THERMAL PHYSICS 15

T l l  l F  YAaT
 
T l l Expansion of Liquids
Here, we l = l a in place of l aT so as to avoid the For heating a liquid it has to be ut in some container. When
confusion with change in time period. Thus, the liquid is heated, the container will also expand. We define
coefficient of apparent expansion of a liquid as the apparent
T l  l a increase in volume per unit origional volume per °C rise in
 1  a 
12

T l temperature. It is represented by g a . Thus,

 1  g r  coefficient of volume expansion of liquid

or T   T 1  a  (if a  1 )
 2  and g g  coefficient of volume expansion of the container

1
or T  T  – T  Ta
2
3. CALORIMETRY
Time lost in time t (by a pendulum clock whose actual time
period is T and the changed time period at some higher When two systems at different temperatures are connected
temperature is T  ) is together then heat flows from higher temperature to lower
temperature till the time their temperatures do not become
 T  same.
t   t
 T  Principle of calorimetry states that, neglecting heat loss to
 At some higher temperature a scale will expand and scale surroundings, heat lost by a body at higher temperature is
reading will be lesser than true value. equal heat gained by a body at lower temperature.

However, at lower temperature scale reading will be more heat gained = heat lost
or true value will be less. Whenever heat is given to any body, either its temperature
 When a rod whose ends are rigidly fixed such as to prevent changes or its state changes.
from expansion or contraction undergoes a change in 3.1 Change in Temperature
temperature, thermal stresses are developed in the rod.
This is because, if the temperature is increased, the rod When the temp changes on heating,
has a tendency to expand with since it is fixed at two Then
ends it is not allowed to expand. So, the rod exerts a force
Heat supplied  change in temp (T)
on supports to expand.
 amount of substance (m/n)
 nature of substance (s/C)
 H = msT
m = Mass of body
s = specific heat capacity per kg
T = Change in temp
Fig. 14.14 or H = nCT
n = Number of moles
l
Thermal strain   aT C = Specific/Molar heat Capacity per mole
l
T = Change in temp
So thermal stress   g  (thermal strain) YaT Specific Heat Capacity : Amount of heat required to raise
the temperature of unit mass of the substance through one
or force on supports F = A (stress) YAaT degree.
Here, Y = Young’s modulus of elasticity of the rod.

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Units  In case any material is not at its B.P or M.P, then on heating
the temperature will change till the time a particular state
SI  J/KgK SH O L  = 1 cal/g°C
2
change temperature reaches.
Common  cal/g°C SH O ice  = 0.5 cal/g°C For Example : If water is initially at –50°C at 1 Atm pressure
2

in its solid state.

Molar Heat Capacity : Amount of heat required to raise the
On heating.
temperature of unit mole of the substance through one degree
Step - 1 : Temp changes to 0°C first
Units
Step - 2 : Ice melts to H2O(l) keeping the temp constant
SI  J/mol K
Step - 3 : Temp. increase to 100°C
Common  Cal/g°C
Step - 4 : H2O(l) boils to steam keeping the temp constant
Heat Capacity : Amount of heat required to raise the
temperature of a system through one degree Step - 5 : Further temp increases
 H = ST
where S = Heat Capacity
Units
SI  J/K
Common  Cal/°C
 For H2O specific heat capacity does change but fairly very
less.
 Materials with higher specific heat capacity require a lot
of heat for same one degree rise in temperature Fig. 14.15
3.2 Change in State  The slope is inversely proportional to heat capacity.
 Length of horizontal line depends upon mL for the process.
When the phase changes on heating
Then 3.3 Pressure dependence on melting point and boiling
point
Heat supplied  amount of substance which changes the state (m)
  nature of substance (L)  For some substance melting point decreases with increase
in pressure and for other melting point increases
 H = mL
 Melting poing increases with increase in temperature. We
Where L = Latent Heat of substance
can observe the above results through phaser diagrams.
Latent Heat : Amount of heat required per mass to change
the state of any substance. P P
B B
(atm) (atm)
Units
C C
Liq Liq
SI  J/Kg Solid
Solid
Common  Cal/g
O Vapour O Vapour
 The change in state always occurs at a constant
A A
temperature. T(°C) T(°C)
For H2O For CO2
For example

