HEAT TRANSFER

Heat transfer is the transfer of thermal energy from a higher temperature object or system to
another system, raising its temperature. This process changes the thermal energy of both
systems involved until thermal equilibrium is reached.
Thermal energy can be transferred from one end of a material to the other or from one material
to another through conduction. It is also transferred indirectly by convection and radiation.
MODES OF HEAT TRANSFER
There are three modes by which heat can be transfer, these modes are: Conduction, Convection
and Radiation.

Figure 1: Mode of Heat Transfer

Heat transfer by a combination of thermal conduction and thermal convection can be analyzed
using conductive thermal resistance and convective thermal resistance. These thermal
resistances are functions of thermal conductivity, convective heat transfer coefficient(s), and
geometrical parameters.
CONDUCTION AND CONDUCTIVITY
Conduction is a specific form of heat transfer during which the heat is transferred from a zone at
high temperature to a zone low temperature due to the presence of temperature gradient within
the system of solids and fluids.
As a mechanism of heat transfers, it primarily requires bodies to be in contact for it to occur.
The mechanism by which conduction occurs depends on the nature of the material concerned.

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Fourier Law of Thermal Conduction
Fourier’s law states that the rate of heat transfer by conduction between two points on a body is
proportional to the surface area of the material and the temperature gradient between the two
points.
dQ dθ
∝ A
dt dL

dQ dθ
dt
= − ΚA
dL
(1)

The negative sign acts as a normalizing sign because naturally heat is transferred from a higher
region to a lower region. K is a constant whose value depends on the material of the body. It is
called the Coefficient of Thermal Conductivity or just Thermal Conductivity.
Thermal Conductivity
Thermal conductivity of a solid is the quantity of heat per second that flows in a steady state
through the opposite face of a unit cube of the material when the temperature when the
temperature difference of 1 K is maintained across the face. Its unit is Wm-1K-1.
It is the measure of the ability of a material to transfer heat. Given two surfaces on either side of
a material with a temperature difference between them, the thermal conductivity is the heat
energy transferred per unit time and per unit surface area, divided by the temperature difference.
It is unique for material and differs from any other material.
Factors that affect Thermal conductivity of a material
1. Surface area of the material
2. Temperature difference between two points
3. Distance between two points
4. Nature of the material
5. Time of heat flow.
Example: The temperature of air-conditioned office is 20 °C while the surrounding temperature
is 28 °C. How much heat will flow per hour through a 5 mm thick glass window measuring 105
cm x 75 cm thermal conductivity of glass is 0.6 Wm-1K-1.
Solution
L = 5 mm = 5 × 10−3 m, A = 105 cm × 75 cm = 7875 cm2 = 0.7875 m2
2
Q θ2 − θ1 28 − 20
= − KA = 0.6 × 0.7875 × = 756 Js−1 = 2.72 × 106 JHr−1
t L 5 × 10−3

Example: A layer of ice 20 cm on a pond. The temperature of air is - 10 °C. Find how long it
will take for the thickness of ice to increase by 1 mm. Given the density and thermal
conductivity of ice are 900 kgm-3 and 2.09 Wm-1K-1 respectively.
Solution
Latent heat of fusion of ice, L = 3.34 × 105 Jkg−1 , increase in length, dl = 1 × 10−3 m
Length, l = 20 cm = 20 × 10−2 m, dθ = − 10 ℃ = 263 K
dQ dθ
dt
= KA
dl
(I)

Q = mL (II)
Divide both sides of equation (II) by dt
dQ mL
dt
=
dt
(III)

Equating (I) and (III),

mL dθ
dt
= KA
dl
(IV)

But m = ρV = ρAl
ρAlL dθ
dt
= KA dl
ρLldl
dt =
Kdθ
(V)
900 × 3.34 × 105 × 20 × 10−2 × 1 × 10−3
dt = = 109.4 secs
2.09 × 263

Conduction Through Composite Walls

Consider two plane walls in contact (called a composite wall) as shown in Figure 2 below. The
individual walls are labeled 1 and 2 as are each the thermal conductivity and thickness. Assume
the wall boundaries conduct heat to environment on both sides

