554 Thermometry, Thermal Expansion and calorimetry

Chapter

12
Thermometry, Thermal Expansion and Calorimetry
Temperature (8) NTP or STP implies 273.15K (0°C = 32°F)

Scales of Temperature

Boiling 212°F 100°C 373 K

water

Freezing 32°F 0°C 273 K

water

Temperature is defined as the degree of hotness or coldness of a body.

The natural flow of heat is from higher temperature to lower temperature.
Fahrenheit Celsius Kelvin
Two bodies are said to be in thermal equilibrium with each other, Fig. 12.1
The centigrade (°C), Farenheite (°F), Kelvin (K), Reaumer (R),
when no heat flows from one body to the other. That is when both the
Rankine (Ra) are commonly used temperature scales.
bodies are at the same temperature.
(1) To construct a scale of temperature, two fixed points are taken.
(1) Temperature is one of the seven fundamental quantities with
First fixed point is the freezing point (ice point) of water, it is called lower
dimension [ ]. It is a scalar physical quantity with S.I. unit kelvin. fixed point (LFP). The second fixed point is the boiling point (steam point)
(2) When heat is given to a body and its state does not change, the of water, it is called upper fixed point (UFP).
temperature of the body rises and if heat is taken from a body its (2) Celsius scale : In this scale LFP (ice point) is taken 0° and UFP
temperature falls i.e. temperature can be regarded as the effect of cause (steam point) is taken 100°. The temperature measured on this scale all in
“heat”. degree Celsius (°C).
(3) According to kinetic theory of gases, temperature (macroscopic (3) Farenheite scale : This scale of temperature has LFP as 32°F and
physical quantity) is a measure of average translational kinetic energy of a UFP as 212°F. The change in temperature of 1°F corresponds to a change
molecule (microscopic physical quantity). of less than 1° on Celsius scale.
(4) Although the temperature of a body can to be raised without limit, (4) Kelvin scale : The Kelvin temperature scale is also known as
it cannot be lowered without limit and theoretically limiting low thermodynamic scale. The triple point of water is also selected to be the
temperature is taken to be zero of the kelvin scale. zero of scale of temperature. The temperature measured on this scale are
(5) Highest possible temperature achieved in laboratory is about 10 K 8 in Kelvin (K).
while lowest possible temperature attained is 10 K. –8

The triple point of water is that point on a P-T diagram where the
(6) Temperature of the core of the sun is 10 K while that of its surface
7 three phases of water, the solid, the liquid and the gas, can coexist in
is 6000 K. equilibrium.

(7) Normal temperature of human body is 310. 15 K (37°C = 98.6°F). Table 12.1 : Different measuring scales

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 Thermometry, Thermal Expansionand and Calorimetry 555

Scale Symbol for LFP UFP Number of

each degree divisions on the
scale
Celsius °C 0°C 100°C 100
Fahrenheit °F 32°F 212°F 180
An instrument used to measure the temperature of a body is called a
Reaumer °R 0°R 80°R 80 thermometer
Rankine °Ra 460 Ra 672 Ra 212 It works by absorbing some heat from the body, so the temperature
Kelvin K 273.15 K 373.15 K 100 recorded by it is lesser than the actual value unless the body is at constant
temperature. Some common types of thermometers are as follows
(5) Temperature on one scale can be converted into other scale by (1) Liquid (mercury) thermometers : In liquid thermometers mercury
using the following identity.
is preferred over other liquids as its expansion is large and uniform and it
Reading on any scale  LFP has high thermal conductivity and low specific heat.
= Constant for all scales
UFP  LFP
(i) Range of temperature :  50 C to 350 C
(freezing point) (boiling p oint)
(6) All these temperatures are related to each other by the following
relationship (ii) Upper limit of range of mercury thermometer can be raised upto
550°C by filling nitrogen in space over mercury under pressure (which
C0 F  32 K  273 . 15 R0 Ra  460
    elevates boiling point of mercury).
100 212  32 373 . 15  273 . 15 80  0 672  460
(iii) Mercury thermometer with cylindrical bulbs are more sensitive
C F  32 K  273 R Ra  460 than those with spherical bulbs.
or    
5 9 5 4 10 . 6 (iv) If alcohol is used instead of mercury then range of temperature
(7) The Celsius and Kelvin scales have different zero points but the same measurement becomes – 80°C to 350°C
size degrees. Therefore any temperature difference is the same on the Celsius l  l0
and Kelvin scales (T – T)°C = (T – T) K. (v) Formula : t   100 C
2 1 2 1
l100  l0
Thermometry (2) Gas thermometers : These are more sensitive and accurate than
A branch of science which deals with the measurement of temperature liquid thermometers as expansion of gases is more than that of liquids. The
of a substance is known as thermometry. thermometers using a gas as thermoelectric substance are called ideal gas
thermometers. These are of two types
(1) The linear variation in some physical properties of a substance with
change of temperature is the basic principle of thermometry and these (i) Constant pressure gas thermometers
properties are defined as thermometric property (x) of the substance. (a) Principle V  T (if P = constant)
(2) Thermometric properties (x) may be as follows V  V0 V
(i) Length of liquid in capillary (b) Formula : t   100 C or T  273 . 16 K
V100  V0 VTr
(ii) Pressure of gas at constant volume.
(ii) Constant volume gas thermometers
(iii) Volume of gas at constant pressure.
(a) Principle P  T (if V = constant)
(iv) Resistance of a given platinum wire.
P  P0 P
(3) In old thermometry, freezing point (0°C) and steam point (100°C) (b) Formula : t   100 C or T  273 .16 K
are taken to define the temperature scale. So if the thermometric property P100  P0 PTr
at temperature 0°C, 100°C and t°C are x , x and x respectively then
0 100
(c) Range of temperature :
t0 x  x0 x  x0 Hydrogen gas thermometer : – 200°C to 500°C
  t C   100 C
100  0 x 100  x 0 x 100  x 0 Nitrogen gas thermometer : – 200°C to 1600°C
Helium gas thermometer : – 268°C to 500°C
(4) In modern thermometry instead of two fixed points only one
reference point is chosen (triple point of water 273.16 K) the other is itself (3) Resistance thermometers : Usually platinum is used in resistance
0 K where the value of thermometric property is assumed to be zero. thermometers due to high melting point and large value of temperature
coefficient of resistance.
So if the value of thermometric property at 0 K, 273.16 K and TK are
0, x and x respectively then Resistance of metals varies with temperature according to relation.
Tr