Solid  Liq Lf Fig. 14.16 Fig. 14.17

Line AO  Sublimation curve
Liq  Gas Lv
Line OB  Fusion curve
Lf = Latent Heat of fusion Line OC  Vapourization curve
Lv = Latent heat of vaporization Point O  Triple Point

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Point C  Critical temperature

Triple Point : The combination pressure and temperature at
which all three states of matter (i.e. solids, liquids gases co-
exist.
For H2O it is at 273.16 K and 0.006 Atm.
Critical Point : The combination of pressure & temp beyond
which a vapour cannot be liquified is called as critical point.
Corresponding temperature, pressure are called as critical
temperature & critical pressure.
From the phasor diagram, we can see that melting point
decreases with increases in pressure for H2O.
Based on this is the concept of regelation.
Regelation : The phenomena of refreezing of water melted
below the normal melting point due to increase in pressure.
 It is due to this pressure effect on boiling point that cooking
is tough on mountains and easier in pressure cooker.
3.4 Mechanical equivalent of heat
In the history of science, the mechanical equivalent of heat
states that motion and heat are mutually interchangeable and
that in every case, a given amount of work would generate
the same amount of heat, provided the work done is totally
converted to heat energy. Fig. 14.18
Note:

HEAT TRANSFER
Water equivalent of a container
Normally, a liquid is heated in a container. So, some heat is
wasted in heating the container also. Suppose water
equivalent of a container is 10 g, then it implies that heat 1. HEAT TRANSFER
required to increase the temperature of this container is equal
to heat required to increase the temperature of 10 g of water. 1.1 Introduction to heat transfer
3.5 Calorimeter Heat transfer is the process of the movemnet of energy dut to
Calorimeter, device for measuring the heat developed during a temperature difference. The calculations we are interested
a mechanical, electrical, or chemical reaction, and for in include determining the final temperatures of materials and
calculating the heat capacity of materials. how long it takes for these materials to reach these
A calorimeter consists of an insulated container, water, a temperatures.
thermometer, a stirring rod, and an object that will either
1.2 Modes of Heat Transfer
absorb or emit heat. To do a Calorimetry experiment, an object
with a certain mass and temperature is placed in the water There are three modes of heat transfer.
and the change in the temperature measured .
 Conduction
A calorimeter is a device that is in use for measuring the
warmth of chemical reactions or physical changes also as heat  Convection
capacity. The most common types of calorimeters are
 Radiation
differential scanning calorimeters, titration calorimeters,
isothermal micro calorimeters, and accelerated rate
calorimeters.

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2. CONDUCTION AND CONVECTION 2.2 Electrical Analogue of Thermal Conduction

Thermal Resistance (R)
2.1 Conduction

Thermal conduction is the process in which thermal energy is dQ T T

We know dt  H  l KA  R
transferred from the hotter part of a body to the colder one or
from hot body to a cold body in contact with it without any Here, T  temperature difference (TD) and
transference of material particles.
l
TC > T D R  thermal resistance of the rod.
TC TD KA
L
Consider a section ab of a rod as shown in figure. Suppose
A
Q1 heat enters into the section at ‘a’ and Q2 leaves at ‘b’, then
Direction of
heat flow Q2 < Q1. Part of the energy Q1 – Q2 is utilized in raising the
temperature of section ab and the remaining is lost to
Fig. 14.19 atmosphere through ab. If heat is continuously supplied fro
At steady state, the left end of the rod, a stage comes when temperature of
the section becomes constant. In that case Q1 = Q2 if rod is
The rate of heat energy flowing through the rod becomes insulated from the surroundings (or loss through ab is zero).
constant. This is called the steady state condition. Thus, is steady state
temperature of different section of the rod becomes constant
T  TD 
This is rate Q  KA
C
...(i) (but not same). Hence, in the figure:
L