Figure 2: Heat flow through composite walls

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Let the temperature at the interface point be θ
For material 1
Q θ − θ1
= − Κ1 A1
t L1
Q
But power, P =
t
θ − θ1
P1 = − Κ1 A1
L1
(2)

For material 2
Q θ2 − θ
= − Κ2 A2
t L2
θ2 − θ
P2 = − Κ2 A2 (3)
L2

Since same quantity of heat passes through the material, �1 = �2

θ − θ1 θ2 − θ
− Κ1 A1 L1
= − Κ 2 A2
L2
Κ1 A1 Κ2 A2
Let R1 = − and R2 = −
L1 L2

R1 θ − θ1 = R2 θ2 − θ
R2 θ2 +R1 θ1
θ =
R2 +R1
(4)

Example: A composite rod consists of 100 cm of steel and an equal length of copper joined
end-to-end. The steel end is immersed in boiling water (100 °C) and the copper end in ice-water
(0 °C). If both rods have a cross-sectional area of 10 cm2, determine
(a) The temperature of the steel-copper junction
(b) The rate of heat conduction from the boiling water to the ice water.
Κsteel = 50 Wm−1 K−1 , Κcopper = 385 Wm−1 K−1

Solution
(a) Let the temperature at the steel-copper junction point be θ
Heat conducted through steel = Heat conducted through copper
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 Q Q
=
t steel t copper

For steel
Q θ − θS
= − KS AS
t steel LS

Q θ − 100
= − 50 × 0.001 × = − 0.05 θ − 100
t steel 1

For copper
Q θC − θ
= − K C AC
t copper LC

Q 0−θ
= − 385 × 0.001 × = − 0.385 0 − θ
t copper 1

−0.05 θ − 100 = − 0.385 0 − θ

θ = 11.6 ℃
(b) Rate of heat loss
Q θ − θS
= − KS AS
t steel LS

Q 11.6 − 100
t steel
= − 50 × 0.001 × 1
= 4.42 Js−1

Heat Flux and Thermal Resistance

Heat flux is the measure of the amount of heat transfer to or from a surface per unit area per unit
time.
Q
q =
A
(5)

Thermal resistance is the measurement of the temperature difference by which an object or

material resist heat flows.
Thermal resistance is the reciprocal of thermal conductance.
L
Rt =
KA
(6)

CONVECTION
Convection is a mode of heat transfer that is due to the bulk movement of molecules within
fluids. Convective heat transfer is one of the major types of heat transfer, and convection is also
a major mode of mass transfer in fluids. Convective heat and mass transfer take place both by

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diffusion the random Brownian motion of individual particles in the fluid – and by advection, in
which matter or heat is transported by the larger-scale motion of currents in the fluid.
For a fluid flowing over a solid surface, the rate of heat flow by convection QC, from surface to
the fluid (or vice-versa) is given as:
dθC
dt
= hAs θs − θf (7)
dθC
where, = rate of heat tranfer
dt

h = convection coefficient
AS = Area of contact between the fluid and the surface
θS = temperature of the surface

θf = temperature of the main body of the fluid

N/B: The above expression is similar to Newton’s law of cooling.

Example: Forced air flows over a convective heat exchanger in a room heater, resulting in a
convective heat transfer of thermal coefficient 1.136 kWm-2K-1. The surface temperature of heat
exchanger may be considered constant at 65 °C, and the air is at 20 °C. Determine the heat
exchanger surface area required for 8.8 kW of heating.
Solution
Q = hA θs − θf
Q 8.8 × 103
A = = = 1.72 m2
h θs − θf 1.136 × 103 × 65 − 20

THERMAL RADIATION
The transfer of heat by conduction and convection requires direct physical contact between the
hot and the cold matter. Heat can however, also be transferred between two bodies which are
not in direct physical contact. This mode of heat transfer is referred to as radiation.
Thermal radiation is the emission of electromagnetic waves from all matter that has a
temperature greater than absolute zero.
Properties of Thermal Radiations
1. Thermal radiations can travel through vacuum.

