing or melting point of water to -6°C. In cold countries, methyl alcohol and
glycerine are mixed with water in car radiators to prevent it from freezing. This is because they lower the freezing
point of water.
(ii) Pressure changes (or alters) the melting point and freezing points of a substance: The melting point of a
substance is altered if the pressure on it is changed. For substances which expand on freezing (contracts on
melting), their melting point is lowered if the pressure on them is increased. Ice for example, expands on freezing;
therefore, its melting point is less than 0°C when the pressure on it is increased.
Effect of pressure of the melting point of ice
We can demonstrate the effect of pressure on melting point of ice as follows:
• Place a big block of dry ice on two supports.
• Apply pressure at a point on the ice by placing a thin wire with two big weights at its ends on the ice.
• Observe that the wire passes through the ice without cutting it into two halves.
The wire exerts high pressure on the ice at the point of contact. The ice under the wire melts because the high
pressure at the point of contact lowers its melting point. The wire passes through the water formed, pressure is
reduced and the water above the wire refreezes. The melting of ice and refreezing of water is called regelation.
Regelation explains why the wire could pass through the ice without cutting it into two parts.
Figure 5: Demonstrating pressure effect on melting point
Two small pieces of ice can fuse or stick together by exerting pressure on them. The high pressure at the point of
contact reduces the melting point of the ice making the ice to melt. If the pressure at the point of contact is
removed, the water formed between the ice blocks refreezes to join the two pieces of ice together.

Latent heat of vaporization

If energy is constantly supplied to water, it boils when the temperature is 100°C. The temperature of the boiling
water remains constant at 100°C as water changes to steam. If enough energy is supplied to the water, it changes
to steam in a process known as vaporization. Vaporization of water occurs at a constant temperature of 100°C.
The heat energy supplied as water changes to steam is not registered by the thermometer and is called latent heat
of vaporization.

Specific latent heat of vaporization

The specific latent heat of vaporization is the heat energy needed to convert 1 kg of a liquid to vapour at
constant temperature.

The specific latent heat of vaporization of steam is about 2,260,000 J kg-1; that is the energy needed to convert
completely 1 kg of water at boiling point to steam at the same temperature is 2,260,000 J. On the other hand, if
1 kg of steam is condensed to water without change of temperature 2 260 000 J is given out. The energy given out
by 1 kg of steam to condense to water is much greater than the energy absorbed by 1 kg of water to increase its
temperature by 100°C. This is why steam burns is more dangerous than hot water burns.
The quantity of heat energy absorbed or radiated by a substance to change its state is given by:
Heat energy = Mass × specific latent heat of vaporization
Q = mlv
The unit of specific latent heat of vaporization is joule per kilogram (J kg -1) or joule per gram (J g-1).
Worked examples
1. Calculate the heat absorbed in each of the following cases:
(a) 300 g of water at 100°C is vaporized completely;
(b) 800 g of water at 70°C is converted to steam at 100°C;
(c) 1 kg of ice at 0°C is converted to steam at 100°C.
{Specific heat capacity of water = 4 200 J kg-1 K-1, specific latent heat of fusion of ice = 336 000 J kg-1 and specific
latent heat of vaporization of steam = 2 260 000 J kg-1}
Solution
(a) Q = mlv
lv = 2 260 000 J kg-1, m = 300 g = 0.3 kg
 Q = 0.3 × 2 260 000 = 678000 J
(b) The total heat absorbed is the sum of heat absorbed by water from 30°C to 100°C and the heat absorbed by
water to change to steam at 100°C.
Q = mwcwΔθw + mlv
Q = 0.8×4200×30 + 0.8×2 260 000
Q = 100 800 J + 1 808 000 J
Q = 1 908 800 J
(c) The heat is absorbed in three stages: heat absorbed to melt the ice at 0°C, the heat absorbed from 0°C to
100°C and the heat absorbed to change the water to steam at 100°C.
Q = mili + mwcw Δθw + mlv
Q = 1 ×336 000+1 ×4 200x 100+1 ×2 260 000
Q = 336 000+420 000+2 260 000
Q = 3 016 000 J.
2. Solid copper of mass 500 g at a temperature of θ°C is dropped into a vessel containing 300 g of water at 40°C.
If the mass of water that boils away is 20 g, calculate the initial temperature of the copper solid. Neglect the heat
capacity of the vessel.
{Specific heat capacity of water = 4200 J kg-1 K-1, specific heat capacity of copper = 400 Jkg-1 and specific latent heat
of vaporization of steam = 2 260 000 J kg-1}
Solution
Mass of copper solid mc = 500 g = 0.5 kg
Mass of water mw = 300 g = 0.3 kg
Mass of steam ms= 20 g = 0.02 kg
Heat lost by copper = mcccΔ θc = 0.5×400(θ- 100)