R  R0 (1  t ) where  is the temperature coefficient of resistance and t

T x  x 
  T  273 . 16  K is change in temperature.
273 . 16 x Tr  x Tr 
R  R0 R
(i) Formula : t   100 C or T  273 .16 K
Thermometers R100  R0 RTr
(ii) Temperature range : For Platinum resistance thermometer it is –
200°C to 1200°C
For Germanium resistance thermometer it is 4 to 77 K.

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 556 Thermometry, Thermal Expansion and calorimetry
(4) Thermoelectric thermometers : These are based on “Seebeck
effect” according to which when two distinct metals are joined to form a
closed circuit called thermocouple and the difference in temperature is
maintained between their junctions, an emf is developed. The emf is called
thermo-emf and if one junction is at 0°C, thermoelectric emf varies with
temperature of hot junction (t) according to e = at + bt ; where a and b are
2

constants.
G

Fe When matter is heated without any change in it’s state, it usually

Cu expands. According to atomic theory of matter, a symmetry in potential
energy curve is responsible for thermal expansion. As with rise in
temperature the amplitude of vibration and hence energy of atoms
increases, hence the average distance between the atoms increases. So the
Hot Ice
matter as a whole expands.
(1) Thermal expansion is minimum in case of solids but maximum in
case of gases because intermolecular force is maximum in solids but
Fig. 12.2
Thermoelectric thermometers have low thermal capacity and high minimum in gases.
thermal conductivity, so can be used to measure quickly changing
temperature (2) Solids can expand in one dimension (linear expansion), two
dimension (superficial expansion) and three dimension (volume expansion)
Table 12.2 : Different temperature range while liquids and gases usually suffers change in volume only.
Thermo couple Temperature range (3) Linear expansion : When a solid is heated and it's length increases,
Copper-iron thermocouple 0°C to 260°C
then the expansion is called linear expansion.
L0 L0 + L = L
Iron-constantan thermocouple 0°C to 800°C
o
Tungsten-molybdenum thermocouple 2000 C to 3000°C
(A) Before heating (B) After heating

Fig. 12.4
(5) Pyrometers : These are the devices used to measure the
(i) Change in length  L = L T
temperature by measuring the intensity of radiations received from the 0

body. They are based on the fact that the amount of radiations emitted (L = Original length, T = Temperature change)
0

from a body per unit area per second is directly proportional to the fourth
(ii) Final length L = L (1 + T)
power of temperature (Stefan’s law). 0

L
Telescope (iii) Co-efficient of linear expansion  
L0 T
Radiations
Lamp (iv) Unit of  is C 1 or K 1 . It's dimension is [ 1 ]
(4) Superficial (areal) expansion : When the temperature of a 2D
Red glass object is changed, it's area changes, then the expansion is called superficial
filter expansion.
A

L0
Fig. 12.3
(i) These can be used to measure temperatures ranging from 800°C to
L
6000°C. L0