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2. They travel along straight lines with the speed of light.
3. They can be reflected and refracted. They exhibit the phenomenon of interference and
diffraction.
4. They do not heat the intervening medium through which they pass.
5. They obey inverse square law.
Absorptive power
Absorptive power of a body for a given wavelength and temperature is defined as the ratio of
the radiant energy absorbed per unit area per unit time to the total energy incident on it per unit
area per unit time. It is denoted by �� .
Emissive power
Emissive power of a body at a given temperature is the amount of energy emitted per unit time
per unit area of the surface for a given wavelength. It is denoted by �� , its unit is Wm-2.
Concept of a Black Body
A Black body is a surface that absorbs all radiant energy falling on it. The term arises because
incident visible light will be absorbed rather than reflected, and therefore the surface will appear
black. The concept of such a perfect absorber of energy is extremely useful in the study of
radiation phenomena.

Figure 3: Black Body Radiation

The best practical black body is a small hole in a box with a blackened interior, because
practically none of the radiation entering such a hole could escape again, and it would be
absorbed inside.
A perfect black body is the one which absorbs completely heat radiations of all wavelengths

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which fall on it and emits heat radiations of all wavelengths when heated. Since a perfect black
body neither reflects nor transmits any radiation, the absorptive power of a perfectly black body
is unity.
Laws of Thermal Radiation
1. Kirchhoff’s Law of thermal radiation
For an arbitrary body emitting and absorbing thermal radiation in thermodynamic equilibrium,
the emissivity is equal to the absorptivity.
emissivity = absorptivity

2. Prevost Theory
Prevost theory states that all bodies emit thermal radiation at all temperatures above absolute
zero, irrespective of the nature of the surroundings.
3. Wien’s Law
Wien’s displacement law states that the black body radiation curve for different temperature
peaks at a wavelength is inversely proportional to the temperature.
b
λmax =
T
(8)

where
T = absolute temperature in Kelvin

b = constant of proportionality called Wien's displacement constant

b = 2.8977729 × 10−3 mK

Figure 4: Variation of a Black body's radiation as a function of wavelength

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A plot of the variation of a Black body's radiation as a function of wavelength for various
absolute temperatures. Each curve is seen to peak at a somewhat different wavelength; Wien's
law describes the shift of that peak in terms of temperature.
4. Stefan's Law
Stefan's Law states that the total amount of heat radiated by a perfectly black-body per second
per unit area is directly proportional to the fourth power of its absolute temperature.
E ∝ T4

E = σT4 (9)
Where, σ is a constant called Stefan's constant. Its value is experimentally found to be 5.67 X
10-8 Jm-3s-1K-4.
Example: Calculate the steady temperature of a tungsten filament of 1000 W if the filament
length is 0.5 m and radius 0.1 mm assuming it behaves like a black body.
Solution
P
= σT4
A

4 P 4 P 4 1000
T = Aσ
= 2πrlσ
= 2π × 0.1 × 10−3 × 0.5 × 5.67 × 10−8
= 2737.2 K

Example: A black body at 2000 K emits radiation with �� = 1250 �� . Use this result to
evaluate the surface temperature of the star Sirius if �� for Sirius is 71 nm. Assume that the star
behaves like a black body.
Solution
λ1 T1 1250 × 200
λ1 T1 = λ2 T2 ⟹ T2 = = = 3521.1 K
λ2 71

APPLICATION OF HEAT TRANSFER: Thermos Flask (Vacuum Flask)

A vacuum flask is a storage vessel which maintains the temperature of the material stored in it
for a longer time in comparison to the same kept in the open. It is primarily meant to enhance
the storage period of the liquid by maintaining uniform temperature and avoiding possibilities
of off taste and odour. It is an appliance, intended for maintaining the temperature of hot liquid
materials such as soup, milk, tea or coffee for longer period. Vacuum flasks are made of double

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coated with nickel/cadmium, a highly shining material. The vacuum flasks are vacuumed before
sealing from an end. The vacuum inside the two walls of the flask acts as a very good insulator
which does not allow heat dissipation even for hours.

