Liberia Secondary School System (LSSS)
Grade 11, Physics 4th Marking Period Lecture Note
Instructor: Mr. N. W. Keeta (0770772276/0886419872)
Topic: Heat
Heat Transfer
Heat is the transfer of energy from a warmer object to a cooler object. Heat can be transferred
in three ways: by conduction, by convection, and by radiation.
Conduction is the transfer of energy from one molecule to another by direct contact. This
transfer occurs when molecules hit against each other, similar to a game of pool where one
moving ball strikes another, causing the second to move. Conduction takes place in solids,
liquids, and gases, but works best in materials that have simple molecules that are located
close to each other. For example, metal is a better conductor than wood or plastic.
Convection is the movement of heat by a fluid such as water or air. The fluid (liquid or gas)
moves from one location to another, transferring heat along with it. This movement of a mass
of heated water or air is called a current.
Radiation is the transfer of heat by electromagnetic waves. When you stand in the sun, you
are warmed by the electromagnetic waves, mainly infrared radiation (and to a lesser extent,
visible light), that travels from the sun to Earth. In addition to the sun, light bulbs, irons, and
toasters also transfer heat via radiation. Note that, unlike conduction or convection, heat
transfer by radiation does not need any matter to help with the transfer.
Heat capacity (Thermal capacity)
The heat capacity of an object is the quantity
of heat energy needed to raise the
temperature of the object by one kelvin (1K)
or one degree Celsius. The SI unit of heat
capacity is the JK -1 (J/K).
Where Q is the quantity of heat needed to
produce a change in the temperature of the body, ΔT. The unit we shall use for heat capacity
are J/ or J/K.
Specific heat capacity (c)
Specific heat capacity is the amount of heat required to raise the temperature of 1kg of a
substance by 1K. The SI unit for specific heat capacity is Joules per kilogram Kelvin
(J/kg.K). Different substances have different specific heat capacities. The specific heat
capacity of aluminum is 900J/kgK or 0.900J/gºC. This implies that 900 J of heat energy are
needed to raise the temperature of 1kg of aluminum by 1K.
Which when simplified yields?
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Where c is the specific heat capacity, Q is the quantity of heat needed, m is the mass of the
substance and ΔT is the change in temperature.
Quantity of heat
Quantity of heat is the amount of heat absorbed or given out by a body. The SI unit is Joules
(J). From the definition, it implies that the amount of head absorbed by a body of mass,
mover a temperature change of θ ºC is
Where Q is the quantity of heat, m is the mass of the substance, c is the specific heat capacity
and θ or ΔT is the temperature in degree Celsius.
Example: Calculate the quantity of heat released when a piece of aluminum of mass 600g
and specific heat capacity of 900 J/kgK cools from 65 ºCto 45ºC.
Given
Unknown
Basic equation
m = 600g = 0.6kg
Q
Q = mcθ
c = 900J/kgK
θ1 = 45K
θ2 = 65K
Solution
Working equation:
Q = mc (θ2 – θ1 )
Q = (0.6kg) (900J/kgK) (65 – 45) K = (540J) (20) = 10800J or 10.8kJ ans.
Law of heat exchange
Hot water can be cooled by the addition of cool water. As the two mixes, the temperature of
the hot water is lowered while the temperature of the added cold water is raised. The final
temperature of the mixture lies between the original temperatures of the hot and cold water.
Each time two substances of unequal temperature are mixed the warmer one loses heat
and the cooler one gains heat until both finally reach the same temperature. A process that
absorbs heat as it progresses is endothermic. An endothermic process gives off heat as it
progresses.
No thermal energy is lost when substances of unequal temperatures are mixed. In any heat
transfer system, the heat lost by hot substances equal the heat gained by cold substances.
This is known as the law of heat exchange. This can be expressed as
This equality is the basis of a simple technique known as the method of mixtures for
measuring a quantity of heat in transit from one substance to another.
Example: Calculate the final temperature of a mixture if 200 g of warm water at 80ºC is
added to 300 g of cold water at 10ºC and well stirred. Assume that the heat absorbed by the
container and it surroundings is negligible (unimportant).
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Given
Unknown
Basic equation
mc = 300g = 0.3kg
Tf
Qlost = Qgained
mh = 200g = 0.2kg
Q = mcΔT
ΔTc = (80 – θ)K
ΔTh = (θ – 10)K
c = 4200J/kgK
Solution
Heat lost by warm water = heat gained by cold water
Let the final temperature of the mixture be θ , then the change in temperature of warm water
= (80 – θ) and the change in temperature of cold water = (θ – 10)ºC.