(ii) They cannot measure temperature below 800°C because the

amount of radiations is too small to be measured.
L
(6) Vapour pressure thermometer : These are used to measure very
low temperatures. They are based on the fact that saturated vapour Fig. 12.5
pressure P of a liquid depends on the temperature according to the relation (i) Change in area is A = A T 0

c (A = Original area, T = Temperature change)

log P  a  bT  0

T
(ii) Final area A = A (1 + T)
The range of these thermometers varies from 0.71 K to 120 K for
0

different liquid vapours. A

(iii) Co-efficient of superficial expansion  
Thermal Expansion A0 T

(iv) Unit of  is °C or K .
–1 –1

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 Thermometry, Thermal Expansionand and Calorimetry 557

(5) Volume or cubical expansion : When a solid is heated and it's heating due to unequal linear expansion of the two metal. The strip will
volume increases, then the expansion is called volume or cubical expansion. bend with metal of greater  on outer side i.e. convex side.

Steel Brass
L0 L0

L0 V

Room temperature Higher temperature

(A) (B) OFF

ON
Bimetallic
(i) Change in volume is V  V0T strip
Fig. 12.6
(V = Original volume, T = change in temperature)
0

(ii) Final volume V  V0 (1  T )

V At room temperature At high temperature

(iii) Volume co-efficient of expansion   (C) (D)
V0 T Fig. 12.7
(2) Effect of temperature on the time period of a simple pendulum : A
(iv) Unit of  is °C or K .
–1 –1

pendulum clock keeps proper time at temperature . If temperature is

(6) More about ,  and  : The co-efficient ,  and  for a solid are increased to  (  ) then due to linear expansion, length of pendulum and
related to each other as follows hence its time period will increase.
  T 1
    :  :  1 : 2 : 3 Fractional change in time period   
2 3 T 2
(i) Hence for the same rise in temperature (i) Due to increment in its time period, a pendulum clock becomes
Percentage change in area = 2  percentage change in length. slow in summer and will lose time.

Percentage change in volume = 3  percentage change in length. 1

Loss of time in a time period T    T
2
(ii) The three coefficients of expansion are not constant for a given
(ii) Time lost by the clock in a day (t = 86400 sec)
solid. Their values depend on the temperature range in which they are
measured. 1 1
t    t    (86400 )  43200   sec
(iii) The values of ,  ,  are independent of the units of length, area 2 2
and volume respectively. (iii) The clock will lose time i.e. will become slow if     (in
summer)
(iv) For anisotropic solids    x   y   z where  ,  , and 
and will gain time i.e. will become fast if     (in winter).
x y z

represent the mean coefficients of linear expansion along three mutually

perpendicular directions. (iv) The gain or loss in time is independent of time period T and
depends on the time interval t.
(7) Contraction on heating : Some rubber like substances contract
with rising temperature, because transverse vibration of atoms of substance (v) Since coefficient of linear expansion () is very small for invar,
hence pendulums are made of invar to show the correct time in all seasons.
dominate over longitudinal vibration which is responsible for expansion.
(3) Thermal stress in a rigidly fixed rod : When a rod whose ends are
Table 12.3 :  and  for some materials rigidly fixed such as to prevent expansion or contraction, undergoes a
Material change in temperature, due to thermal expansion or contraction, a
[K–1 or (°C)–1] [K–1 or (°C)–1]
compressive or tensile stress is developed in it. Due to this thermal stress
Steel 1.2  10–5 3.6  10–5 the rod will exert a large force on the supports. If the change in
Copper temperature of a rod of length L is  then
1.7  10–5 5.1  10–5
Brass 2.0  10 –5
6.0  10–5
Aluminium 2.4  10–5 7.2  10–5

Application of Thermal Expansion in Solids (A) Heating (B) Cooling

Fig. 12.8
(1) Bi-metallic strip : Two strips of equal lengths but of different
materials (different coefficient of linear expansion) when join together, it is L  L 1 
Thermal strain     As   L   
called “bi-metallic strip”, and can be used in thermostat to break or make L  
electrical contact. This strip has the characteristic property of bending on

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 558 Thermometry, Thermal Expansion and calorimetry

 stress  expands less) but later on, it starts rising due to faster expansion of the
So Thermal stress  Y   As Y  strain  liquid.
 