3. If the specific heat capacity of water = 4200 J kg-1 K-1, specific heat capacity of copper = 400 J kg-1, specific latent
heat of ice = 336 000 J kg-1 and vaporization of steam = 2 260 000 J kg-1, calculate the mass of steam that will pass
into the calorimeter of mass 60 g containing 80 g of water and 10 g of ice at 0°C to raise the temperature to 30°C.
Neglect heat lost to the surroundings.
Solution

Determination of specific latent heat of steam

Figure 6: Determination of latent heat of vaporisation of steam
Apparatus: Calorimeter, thermometer, beaker of water, steam can, steam trap, beam balance, tripod stand,
screen and Bunsen burner.
Method
(i) Clean the calorimeter and put the stirrer inside, weigh and record the mass of the empty calorimeter and
stirrer (mc);
(ii) Pour water into the calorimeter until it is half filled, 5°C below room temperature record the temperature (θ1)
and weigh the calorimeter a second time to determine the mass of water(m1).
(iii) Place the calorimeter in its lagging material and pass dry steam into the water in the calorimeter until the
temperature rises by 5°C above the room temperature.
(Steam gives up its latent heat as it condenses to water.)
(iv) Stir the water in the calorimeter and record its temperature (θ2); weigh the calorimeter a third time and
record its mass (m2).
Calculations
Mass of water mw = m1 – mc
Mass of condensed steam ms = m2 – mc
Fall in temperature of steam = 100 – θ2
Rise in temperature of water and cal. = θ2 – θ1
Heat lost by steam = mslv + mscw (100 - θ2)
Heat gained by water = mwcw (θ2 – θ1)
Heat gained by calorimeter = mccc (θ2 – θ1 )
Heat lost by steam = Heat gained by water and calorimeter
mslv + mscw (100 - θ2)= mwc(θ2 – θ1 )+mccc (θ2 – θ1 )
Precautions
1. Only dry steam is allowed to enter the calorimeter. The steam trap is used to remove condensed water (wet
steam) from passing into the calorimeter.
2. The calorimeter and its content are stirred continuously to ensure even temperature.
3. The calorimeter and its content are cooled below the room temperature and warmed above the room
temperature by the same amount.
4. A screen or shield is used to stop the heat from the burner from affecting the experiment.

Evaporation
When a liquid change to vapour (gas) at ordinary temperature, it is said to evaporate.
Evaporation is the breaking away of liquid molecules from the surface of the liquid and remaining outside the
liquid as vapour.
It occurs at any temperature above the absolute zero without reaching the boiling point of the liquid.

Cooling effect of evaporation

Carry out the following activities and state what you observe:
• Put some drops of methylated spirit on the back of your hand. What happens to your hand as the methyl spirit
evaporates?
• Choose two thermometers and record their readings, cover the bulb of one of the thermometer with cotton
soaked in methyl spirit and the bulb of the second thermometer uncovered.
• Place the two thermometers under a fan, observe and record the reading of the thermometer after few
minutes. What do you observe?
Observations
1. In the first activity, the hand feels cold because the methyl spirit extracts latent heat of vaporization from it.
The removal of heat from the hand makes its temperature to drop.
2. In the second activity, the temperature of the thermometer covered with cotton soaked in methylated spirit
falls rapidly. The fall in temperature is caused by the evaporation of the methyl spirit around the bulb.
We conclude that evaporation causes cooling. Liquids like methyl spirit, ether, petrol, etc., which evaporates with
ease at ordinary temperature, are called volatile liquids.
Figure 8.7 is a simple demonstration to prove that evaporation of liquids produces cooling.

