LATENT HEAT
Latent Heat
When a substance is
heated, the temperature of the substance increases gradually until it reaches
its melting point where the substance start to melt. As the substance start to
melt, the temperature of the substance remains constant until the melting is completed.
During this time that the object start
to melt to the time that the melting is completed. The thermometer will not
indicate any increase in temperature. The heat that is supplied from the time
the object stated to melt to the time the melting is completed , is used to
overcome the intermolecular force that binds the molecules of the substance
together and make the substance change from solid state to liquid state.
Also, when a liquid is heated, the temperature
of the liquid gradually increases until it reaches the boiling point of the
liquid, at which the liquid starts to boil. This temperature at which the
liquid start to boil is called saturation temperature. During this time that
the liquid is boiling, the temperature of the boiling water remain constant at
100°C, though more heat is added. The water changes to vapour and escape to the
atmosphere. The thermometer did not read any increase in temperature because
the additional heat is hidden. The additional heat that is applied to the
boiling water is used to break the intermolecular force of attraction between
the water molecules and make the water changed to vapour / gas.
Definition
Of Latent Heat:
Latent heat is the quantity of heat energy
that is required to change a substance from one state to another at constant
temperature and pressure. The unit of latent heat is Joule (J), and the symbol
used for latent heat is Q.
Latent Heat
Capacity:
The quantity of heat that is required to
change a substance from one state to another is directly proportional to the
mass of the substance.
Mathematically: Quantity of
heat energy
L is the constant of proportionality. It is
called the specific latent heat of the substance. M is the
mass of the substance measured in kilogram. Q is the
quantity of heat measured in Joule.
Factors That Determine The Quantity Of Heat
Required To Change The State Of A Substance:
The following factors determine the quantity of
heat energy that is required to change a substance from one state to another:
I. The nature
of the substance. II. The mass of the
substance. III. The quantity of
energy supplied.
IV. The volume
of the substance V. The area of the
substance VI. The state of the substance as
the time the heat energy is supplied.
Specific Latent Heat Of A Substance:
Specific latent heat of a substance is
the quantity of heat energy that is required to change 1kg mass of the
substance from one state to another at constant temperature.
Specific Latent Heat Of Fusion of Ice Lice :
Specific latent heat of fusion of ice is
the quantity of heat energy that is required to change 1kg mass of ice block
into liquid at constant temperature Or melting point of the ice block. It is a
scalar quantity. Specific Latent Heat Of
Fusion of Ice L ice = 336 J/g
Formula Of Specific Latent Heat Of Fusion Of
Ice:
Quantity of Heat energy Q = mass of ice
block x specific latent heat of ice block; Q = m x Lice.
Specific latent beat of ice Lice = Q/m
Worked Examples:
I. Calculate
the quantity of heat energy that is required to change 25.6 kg mass of ice to
liquid at constant temperature.
Solution:
Mass of ice block m = 25.6 kg = 25.6 x 1000 = 25600g, Lice = 336J/g, Q = ?; Q = m x Lice = 25600 x 336 = 8601600 Joules
II. The
quantity of heat that is required to change X g of ice to liquid is 235 J. If
the Lice is 336J/g, calculate the mass of the object.
Solution:
Mass = ?, Q = 235 J, Lice = 336J/g; Q = m x
Lice ; 235 = m x 336; m = 235/336 ; M = 0.699 g
Experiment To Determine Latent Heat Of Fusion Of
Ice:
Aim:
To determine the latent heat of fusion of
ice.
Apparatus: Beaker,
funnel, heater, ice block,
Setup Diagram:
Procedures:
Insert the heater into ice block contained in a
funnel. Switch on the heater so that current flow for some times t. Adjust the
rheostat so that steady current flow. Collect
the melted ice in a beaker and measure its mass.
Data:
Current that flow in the heater = I, Voltage
supplied = V, Time for which current flow = t
Resistance of the heater = R, Mass of melted
ice = Mice, Specific latent heat of fusion of ice = L
Theory Of Calculation:
Quantity of Heat Supplied by heater = Quantity
of Heat Gained by melted ice
Formula of Calculation I²Rt = M x Lice
Precautions:
1. Use dry ice
only 2. Add small
quantity of ice at the a time
3. Stir the mixture gently and continuously for equal
temperature. 4. Lag the calorimeter to prevent heat loss.