R
or Force on the supports F  YA   P
Q PQ  represents expansion of vessel
(4) Error in scale reading due to expansion or contraction : If a scale QR  represents the real
expansion of liquid
gives correct reading at temperature , at temperature  (  ) due to PR  Represent the apparent
linear expansion of scale, the scale will expand and scale reading will be expansion of liquid
lesser than true value so that,
True value = Scale reading [1   (    )]

i.e. TV  SR [1    ] with   (    ) Fig. 12.11

(3) The actual increase in the volume of the liquid = The apparent
0 a 0 SR a 0 a SR increase in the volume of liquid + the increase in the volume of the vessel.
(4) Liquids have two coefficients of volume expansion.
(i) Co-efficient of apparent expansion (  ) : It is due to apparent (that
a

appears to be, but is not) increase in the volume of liquid if expansion of

at  at  >  at  <  vessel containing the liquid is not taken into account.
TV = SR TV > SR TV < SR
Apparent expansion in volume (V )a
Fig. 12.9 a  
However, if     , due to contraction of scale, scale reading will be Initial vo lume   V  
more than true value, so true value will be lesser than scale reading and will (ii) Co-efficient of real expansion ( ) : It is due to the actual increase
r

still be given by above equation with   (    ) negative. in volume of liquid due to heating.
Real increase in volume (V )r
(5) Expansion of cavity : Thermal expansion of an isotropic object may r  
be imagined as a photographic enlargement. So if there is a hole A in a Initial vo lume   V  
plate C (or cavity A inside a body C), the area of hole (or volume of cavity) (V )Vessel
(iii) Also coefficient of expansion of flask  Vessel 
will increase when body expands on heating, just as if the hole (or cavity) V  
were solid B of the same material. Also the expansion of area (or volume) of (iv)  Real   Apparent   Vessel
the body C will be independent of shape and size of hole (or cavity), i.e.,
(v) Change (apparent change) in volume in liquid relative to vessel is
will be equal to that of D.
Vapp  V  app  = V ( Real   Vessel )  V ( r  3 )

 = Coefficient of linear expansion of the vessel.

a A r r a
B D Table 12.4 : Different level of liquid in vessel

C  V Level
b b Level of liquid in
Expansion of A = Expansion of B Expansion of C = Expansion of D  Real   Vessel (=3)  app  0 Vapp is positive vessel will rise on
heating.
Fig. 12.10
Level of liquid in
(6) Some other application  Real   Vessel (=3)  app  0 Vapp is negative vessel will fall on
heating.
(i) When rails are laid down on the ground, space is left between the
ends of two rails. level of liquid in
 Real   Vessel (=3)  app  0 Vapp  0 vessel will remain
same.
(ii) The transmission cable are not tightly fixed to the poles.
(iii) Test tubes, beakers and crucibles are made of pyrex-glass or silica (5) Anomalous expansion of water : Generally matter expands on
because they have very low value of coefficient of linear expansion. heating and contracts on cooling. In case of water, it expands on heating if
its temperature is greater than 4°C. In the range 0°C to 4°C, water
(iv) The iron rim to be put on a cart wheel is always of slightly smaller contracts on heating and expands on cooling, i.e.  is negative. This
diameter than that of wheel. behaviour of water in the range from 0°C to 4°C is called anomalous
expansion.
(v) A glass stopper jammed in the neck of a glass bottle can be taken
out by warming the neck of the bottle This anomalous behaviour of water causes ice to form first at the
surface of a lake in cold weather. As winter approaches, the water
Thermal Expansion in Liquids temperature increases initially at the surface. The water there sinks because
of its increased density. Consequently, the surface reaches 0°C first and the
(1) Liquids do not have linear and superficial expansion but these only lake becomes covered with ice. Aquatic life is able to survive the cold winter
have volume expansion. as the lake bottom remains unfrozen at a temperature of about 4°C.
(2) Since liquids are always to be heated along with a vessel which At 4°C, density of water is maximum while its specific volume is
contains them so initially on heating the system (liquid + vessel), the level of minimum.
liquid in vessel falls (as vessel expands more since it absorbs heat and liquid
max
min
Anomalous
behaviour

Anomalous
behaviour
vol/mass

Density

0°C 4°C Temperature 0°C 4°C Temperature

(A) (B)
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 Thermometry, Thermal Expansionand and Calorimetry 559

(3) The amount of heat (Q) is given to a body depends upon it's mass
(m), change in it's temperature ( ° = ) and nature of material i.e.
Q  m .c.  ; where c = specific heat of material.