for uniform cross-section rods

where Q = Rate of heat energy flow (J/s or W)
2
A = Area of cross-section (m )
TC,TD = Temperature of hot end and cold end respectively
(°C or K)
L = Length of the rod (m)
K = coefficient of thermal conductivity
Coefficient of Thermal Conductivity :
It is defined as amount of heat conducted during steady state
in unit time through unit area of any cross-section of the Fig. 14.20
substance under unit temperature gradient, the heat flow being T1 = constant, T2 = constant etc. and T1 > T2 > T 3 > T4
normal to the area.
Now, a natural question arises, why the temperature of
Units whole rod not becomes equal when heat is being
SI  J/msK or W/mK. continuously supplied? The answer is: there must be a
 Larger the thermal conductivity, the greater will be rate of temperature difference in the rod for the heat flow, same as
heat energy flow for a given temperature difference. we require a potential difference across a resistance for the
current flow through it.
 Kmetals > Knon metals
In steady state, the temperature varies linearly with
 Thermal conductivity of insulators is very low. Therefore,
distance along the rod if it is insulated.
air does not let the heat energy to be conducted very easily.

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2.3 Series and Parallel Combination of Rods

Series Combination:
We can compute the effective coefficient of heat conductivity
when different materials are linked with varying coefficients
of thermal conductivity. When the rods are linked in series,
the flow rate through both sections is the same. However,
because the different connections get varying amounts of heat
energy, the temperature differential between them will be
varied.
The total of the temperature differences between the junctions
Fig. 14.21 equals the temperature difference between the first and last
ends. When two materials are linked in series and their
 Comparing equation number (iii), i.e. heat current physical dimensions are the same.
dQ T  I 
H   where R   2K1 K 2
dt R  KA  KS 
K1  K 2
with the equation, of current flow through a resistance,
dQ V  I  Parallel Combination:
I   where R  
dt R  A  When two materials are linked in a parallel configuration,
We find the following similarities in heat flow through a the entire available heat energy per second is divided between
rod and current flow through a resistance. them. In any case, the temperature will be the same at the
end. When the rods are linked in parallel, the effect on thermal
S.No. Heat flow Current flow
resistance is identical to the effect on electric resistance.
through a through a
conducting rod resistance We may construct a simple equation for the influence on
1. Conducting rod Electrical coefficient of heat conductivity when two distinct rods have
resistance the same length and cross-section area, as shown below.
2. Heat flows Charge flows
3. TD is required PD is required K1  K 2
4. Heat current Electric current Kp 
2
dQ dQ
H  rate of I  rate of Note:
dt dt
heat flow charge flow Low thermal conductivity materials transmit heat at a slower
5. T TD V PD pace than high thermal conductivity materials. Metals, for
H  I 
R R R R example, have a high thermal conductivity and are extremely
6. l l effective at transferring heat, whereas insulating materials
R R
KA A like Styrofoam are the polar opposite. High thermal
7. K=Thermal   electrical conductivity materials are commonly utilised in heat sink
conductivity conductivity applications, while low thermal conductivity materials are
used as thermal insulation. Thermal resistance is the
counterpart of thermal conductity.
From the above table it is evident that flow of heat through
rods in series and parallel is analogous to the flow of current 2.4 Growth of Ice in Lakes
through resistance in series and parallel. This analogy is of Warm water generally gets more dense as it gets colder, and
great importance in solving complicated problems of heat therefore sinks. This fact may lead you to believe that ice
conduction. should form on the bottom of a lake first. But a funny thing
happens to water as it gets even colder. Colder than 4° Celsius
(39° Fahrenheit), water begins expanding and becomes less

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