5. The amount
of temperature below and above room temperature for the final and initial
temperature should be the same.
Effect Of Expansion And Contraction On Fusion:
I. Glass bottle of water crack on freezing
because water expands on freezing and because of unequal expansion of the
bottle.
II. When water freezes into ice, it expands,
become less dense and float on water with nine - tenth (9/10 ) of its volume
submerged in water. Ice contracts on melting.
Effect Of Pressure on Freezing Point of a
Substance:
I. Increase in
pressure lower the melting point of ice or the freezing point of water. This is
called regelation of ice.
II. An
increase in pressure lowers the freezing point of any liquid which expands on
solidifying. E.g water.
III. Increase
in pressure increases the freezing point of any substance which contract on
solidifying. Eg paraffin-wax.
Regelation of Ice:
When a wire to whose ends two weights are tied
is hung on ice block, the pressure of the wire on the ice block, due to the
weights, reduces the melting point of the ice block and the ice block starts to
melt as the wire cuts through the ice. When the pressure is removed, the
freezing point of the melted ice block increases and the water refreezes again.
On refreezing, it gives out latent heat which is conducted by the wire to melt more
ice underneath it. The process is repeated till the wire passes through the ice
block without cutting it into two.
When two ice blocks are pressed together, they
stick together. The pressure between the contact of the ice blocks caused the
ice block to melt. When the pressure is removed, the melt ice refreeze and the
two ice blocks join together.
Regelation of ice is the process whereby when
ice block is subjected to high pressure, its melting point increases and the
ice block melts, but when the pressure is removed, its freezing point increase
and the ice refreezes.
Effect Of Impurities On Freezing Point Of
Substance
The presence of impurities in a substance lower
the melting point of pure substance
Specific Latent Heat Of Vaporization Of Steam
Formula Of Specific Latent Heat Of Vapourization
Of Steam (Lsteam):
Specific latent heat of vaporization of steam is
the quantity of heat energy that is required to change 1kg mass of a liquid
into steam / gas at constant temperature ( boiling point) and pressure.
It is a scalar quantity. The unit of specific latent heat of vaporization of
steam is J/Kg or J/g.
Relationship Between Specific Latent Heat Of
Fusion Of Ice Lice and Specific Latent Heat Of Vaporization Of Steam Lsteam:
Specific latent heat of fusion of ice Lice
= 1/7 of specific latent heat of vaporization of steam
L ice = 1/7 x L steam
Examples:
1. Calculate the total energy required to evaporate completely 1 Kg of ice that is
initially at -10oC ( Specific heat capacity of ice = 2.2x103JKg-1K-1,
specific capacity of water = 4.18x 103 JKg-1K-1,
Specific latent of fusion of ice =3.34x105 JKg-1,
Specific latent heat of vapourization of water = 2.26x106 JKg-1.)
Solution
Heat required to raise the
temperature of ice from -10oC to 0oC: M x Cice x
Heat required to melt the ice at 0oC
= M x Lice = 1 x 3.34x105 = 334,000 J
Heat required to raise water from 0oC
to 100oC = M
x Cw x
Heat required to vapourize liquid = M
x Lvap = 1 x 2.26x106 = 2,260,000 J
Total heat = 22,000 + 334,000 +
418,000 + 2,260,000 = 3,034,000 J =
3.034 x 106 J
2. 150g of ice at 0oC is
mixed with 300g of water at 50oC. Calculate the temperature of the
mixture.
Solution
Given: M ice = 150g = 0.15Kg; Mw = 300g =
0.30Kg;
Heat required to melt the ice at 0oC
= M x Lice = 0.15 x 3.36x105 = 50,400 J
Heat required to raise water from 0oC
to mixture = M x Cw x
Heat loss by water in cooling from 50oC
to mixture = M x Cw x
Total heat gained = Total heat lost
630
1890
Assignment
1. What mass of ice at -14oC will be needed to cool 200cm3 of an orange drink (essentially water) from 25oC to 10oC? (Specific heat capacity of ice = 2100 JKg-1K-1, specific capacity of water = 4200 JKg-1K-1, Specific latent of fusion of ice =3.36x105 JKg-1 ).
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