(4) Heat is a scalar quantity. It's units are joule, erg, cal, kcal etc.
(5) The calorie (cal) is defined as the amount of heat required to raise
(6) Effect of temperature on upthrust : The thrust on V volume of a the temperature of 1 gm of water from 14.5°C to 15.5°C.
body in a liquid of density  is given by Th  Vg Also 1 kcal = 1000 cal = 4186 J and 1 cal = 4.18 J
Now with rise in temperature by  °C, due to expansion, volume of (6) British Thermal Unit (BTU) : One BTU is the quantity of heat
the body will increase while density of liquid will decrease according to the required to raise the temperature of one pound ( 1 lb ) of water from 63°F
relations V   V (1   S  ) and     /(1   L  )
to 64°F
T h  V  g (1   S  ) 1 BTU = 778 ft. lb = 252 cal = 1055 J
So the thrust T h  V  g   
Th Vg (1   L  )
(7) In solids thermal energy is present in the form of kinetic energy, in
and apparent weight of the body W = Actual weight – Thrust
app
liquids, in the form of translatory energy of molecules. In gas it is due to
the random motion of molecules.
As  S   L  T h   Th with rise in temperature thrust also
decreases and apparent weight of body increases. (8) Heat always flows from a body of higher temperature to lower
temperature till their temperature becomes equal (Thermal equilibrium).
Variation of Density with Temperature
(9) The heat required for a given temperature increase depends only
Most substances (solid and liquid) expand when they are heated, i.e., on how many atoms the sample contains, not on the mass of an individual
volume of a given mass of a substance increases on heating, so the density atom.
 1 1
should decrease  as    . For a given mass   Specific Heat
 V V
 V V V 1 When a body is heated it's temperature rises (except during a change
    in phase).
 V  V  V V  V 1   
(1) Gram specific heat : The amount of heat energy required to raise
 the temperature of unit mass of a body through 1°C (or K) is called specific
     (1    )1 =  (1    )
1    heat of the material of the body.

Expansion of Gases If Q heat changes the temperature of mass m by  then specific heat
Gases have no definite shape, therefore gases have only volume Q
c
expansion. Since the expansion of container is negligible in comparison to m 
the gases, therefore gases have only real expansion.
(i) Units : Calorie/gm  °C (practical), J/kg  K (S.I.) Dimension :
(1) Coefficient of volume expansion : At constant pressure, the unit 2 2 1
[L T  ]
volume of a given mass of a gas, increases with 1°C rise of temperature, is
called coefficient of volume expansion. (ii) For an infinitesimal temperature change d and corresponding
V 1 quantity of heat dQ.
   Final volume V   V (1   )
V0  1 dQ
Specific heat c  .
m d
P 1
(2) Coefficient of pressure expansion :   
P  (2) Molar specific heat : Molar specific heat of a substance is defined
as the amount of heat required to raise the temperature of one gram mole
 Final pressure P   P (1    )
of the substance through a unit degree it is represented by (capital) C.
For an ideal gas, coefficient of volume expansion is equal to the
1 Molar specific heat (C)  M  Gram specific heat (c)
coefficient of pressure expansion i.e.     C 1
273
(M = Molecular mass of substance)
Heat
Q 1 Q  m
(1) The form of energy which is exchanged among various bodies or CM   where, Number of moles   
system on account of temperature difference is defined as heat. m     M 

(2) We can change the temperature of a body by giving heat Units : calorie/mole  °C (practical); J/mole  kelvin (S.I.)
(temperature rises) or by removing heat (temperature falls) from body.
Dimension : [ML2 T 2 1 ]

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 560 Thermometry, Thermal Expansion and calorimetry

Specific Heat of Solids (3) The variation of specific heat with temperature for water is shown
in the figure. Usually this temperature dependence of specific heat is
When a solid is heated through a small range of temperature, its neglected.
volume remains more or less constant. Therefore specific heat of a solid
may be called its specific heat at constant volume C . V

1.0089

Sp. heat cal/g C°

Y 3R 1.004
CV 1.000
0.996
20 40 60 80 100
Temp. in °C
X
T Debye temp. (4) As specific heat of water is Fig.
very12.14
large; by absorbing or releasing
large amount of heat its temperature changes by small amount. This is why,
Fig. at
(1) From the graph it is clear that 12.13
T = 0, C tends to zero it is used in hot water bottles or as coolant in radiators.
V

(2) With rise in temperature, C increases and at a particular Specific Heat of Gases
V

temperature (called Debey's temperature) it becomes constant = 3R = 6

(1) In case of gases, heat energy supplied to a gas is spent not only in
cal/mole  kelvin = 25 J/mole  kelvin raising the temperature of the gas but also in expansion of gas against
(3) For most of the solids, Debye temperature is close to room atmospheric pressure.
temperature. (2) Hence specific heat of a gas, which is the amount of heat energy
required to raise the temperature of one gram of gas through a unit degree
(4) Dulong and Petit law : Average molar specific heat of all metals at shall not have a single or unique value.
room temperature is constant, being nearly equal to 3R = 6 cal. mole K = –1 –1