Figure 7: Evaporation produces cooling effect

A watch glass with liquid ether is placed on a few drops of water. Dry air is blown through the ether to increase the
rate of evaporation. The evaporating ether takes latent heat of vaporization from the drops of water under the
watch glass. The temperature of the water drops falls and some minutes later, the water under the glass freezes.
The watch glass sticks to the frozen ice.

Application of evaporation
1. The refrigerator
The refrigerator is a machine which cools food items and other substances kept in them by extracting latent heat
from them to vaporize the volatile liquid or refrigerant. It consists of the following essential parts:
• A compressor or pump and expansion valve: The compressor is used to compress and condense vapour of the
refrigerant to liquid while the expansion valve allows for the expansion or evaporation of highly pressurized liquid.
• The cold compartment and the cooling fins: The cold compartment gives up latent heat to vaporize the Freon
liquid (the refrigerant) while the cooling fins conduct heat away from the condensed Freon liquid.
• The refrigerant: The volatile liquid that produces the cooling effect in the refrigerator is called the refrigerant.

Figure 8: The refrigerator

Working of refrigerators
Refrigerants like Freon, liquid ammonia and liquid carbon (IV) oxide are made to evaporate at low pressure. These
liquids obtain their latent heat from the items kept in the cold compartment. The Freon vapour is forced through
compressor to the high-pressure pipe where it condenses to liquid at high pressure. The refrigerant gives up its
latent heat to the high-pressure pipe and is conducted through the cooling fins to the surroundings. The highly
pressurized liquid is forced through the expansion valve into the cold compartment where it evaporates. The
process is repeated continuously until the refrigerant is used up.
2. The air conditioner
The air conditioner, like the refrigerator, works on the principle that evaporation produces cooling. Warm air is
drawn into the cooling unit where it gives up latent heat to evaporate the refrigerant in the coiled pipe.
NOTE:
Explain why it is desirable to install an air conditioner near the ceiling of a room and not close to the floor.
WASSCE, June 2012, No 7.
Solution
Air molecules near the ceiling get cooled, become denser and sink. Warmer air molecules below being lighter
rise to the top, get cooled, become denser and sink again; thus conventional current is set up.

Vapour pressure
A liquid left exposed will evaporate. Molecules of the liquid with enough energy break away from the liquid and
stay outside the liquid as vapour. These free molecules (vapour) are moving constantly and exert pressure on the
liquid surface. The pressure due to these molecules on the liquid is called vapour pressure. The two types of
vapour pressure are unsaturated vapour pressure and saturated vapour pressure.
(i) Unsaturated vapour pressure
The vapour which is not in contact with its own liquid in a confined space, is called unsaturated vapour. An
unsaturated vapour is not in dynamic equilibrium with its own liquid. The rate at which the liquid evaporates is
greater than the rate at which it condenses.
The pressure exerted by a vapour which is not in contact with its own liquid in a confined space, is called
unsaturated vapour pressure.
(ii) Saturated vapour pressure (S.V.P.)
The vapour which is in contact with its own liquid in a confined space, is called saturated vapour. A saturated
vapour is in dynamic equilibrium with its own liquid; that is, the number molecules leaving liquid is equal to the
number of molecules returning to the liquid. Saturated vapour exerts. Saturated vapour pressure on the liquid
called saturated vapour pressure (S.V.P.)
The pressure exerted by a vapour which is in contact with its own liquid in a confined space, is called saturated
vapour pressure (S.V.P.).

Measurement of unsaturated and saturated vapour pressures

The vapour pressure of liquid can be studied at room temperature with the help of simple barometers as shown in
Figure 9. Three barometric tubes P, Q and R are filled with mercury and inverted in a mercury trough B. The
barometer Q is used as a control; a small liquid is introduced at the base of barometer P and R using a pipette.