Working equation: mccΔTc = mh cΔTh
mcc (θ – 10) = mh c (80 – θ)
mccθ – 10mcc = 80mh c – mh cθ ↔ mccθ + mh cθ = 80mh c +10mcc
θ(mcc +mh c) = 10c(8mh + mc) ↔ θ = 10c (8mh +mc)/(mcc + mh c)
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Example: A copper calorimeter with a mass of 150.0g contains 350.0g of water; the
temperature of both the water and the calorimeter is 20.0 ºC A metal cylinder, with a mass of
200.0g and a temperature of 99.5 , is placed in the calorimeter. The final temperature of the
calorimeter, water and metal cylinder is 26.7 . Find the specific heat of the metal cylinder.
(The specific heat capacity of copper is 0.387J/g.
and specific heat capacity of water is
4.19J/g ).
Given
Unknown
Basic equation
mc = 150.0g
CM
Qlost = Qgained
cc = 0.387J/g.
Q = mcΔT
mM =200.0g
TM = 99.5
mW = 350.0g
cW =4.19J/g.
TW = TC = 20.0
Tf = 26.7
Solution: Working equation: ⏞
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Example: An aluminum calorimeter with a mass of 125 g contains 110.0g of water at a
temperature of 15.0ºC. A 200.0 g mass of lead buckshot, at a temperature of 90.0ºC, is added
to the calorimeter. Calculate the final temperature of the mixture. (Specific heat capacity of
aluminum is 0.909J/g.ºC, specific heat capacity of water is 4.19J/gºC and specific heat
capacity of lead is 0.128J/gºC).
Given
Unknown
Basic equation
mc = 125.0g
Tf
Qlost = Qgained
cc = 0.909J/g.
Q = mcΔT
mW =110.0g
cW = 4.19J/g
TW = 15.0
mL =200.0g
cL = 0.128J/g.
TL = 90.0
Solution
Working equation: ⏞
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Practice problems
1. An iron calorimeter has a mass of 150.0g. It contains 300.0g of water. Temperature of the
calorimeter and the water is 21.5 . A metal cylinder with a mass of 450.0g and a
temperature of 99.5 is placed in the calorimeter. The final temperature of the mixture is
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31.0 . Calculate the specific heat capacity of the metal cylinder. (Use specific heat capacity
of water = 4.19J/g , specific heat capacity of is 0.450J/g°C). Ans. 0.408J/g
2. A copper calorimeter with a mass of 170.0g contains 145.0g of water. The temperature of
the calorimeter and water is 19.5 . A 75.0g mass of water, at a temperature of 90.5 , is
poured into the calorimeter. Calculate the final temperature of the mixture. (Use 4.19J/g as
the specific heat capacity for water and 0.387J/g for copper). Ans. 42.1
3. An aluminum calorimeter with a mass 199.0g contains 145.0g of water. The temperature of
both the calorimeter and the water is 20.2 . A 32.0g mass of water, at a temperature of
92.5 , is poured into the calorimeter. Calculate the final temperature of the mixture. (Use
4.19J/g as the specific heat capacity for water and 0.909J/g for aluminum). Ans. 30.7
Specific Latent Heat of Fusion
Specific latent heat of fusion (LF) of a substance is the quantity of heat required to change a
kilogram mass of a substance from the solid state to the liquid state at melting point. The SI
unit is Joule per kilogram (J/kg), the LF for ice 334J/g at 0 , meaning that this quantity of
heat must be added to each gram of ice at 0 to convert it to water at 0 .
The method of mixture can be used to determine the latent heat of fusion of a solid. Suppose
we wish to determine experimentally the latent heat of fusion of ice. Since hot water melts
ice, we find the mass of ice at a known temperature that can be melted by a known mass of
water at a known temperature.
Example: An aluminum calorimeter, mass 150.0g, contains 240.0g of water at 35.0 . The
specific heat of the calorimeter is 0.909J/g . An 83.5g amount of ice at 0.0 is place in the
calorimeter and melts completely. The final temperature is 17.0 . Calculate ice’s latent heat
of fusion.