(3) If the gas is compressed suddenly and no heat is supplied from

25 J mole K , where R is gas constant for one mole of the gas. This
–1 –1

outside i.e. Q = 0, but the temperature of the gas raises on the account of
statement is known as Dulong and Petit law. compression.
(5) Debey's law : It was observed that at very low temperature molar Q 0
 c  0
specific heat  T 3 (exception are Sn, Pb and Pt) m ( ) m 
(4) If the gas is heated and allowed to expand at such a rate that rise
cal
(6) Specific heat of ice : In C.G.S. cice  0 . 5 in temperature due to heat supplied is exactly equal to fall in temperature
gm  C
due to expansion of the gas. i.e.  = 0
cal Joule Q Q
In S.I. c  500
ice
 2100 .  c  
kg  C kg  C m ( ) 0
Table 12.5 : Specific heat of some solids at room temperature and atmospheric (5) If rate of expansion of the gas were slow, the fall in temperature of
pressure the gas due to expansion would be smaller than the rise in temperature of
Substance Specific heat Molar specific heat the gas due to heat supplied. Therefore, there will be some net rise in
( J-kg–1 K–1) (J-g mole–1 K–1) temperature of the gas i.e. T will be positive.
Aluminium 900.0 24.4 Q
 c  Positive
Copper 386.4 24.5 m ( )
Silver 236.1 25.5 (6) If the gas were to expand very fast, fall of temperature of gas due
Lead 127.7 26.5 to expansion would be greater than rise in temperature due to heat
Tungsten 134.4 24.9 supplied. Therefore, there will be some net fall in temperature of the gas i.e.
 will be negative.
Specific Heat of Liquid (Water) Q
 c  Negative
(1) Among all known solids and liquids specific heat of water is m ( )
maximum i.e. water takes more time to heat and more time to cool w.r.t. Hence the specific heat of gas can have any positive value ranging
other solids and liquids. from zero to infinity. Further it can even be negative. The exact value
(2) It is observed that by increasing temperature, initially specific heat depends upon the mode of heating the gas. Out of many values of specific
of water goes on decreasing, becomes minimum at 37°C and then it start heat of a gas, two are of special significance, namely C and C , in the P V

increasing. Specific heat of water is – chapter “Kinetic theory of gases” we will discussed this topic in detail.
1 cal cal J Specific heat of steam : c steam  0 . 47 cal / gm  C
 1000  4200
gm  C kg  C kg  C
Phase Change and Latent Heat
(This value is obtained between the temperature 14.5°C to 15.5°C)

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 Thermometry, Thermal Expansionand and Calorimetry 561

(1) Phase : We use the term phase to describe a specific state of But when a solid gets converted to a liquid, then the increase in volume is
matter, such as solid, liquid or gas. A transition from one phase to another negligible. Hence very less amount of heat is required. So, latent heat of
is called a phase change. vaporisation is more than the latent heat of fusion.
(i) For any given pressure a phase change takes place at a definite Thermal Capacity and Water Equivalent
temperature, usually accompanied by absorption or emission of heat and a
change of volume and density. (1) Thermal capacity : It is defined as the amount of heat required to
raise the temperature of the whole body (mass m) through 0°C or 1K.
(ii) In phase change ice at 0°C melts into water at 0°C. Water at 100°C
boils to form steam at 100°C. Q
Thermal capacity  mc  C 

Heat taken (–  Q)
The value of thermal capacity of a body depends upon the nature of
Solid the body and its mass.
at 0°C
Dimension : [ML2 T 2 1 ] , Unit : cal/°C (practical) Joule/k (S.I.)
Liquid at 0°C
Heat given (+  Q) (2) Water Equivalent : Water equivalent of a body is defined as the
(A) mass of water which would absorb or evolve the same amount of heat as is
Heat taken (–  Q) done by the body in rising or falling through the same range of
temperature. It is represented by W.