Figure 9: Saturated and unsaturated vapour pressures

The liquid rises to the top where it vaporizes and fills the space above the mercury in the confined space. The
pressure of the vapour depresses the mercury by h1 in the barometer P. The depression h1 is the unsaturated
vapour pressure of the liquid.
Addition of liquid in barometer R increases the vapour pressure and the mercury is depressed more. Adding more
liquid increases the vapour pressure until the vapour is saturated; a little liquid remains above the mercury. When
this happens, the vapour in the space above the liquid is said to be saturated. The saturation vapour pressure
(S.V.P.) is the depression h2 of the barometer R. Saturation vapour pressure (S.V.P.) of a liquid is constant if the
temperature of the liquid remains constant.

The effect of temperature on saturation vapour pressure (S.V.P.)

The saturation vapour pressure varies with temperature. At constant temperature, it is constant but increases
progressively as the temperature increases. Close to the boiling point of the liquid, the increase in the saturation
vapour pressure (S.V.P.) becomes rapid. We can study the variation of SVP using the setup in Figure 8.10.
• Two simple barometers P and Q are placed in a water bath. Barometer P contains saturated water vapour while
barometer Q contains no vapour and is used as a control.
• The initial temperature and saturation vapour pressure (S.V.P.) h are measured and recorded.
• Steam is passed into the water bath; the temperature rises gradually. The saturated vapour pressures are
measured at different temperatures.
• The values of S.V.P. are measured at various temperatures and the graph of S.V.P. against temperature shows
that the saturation vapour pressure (S.V.P.) increases as the temperature rises.

Figure 10: Effect of temperature on S.V.P.

Figure 11: Bubbles formation in boiling water

When a liquid is heated in a beaker, bubbles of vapour begin to form at the bottom and rises to the surface. Close
to the boiling point of the liquid, the rate of formation of bubbles increases and it forms throughout the volume of
the liquid. The formation and rising of the bubbles from the bottom to the surface is called boiling.

Boiling and saturation vapour pressure

A liquid boil when bubbles are formed throughout the volume of the liquid. The water above the mercury in Figure
10 boils when the S.V.P. is 760 mmHg (external pressure). Boiling takes place when the S.V.P. of the liquid is equal
to the external pressure.
The pressure inside each bubble at the surface of the liquid is equal to the external pressure on the liquid surface.
Inside the liquid, the pressure inside the bubbles is the sum of the external pressure and the pressure due to the
weight of the liquid on the bubbles. As the bubbles rise to the surface, their inner pressure decreases, making their
size to increase. The bubbles continue to expand until they reach the surface of the liquid where the pressure on
them is only the S.V.P. The bubbles burst open because their inner pressure (S.V.P.) is equal to the external
pressure. For water, this happens at a standard pressure of 760 mm Hg and a temperature of 100°C. The
temperature at which a liquid boil is called its boiling point.
Boiling point of a liquid is the temperature when the saturation vapour pressure (S.V.P.) of the liquid is equal to
the external or atmospheric pressure.
Boiling points of liquids are affected by the pressure exerted on its surface and the presence of impurities.
(a) Effect of pressure on boiling point
The boiling point of a liquid may vary, depending on the magnitude of the atmospheric pressure at the time of
measurement. Increasing the pressure on the liquid surface will increase its boiling point and reducing the
pressure on its surface will lower its boiling point.
Water can be forced to boil at a reduced temperature by reducing the pressure on its surface.
We can make water boil at a temperature far below 100°C using the apparatus in Figure 9.12.
The procedures are as follows:
• Water in a flask is boiled to expel all the air above it. The flask is then covered with a tight stopper with a fixed
thermometer.
• When cold water is poured over the flask, the water begins to boil again.
Pouring cold water on the flask makes the steam above the water to condense and a vacuum is formed above the
water. The pressure above the water is reduced and the water boils at a lower temperature. Boiling continues until
water vapour accumulates above the water surface. If fresh cold water is poured over the flask again, boiling
resumes. If the process is repeated, the water continues to boil until its temperature falls to room temperature.