Figure 5: Parts of a Thermos Flask

A thermos flask has double walls, which are evacuated and the vacuum bottle is silvered on the
inside. The vacuum between the two walls prevents heat being transferred from the inside to the
outside by conduction and convection. With very little air between the walls, there is almost no
transfer of heat from the inner wall to the outer wall by convection. Conduction can only occur
at the points where the two walls meet, at the top of the bottle and through an insulated support
at the bottom. The silvered walls reflect radiated heat back to the inside, the same way a space
blanket does.
The parts of the Thermos flask reduces heat by the following processes
S/N PARTS MODE OF HEAT
1 Plastic cap Conduction and convection
2 Doubled walled container Convection
3 Vacuum Conduction and Convection
4 Silvered surface Radiation

GREEN HOUSE EFFECT AND GLOBAL WARMING

The greenhouse effect is the process by which radiation from a planet's atmosphere warms the
planet's surface to a temperature above what it would be without its atmosphere.
The greenhouse effect increases the temperature of the Earth by trapping heat in our atmosphere.
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This keeps the temperature of the Earth higher than it would be if direct heating by the Sun was
the only source of warming. When sunlight reaches the surface of the Earth, some of it is
absorbed which warms the ground and some bounces back to space as heat. Greenhouse gases
that are in the atmosphere absorb and then redirect some of this heat back towards the Earth.
What causes the greenhouse effect?
The greenhouse effect is caused by the interaction of the sun's energy with greenhouse gases
such as carbon dioxide, methane, nitrous oxide and fluorinated gases in the Earth's atmosphere.
The ability of these gases to trap heat is what causes the greenhouse effect.
Global Warming
Global warming is a gradual increase in the overall temperature of the earth's atmosphere
generally attributed to the greenhouse effect caused by increased levels of carbon dioxide. CFCs,
and other pollutants.
The cause of global warming is the increasing quantity of greenhouse gases in the our
atmosphere produced by human activities, like the burning of fossil fuels or deforestation.
These activities produce large amounts of greenhouse gas emissions which is causing global
warming.
EXERCISES
Question 1:
A sheet of rubber and a sheet of cardboard each 2 mm thick are pressed together and their outer
surfaces are maintained respectively at 0 °C and 25 °C. If the thermal conductivity of rubber
and cardboard are respectively 0.13 and 0.05 Wm-1K-1, find the quality of heat which flows in 1
hour across a piece of composite sheet of area 100 cm2.
Question 2:
The average human body may be considered to have a surface area of 1.25 m2 and a surface
temperature of 32 °C. What is the rate of energy loss by the human body. [assume emissivity of
human body, e = 1]
Question 3:

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The radius of sun is 6.96 x 108 m and its total power output is 3.85 x 1026 W. Assuming the
sun’s surface emits as black body, calculate: (a) Its surface temperature (b) The maximum
wavelength of light from Sun [σ = 5.67 x 10-8 Wm-2K4, b = 2.898 x 10-3mk].
Question 4:
In order to minimize heat loss from a glass container, the wall of the container are made of two
sheets of glass, each 2 mm thick placed 3 cm apart and intervening space being filled with
poorly conducting of heat per unit area through this composite wall to that which would have
occurred had a single sheet of glass being used under the same internal and external temperature
condition. Thermal conductivity of glass and poor conductor are 0.63 and 0.049 Wm-1K-1
respectively.
Question 5:
An electric lamp filament attains a temperature of 2500 °C when dissipating 60 Watts, the
surrounding temperature being 10 °C. Assuming that it radiates as a black body and losses no
heat by conduction or convection, calculate its temperature when it is dissipating 80 Watts.
Question: 6
A kettle with a steel bottom 5 mm thick and 200 cm2 in area is placed on a hot stove. As the
water inside the kettle boils, 200 gmin-1 is evaporated. Determine the temperature of the inner
and outer surfaces of the bottom of the kettle. [Thermal conductivity of steel K = 50 Wm-1K-1,
Latent heat of vaporization of water = 2.26 x 106 Jkg-1, Convection coefficient of kettle = 2 x
104 Wm-2°C].
Question 7:
Suppose 150 W of heat flows by conduction from the blood capillaries beneath the skin to the
body's surface area of 1.5 m2. If the temperature difference is 0.5 °C, estimate the average
distance of capillaries below the skin surface.
Question 8:
A pond of water at 0 °C is covered with a layer of ice 4 cm thick. If the air temperature stays
constant at -10 °C, how long does it take for the ice thickness to increase to 8 cm? [Latent heat
of fusion of ice 3.33 × 105 Jkg−1 , Thermal conductivity of ice 2 Wm-1K-1].

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