Given
mC = 150.0g
cC = 0.909J/g
mW = 240.0g
cW = 4.19J/g
TW = TC = 35.0
mI = 83.5g
TI = 0.0
TF = 17.0
Solution
Working equation:
Unknown
LF
⏞
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Basic equation
QLost = QGained
Q = mcΔT
Q = mLF
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Melting point and freezing point
The melt point is the temperature at which a substance changes from the solid state to the
liquid state or liquid to solid.
Specific latent heat of vaporization
Specific latent heat of vaporization of a substance is quantity of heat required to change a
kilogram mass of a liquid to vapor state at boiling point. If m is the mass of substance, L is
the latent heat and Q is the amount of heat require to effect a change then
Determination of latent heat of vaporization of water (LV)
In determining latent heat of vaporization of water, pass dry steam into the water in a well –
insulated calorimeter for few seconds.
Note the temperature of the mixture and their various masses and the corresponding
temperature. Considering the following:
 Mass of calorimeter = mc
 Mass of water = mw
 Mass of steam = ms
 Steam temperature = 100
 Initial temperature of water and calorimeter = θ 1
 Final temperature of water and calorimeter = θ 2
 Specific heat capacity of calorimeter = cc
 Specific heat capacity of water = cw
If heat lost by steam = heat gained by water and calorimeter, than
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Example: An aluminum calorimeter, mass 130.4g, contains 324.6g of water at a temperature
of 5.0 . Steam at a temperature of 100 is passed into water, after which the total mass of
calorimeter and water is 473.0g. What is the final temperature of the water? (LV= 2260J/g, cW
= 4.19J/g )
Given
Unknown
Basic equation
mC = 130.4g
TF
QLost = QGained
cC = 0.909J/g
Q = mcΔT
mW = 324.6g
Q = mLV
cW = 4.19J/g
TW = TC = 5.0
mS = (130.4g + 324.6g) – 473.0g = 18 g
TS = 100.0
LV = 2260J/g
Solution
Working equation:
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Practice problem
An aluminum calorimeter contains 420g of water at a temperature of 15 . The mass of the
calorimeter is 152g. How many grams of steam at 100 are needed to raise the temperature
of water and the calorimeter to 75 ? (The specific heat capacity of aluminum is 0.909J/g
and water is 4.19J/g ).
Boiling point
Boiling point is the temperature at which saturated vapor pressure is equal to the external
pressure.
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Determination of specific latent heat of fusion of solid ice
To determine the specific latent heat of fusion of ice put a well dry ice into a well-insulated
calorimeter half filled with water which is slightly warmed. Stir the mixture vigorously to
ensure even distribution of heat. Note and record the initial and final temperature of
calorimeter and water.
Electrical method of determination of latent head of vaporization
Electrical heat energy required to boiling water = IVt or I2 Rt. If m = mass of water boiled off
time, t and LV = latent heat of vaporization of water.
Example: Calculate the quantity of heat needed to change 10g of ice at 0 to water at 50 .
(Latent heat of fusion of ice = 336J/g; specific heat capacity of water = 4.19J/g ).
Given
Unknown
Basic equation
M c = 10g
Q
Q = mLf + mcθ
LI = 336j/g
cw = 4.19J/g
ΔT = (50 – 0) = 50
Solution
Working equation: Q = mLI + mcwθ
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Example
A piece of dry ice is added to a 1000cm3 of water at 50 . If the lower temperature obtained
is 25 , calculate the mass of ice that melted.
Given
Unknown
Basic equation
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m
QLost = QGanied
LI = 336j/g
M
cw = 4.19J/g
ΔT = (50 – 25) = 25
Solution
Working equation:
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Mass of water = density of water times volume
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Heat and work
Mechanical Equivalent of heat
Since heat is a form of energy, it can be produced by mechanical work. The study of the
quantitative relationship between heat and other forms of energy is called thermodynamics.
In discussing energy transformation, we shall also look at the term internal energy. Internal
energy is the total available potential and kinetic energy of the particles of a substance. These
particles include molecules, ions, and atoms. When head is added to a substance, its internal
energy increases, and increased its temperature from Ti to some value Tf. The work done on
the water the loss of potential energy of mass m as it falls through a distance d, less kinetic
energy it possesses as it reaches the bottom. The loss in potential energy is mgd. The final
kinetic energy of the mass is
Hence, the amount of work, W, done on the system by the falling mass is
The amount of heat that would be needed to produce the observed temperature change in the
water is
(
)
Where Q is the heat energy in joules, mW is the mass of the water, cW is the specific heat of
water and Tf and Ti are the final and initial temperature of the water respectively.