Figure 12: Boiling water at a temperature below 100°C

We can make water boil without heating it. A vacuum pump is used to pump out air above the water, the water
boils when the pressure of vapour above it is less than the S.V.P. of the room temperature.
(b) Effect of impurity on boiling point
Impurities will change the boiling point of a liquid. Salt added to water will increase its boiling point. A strong
solution of salt will cook food fast than pure water because water boiling at higher temperature contains more
energy. Some impurities like alcohol, lower the boiling point of water.
Determination of boiling point of small quantity of liquid
Apparatus: A beaker of water, stirrer, thermometer, J-tube, and a source of heat.

Figure 13: Determination of boiling of small quantity of a liquid

Method
The boiling point of a liquid can be found using the fact that a liquid boil when its SVP is equal to the external
atmospheric pressure. Some quantity of mercury is poured into a J-tube and air trapped by the mercury is
expelled. A small quantity of liquid whose boiling point is to be determined is introduced into the shorter arm of
the J-tube until a small quantity remains above the mercury. The J-tube and its content are immersed in a beaker
of water and warmed gently. The water is stirred continuously to keep the temperature uniform. As the
temperature rises, the water vaporizes and depresses the mercury until the levels on both arms of the J-tube are
equal. When this happens, the temperature of the thermometer is read and recorded as the boiling point of the
liquid. At this temperature, the S.V.P. of the liquid is equal to the external atmospheric pressure. Another
measurement of temperature is taken when the levels of mercury in both arms are equal as the beaker and its
content are cooling. The average of the two temperatures is the boiling point of the liquid.
Precautions
(i) The water in the beaker is stirred continuously to keep its temperature uniform.
(ii) The readings of the thermometer are taken during the heating and cooling process and the average value
recorded as the temperature of the liquid.
(iii) Only a pure liquid should be used since impurities alter the boiling point of a liquid.
(iv) Parallax error should be avoided when taking the reading of the thermometer.

Differences between boiling and evaporation

S/N Boiling Evaporation

Boiling takes place throughout the volume of Evaporation takes place at the surface of the
1.
the liquid. liquid only.

2. Boiling occurs at a definite temperature. Evaporation occurs at all temperatures.

Wind does not affect the boiling point of a Wind increases the rate of evaporation of a
3.
liquid. liquid.

Temperature need not be steady during

4. Temperature remains steady during boiling.
evaporation.

This is the change from liquid to vapour at the This is the change from liquid to vapour at
5.
boiling. temperature below normal boiling point.

Table 1.1: Differences between boiling and evaporation

Applications of vapour pressure

• Boiling at a reduced pressure
Pressure decreases with altitude (height). As one gains height, pressure reduces and it becomes difficult to obtain
hot water or cook food. This is because water boils at a temperature less than its boiling point at normal
atmospheric pressure; at the top of mountain Everest, the boiling point of water is about 70°C. Boiling at reduced
pressure and temperature is used in the production of condensed milk, sugar and syrups. A condensed milk is
produced by boiling away water from the milk at a reduced temperature to prevent clotting.
• Pressure cooker
A pressure cooker cooks food faster and better than the normal way of cooking. It is made of a vessel which can be
tightly closed and the temperature increased to about 120°C at a high pressure. Water boiling at higher
temperature contains more energy; therefore, food is cooked faster. Mountain climbers and people living in places
with low atmospheric pressure use pressure cookers to boil foods.

NOTE:
Water in a clay pot is cooler than water in a closed plastic container
The clay pot has pores which allow evaporation to take place. In the process of evaporation, molecules take latent
heat from the bulk of the water, leaving the water cooler; but the plastic container does not have pores (to allow
evaporation). [WASSCE]
Food gets cooked faster in a pressure cooker than in an ordinary cooking pot
With the pressure cooker, increased pressure raises the boiling point of water within a short period. This leads to
faster cooking. Ordinarily cooking pot operates at normal atmospheric pressure. [WASSCE]

Humidity
Our atmosphere is not dry but contains some quantity of water vapour due to constant evaporation from open
water sources like oceans, lakes, rivers and evaporation from plants and leaves during photosynthesis. Humidity is
a measure of wetness of our atmosphere. The exact amount of water vapour in the atmosphere at a given
temperature is called absolute humidity. The atmosphere becomes saturated when it contains maximum amount
of water vapour at a particular temperature. Temperature determines the quantity of water vapour in the
atmosphere. At higher temperatures, the atmosphere contains more water vapour compared to water vapour
present at low temperatures. The amount of water vapour in kilogram per cubic metre (Kg m -3) needed to saturate
the atmosphere varies with temperature.