If m = Mass of the body, c = Specific heat of body,  = Rise in

temperature.
Liquid at 100°C Vapours at 100°C
Heat given (+  Q)
Then heat given to body Q  mc  ….. (i)
(B)
Fig. 12.15 If same amount of heat is given to W gm of water and its temperature
(iii) In solids, the forces between the molecules are large and the also rises by . Then
molecules are almost fixed in their positions inside the solid. In a liquid, the
forces between the molecules are weaker and the molecules may move heat given to water Q  W  1   … (ii) [As c water  1 ]
freely inside the volume of the liquid. However, they are not able to come
out of the surface. In vapours or gases, the intermolecular forces are almost From equation (i) and (ii) Q  mc   W  1  
negligible and the molecules may move freely anywhere in the container.
When a solid melts, its molecules move apart against the strong molecular  Water equivalent (W) = mc gm
attraction. This needs energy which must be supplied from outside. Thus,
the internal energy of a given body is larger in liquid phase than in solid (i) Unit : Kg (S.I.) Dimension : [ML0 T 0 ]
phase. Similarly, the internal energy of a given body in vapour phase is
(ii) Unit of thermal capacity is J/kg while unit of water equivalent is
larger than that in liquid phase.
kg.
(iv) In case of change of state if the molecules come closer, energy is
released and if the molecules move apart, energy is absorbed. (iii) Thermal capacity of the body and its water equivalent are
numerically equal.
(2) Latent heat : The amount of heat required to change the state of
the mass m of the substance is written as : Q = mL, where L is the latent (iv) If thermal capacity of a body is expressed in terms of mass of
heat. Latent heat is also called as Heat of Transformation. It's unit is cal/gm water it is called water-equivalent of the body.
or J/kg and Dimension: [L2 T 2 ]
Some Important Terms
(i) Latent heat of fusion : The latent heat of fusion is the heat energy
(1) Evaporation : Vaporisation occurring from the free surface of a
required to change 1 kg of the material in its solid state at its melting point
liquid is called evaporation. Evaporation is the escape of molecules from the
to 1 kg of the material in its liquid state. It is also the amount of heat energy
surface of a liquid. This process takes place at all temperatures and
released when at melting point 1 kg of liquid changes to 1 kg of solid. For
increases with the increase of
water at its normal freezing temperature or melting point (0°C), the latent
temperature. Evaporation leads to
heat of fusion (or latent heat of ice) is
cooling because the faster molecules
L F  Lice  80 cal / gm  60 kJ / mol  336 kilo joule / kg escape and, therefore, the average
kinetic energy of the molecules of
(ii) Latent heat of vaporisation : The latent heat of vaporisation is the
the liquid (and hence the
heat energy required to change 1 kg of the material in its liquid state at
temperature) decreases.
its boiling point to 1 kg of the material in its gaseous state. It is also the
amount of heat energy released when 1 kg of vapour changes into 1 kg of (2) Melting (or fusion)/freezing Evaporation cools hot water produced
by power plants
liquid. For water at its normal boiling point or condensation temperature (or solidification) : The phase change
(100°C), the latent heat of vaporisation (latent heat of steam) is of solid to liquid is called melting or Fig. 12.16
fusion. The reverse phenomenon is
LV  Lsteam  540 cal / gm  40 .8 kJ / mol  2260 kilo joule / kg
called freezing or solidification.
(iii) Latent heat of vaporisation is more than the latent heat of fusion. When pressure is applied on ice,
This is because when a substance gets converted from liquid to vapour, it melts. As soon as the pressure is
there is a large increase in volume. Hence more amount of heat is required.

Fig. 12.17

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 562 Thermometry, Thermal Expansion and calorimetry
removed, it freezes again. This phenomenon is called regelation. S.V.P. at dew point
R.H.(%)   100
(3) Vaporisation/liquefication (condensation) : The phase change from S.V. P. at given temperatu re
liquid to vapour is called vaporisation. The reverse transition is called (10) Variation of melting point with pressure : For those substances
liquefication or condensation. with contract on melting (e.g. water and rubber), the melting point
decreases with pressure. The reason is the pressure helps shrinking and
(4) Sublimation : Sublimation is the conversion of a solid directly into hence melting. Most substances expand on melting. (e.g. max, sulpher etc.)
vapours. Sublimation takes place when boiling point is less than the melting
point. A block of ice sublimates into vapours on the surface of moon An increase of pressure opposes the melting of such substances and
their melting point is raised.
because of very very low pressure on its surface. Heat required to change
unit mass of solid directly into vapours at a given temperature is called heat (11) Variation of latent heat with temperature and pressure : The
of sublimation at that temperature. latent heat of vapourization of a substance varies with temperature and
hence pressure because the boiling point depends on pressure. It
(5) Hoar frost : Direct conversion of vapours into solid is called hoar increases as the temperature is decreased. For example, water at 1 atm
frost. This process is just reverse of the boils at 100°C and has latent heat 2259 Jg but at 0.5 atm it boils at 82°C
–1

process of sublimation, e.g., formation and has latent heat 2310 Jg –1

of snow by freezing of clouds.