First law of Thermodynamics
The first law of thermodynamics states that the quantity of energy supplied to any isolated
system in the form of heat is equal to the work done by the system plus the change in internal
energy of the system. Thus the energy input to an isolated system plus the energy output is in
the form of work. We may also state the first law of thermodynamics as follows: when heat is
converted to another form of energy or when other forms of energy are converted to heat,
there is no loss of energy.
As an application of the first law of thermodynamics, let us assume that an amount of heat,
Q, is added to a substance whose total internal energy is Ei . Generally we would find that the
addition of this heat energy increases the internal energy of the substance of Ef and also
causes the substance to do a quantity of work, W, on its surroundings. This can be stated
algebraically as
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Q is positive when heat is added to the substance, and W is positive when the body does
work on surrounding objects. All quantities must, of course, be in the same units.
We have known that the internal energy of the water is changed because the temperature of
the water is changed.
A process in which no heat is added to or removed from a substance is called an adiabatic
process. In such a case, Q = 0 = (Ef – Ei ) + W. Thus Ef – Ei = - W.
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Isothermal Expansion
To calculate the work done by a gas in expanding, let us imagine the gas enclosed in a
cylinder with a tight-fitting piston. The piston rod is connected to a device on which it may
exert a force. The force, F, acting on the piston due to the pressure, p, exerted by the gas on
area A of the piston head is
Suppose the piston is moved a distance d by the expanding gas in the cylinder. Suppose
further that some heat is applied to the gas while the pressure remains constant during the
expansion. The work done by the expanding gas in moving the piston will be
Where the quantity Ad is the change in volume of the expanding gas, thus the work done by
the gas expanding at constant pressure is
(
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Where Vf and Vi are the final and initial volumes, respectively, of confined gas. A process
that takes place at constant temperature is known as an isothermal process.
Specific Heats of Gases
While the specific heat of a solid or liquid is a fixed value to several significant figures for
the substance at a given temperature, this is not true for gases. When heat is added to a mass
of gas, its pressure, volume, or both may change, and its temperature may rise.
When volume of a unit mass is held constant as head is added, the quantity of heat required to
change the temperature of this unit mass of by 1 is called its specific heat at constant
volume, cV.
When the pressure of the same unit mass of gas is held constant as heat is applied to raise its
temperature 1 , the quantity of heat required is called its specific heat at constant pressure,
cP.
The numerical value of cV and cP differs from that of cP even though the same gas is taken
through the same temperature interval.
Efficiency of Ideal Heat Engines
A device that converts heat energy into mechanical energy is called heat engine. In studying
heat engines, we shall use a theoretical model to represent the actual possesses that take place
in such engines. The model will include the necessary transformation of heat and work.
A quantity of heat, Q1, is delivered to the energy during the beginning of a cycle. Applying
the first law of thermodynamics to this cycle,
The thermal efficiency, e, of the heat engine is define as
Or
Since
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Then
From this equation we can see that the efficiency of an operating engine must always be less
than 100%.
It can be shown that the ratio Q2 /Q1 is also always equal to the ratio of the absolute
temperatures T2 /T1 , provided that the engine is considered frictionless. Therefore we can
write
from this equation we can see that the efficiency of a heat engine may be increased by
making the temperature of the heat source as high as possible and the temperature of the heat
sink as low as possible.
Second law of Thermodynamics
The second law of thermodynamic was first formulated by the German physicist Rudolf
Clausius (1822 – 1888), which state that it is not possible for an engine to transfer heat from
one body to another at a higher temperature unless work is done on the engine.
Entropy
The amount of energy that cannot be converted into mechanical work is related to entropy.
Entropy is a measurable property as important in the study of thermodynamics as energy. As
with potential energy, it is the difference in entropy that is significant rather than the actual
value of entropy. If an amount of heat, ΔQ, is added to a system that is at a kelvin
temperature, T, the change in entropy, ΔS, is
If heat is removed from the system, the quantity ΔQ is negative and the change in entropy is
also negative. Note that this equation defines the change in entropy and not entropy itself.
The units of denoting the change in entropy are joules per degree.
Example
It takes 8.5 x 105 J of heat to melt a given sample of a solid. In the process, the entropy of the
system rises by 2000j/K. find the melting point of the solid in .
Given
Unknown
Basic equation
ΔQ = 8.5 x 105 J TC
ΔS = 2000J/K
Solution
Working equation:
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