Relative humidity
When the mass of water vapour per unit volume present in the atmosphere at a given temperature is compared
with the mass of water vapour per unit volume needed to saturate it at the same temperature is called relative
humidity.
Relative humidity is always expressed in percentage. In terms of saturation vapour pressures, we define relative
humidity as:

A relative humidity of 60% means that the air contains 60% of water vapour at the specified temperature.
Worked examples
1. On a certain day, the mass of water vapour per cubic metre of air at 30°C is 14.8 g while the mass of water
vapour per cubic metre of air required to saturate at the same temperature is 30.04 g. Calculate the relative
humidity of the air on this particular day.
Solution

m = mass of water vapour per unit volume of air at 30°C

M = mass of water vapour per unit volume of air at saturation and temperature of 30°C

2. The temperature of certain day in Lagos is 35°C and the saturation vapour pressure of air at dew point is 45.28
mm Hg. If the pressure of air at saturation is 55.13 mm Hg, calculate the relative humidity on this day.
Solution

An environment with high relative humidity like a coastal area, contains a high percentage of water vapour. People
living in these areas sweat a lot. The sweat remains on the body since evaporation is low; therefore, the body feels
uncomfortable. When the humidity is low, the air is dry and the body may be uncomfortable due to the harsh
weather condition as we have during harmattan or dry season. The dryness of the air encourages evaporation,
therefore, sweat evaporates fast and the body loses a large amount of water. The water lost needs to be replaced
to keep the body fluid constant.

Hygrometer
The hygrometer measures the relative humidity of an environment. The most commonly used hygrometer is the
wet bulb and dry bulb hygrometer. It consists of two thermometers:
• The dry bulb thermometer is an ordinary thermometer to measure the temperature of the air.
• The wet bulb thermometer is covered with wet material dipped in water.
Evaporation of water around the wet bulb thermometer makes its temperature lower than the dry bulb
thermometer. Readings of the two thermometers vary according to the humidity of the place. In a highly humid
environment, water evaporates slowly therefore, the readings of both thermometers differ a little. When the
surrounding air is dry, the difference in the readings will increase because of the increase in the rate of
evaporation of water. We can use the readings of the wet and dry type hygrometer to find the relative humidity of
a place.

Dew and dew point

Dew is the visible water droplets formed when the invisible warm water vapour is adequately cooled. This happens
if a cold object is placed in a warm environment. Evaporation of water and condensation of water vapour depend
on the temperature and the vapour pressure above the water. If the pressure of the vapour is less than its
saturation vapour pressure, more water evaporates. When the pressure of the vapour is more than its saturation
vapour pressure at a particular temperature, excess water vapour condenses into visible drops of water. The
condensation of water vapour to visible drops of water begins at a temperature when the pressure of water
vapour is equal to its saturation vapour pressure. The temperature when water vapour begins to form visible drops
of water on the cold surface is called the dew point.

Figure 14: Wet – and – dry bulb hygrometer

The dew point is the temperature at which the pressure of the water vapour is equal to the saturation vapour
pressure.
When a cold bottle of soft drink or a sachet of water is brought out from the refrigerator, warm water vapour in air
around the bottle are cooled below their dew point and condenses into visible drops of water on the soft drink
bottle or the sachet of water.

Mist, fog and clouds

Mist occurs in wet air with high relative humidity above 75% when water vapour in the air is cooled below its dew
point. Visible tiny drops of water are condensed from the water vapour. These tiny drops of water float about like
cloud near the ground. Mist limits visibility to about 1000 m or less.
Fog is formed when water vapour in the air is cooled down to its dew point. Water vapour condenses as visible
drops of water on suspended dust and smoke particles forming dense cloud close to the ground. Fog is more
serious than a mist as visibility is reduced to less than 200m.
The atmosphere high above is cooler than the ones below. Water vapour rising above the ground level cools as it
expands. At a certain height, the temperature of water vapour falls below its dew point and condenses into clouds.