P P Critical point

Fusion
Critical point L
L
Vaporization Vaporization
(6) Vapour pressure : When the S
Fig. 12.18
space above a liquid is closed, it soon S Triple point
Triple point
becomes saturated with vapour and a dynamic equilibrium is established. V
The pressure exerted by this vapour is called Saturated Vapour Pressure V
The latent heat of fusion shows similar but less pronounced pressure
(S.V.P.) whose value depends only on the temperature – it is independent of T T
dependence.
any external pressure. If the volume of the space is reduced, some vapour Fig. 12.20
The figures show the P-T graphs for (a) a substance (e.g., water)
liquefies, but the pressure is unchanged.
which contracts on melting an (b) a substance (e.g. wax) which expands on
A saturated vapour does not obey the gas law whereas the unsaturated melting. The P-T graph consists of three curves.
vapour obeys them fairly well. However, a vapour differs from a gas in that (i) Sublimation curve which connects points at which vapour (V) and
the former can be liquefied by pressure alone, whereas the latter cannot be solid (S) exist in equilibrium.
liquefied unless it is first cooled. (ii) Vapourization curve which shows vapour and liquid (L) existing in
(7) Boiling : As the temperature of a liquid is increased, the rate of equilibrium.
evaporation also increases. A stage is (iii) Fusion curve which shows liquid and solid existing in equilibrium.
reached when bubbles of vapour start The three curves meet at a single point which is called the triple point.
forming in the body of the liquid It is that unique temperature-pressure point for a substance at which all the
which rise to the surface and escape. A three phases exist in equilibrium.
liquid boils at a temperature at which (12) Freezing mixture : If salt is added to ice, then the temperature of
the S.V.P. is equal to the external mixture drops down to less than 0°C. This is so because, some ice melts
pressure. down to cool the salt to 0°C. As a result, salt gets dissolved in the water
formed and saturated solution of salt is obtained; but the ice point (freeing
It is a fast process. The boiling point Fig. 12.19 point) of the solution formed is always less than that of pure water. So, ice
changes on mixing impurities. cannot be in the solid state with the salt solution at 0°C. The ice which is in
(8) Dew point : It is that temperature at which the mass of water contact with the solution, starts melting and it absorbs the required latent
vapour present in a given volume of air is just sufficient to saturate it, i.e. heat from the mixture, so the temperature of mixture falls down.
the temperature at which the actual vapour pressure becomes equal to the Joule's Law (Heat and Mechanical Work)
saturated vapuor pressure.
(9) Humidity : Atmospheric air always contains some water vapour.
The mass of water vapour per unit volume is called absolute humidity.
The ratio of the mass of water vapour (m) actually present in a given
volume of air to the mass of water vapour (M) required to saturate the
same volume at the same temperature is called the relative humidity (R.H.).
m
Generally, it is expressed as a percentage, i.e., R.H.(%)   100 (%)
M
R.H. May also be defined as the ratio of the actual vapour pressure (p)
Whenever heat is converted into mechanical work or mechanical work is
p
of water at the same temperature, i.e. R.H.(%)   100 (%) converted into heat, then the ratio of work done to heat produced always
P
W
Thus R.H. may also be defined as remains constant. i.e. W  Q or J
Q

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 Thermometry, Thermal Expansionand and Calorimetry 563

This is Joule’s law and J is called mechanical equivalent of heat. Heat lost = Heat gained
(1) From W = JQ if Q = 1 then J = W. Hence the amount of work done i.e. principle of calorimetry represents the law of conservation of heat
necessary to produce unit amount of heat is defined as the mechanical energy.
equivalent of heat. (1) Temperature of mixture (  ) is always lower temperature ( ) and
mix L

(2) J is neither a constant, nor a physical quantity rather it is a higher temperature ( ), i.e.,  L   mix   H .
H

conversion factor which used to convert Joule or erg into calorie or kilo It means the temperature of mixture can never be lesser than lower
calories vice-versa. temperatures (as a body cannot be cooled below the temperature of cooling
Joule erg body) and greater than higher temperature (as a body cannot be heated
(3) Value of J  4 . 2  4 . 2  10 7 above the temperature of heating body). Furthermore usually rise in
cal cal
temperature of one body is not equal to the fall in temperature of the other
Joule body though heat gained by one body is equal to the heat lost by the other.
 4 . 2  10 3 .
kcal (2) Mixing of two substances when temperature changes only : It
(4) When water in a stream falls from height h, then its potential means no phase change. Suppose two substances having masses m 1 and
energy is converted into heat and temperature of water rises slightly. m 2 , gram specific heat c1 and c 2 , temperatures  1 and  2 ( 1   2 )
From W = JQ  mgh = J (mc  ) are mixed together such that temperature of mixture at equilibrium is  mix

[where m = Mass of water, c = Specific heat of water,  = Hence, Heat lost = Heat gained
temperature rise]  m 1 c1 (1   mix )  m 2 c 2 ( mix   2 ) 
gh m 1 c1 1  m 2 c 2 2
 Rise in temperature   C  mix 
Jc m 1 c1  m 2 c 2
(5) The kinetic energy of a bullet fired from a gun gets converted into
Table 12.6 : Temperature of mixture in different cases
heat on striking the target. By this heat the temperature of bullet increases
by . Condition Temperature of mixture
If bodies are of same material m 11  m 2 2
1  mix 
From W = JQ  mv 2  J ( m s  ) i.e. c 1 = c 2 m1  m 2
2
If bodies are of same mass  1 c1   2 c 2
[where m = Mass of the bullet, v = Velocity of the bullet, c = Specific  mix 
heat of the bullet] m1 = m2 c1  c 2

v2 If m1 = m2 and c1 = c2  mix 
1   2
 Rise in temperature t  C
2 Jc 2

If the temperature of bullet rises upto the melting point of the bullet
(3) Mixing of two substances when temperature and phase both
and bullet melts then.
changes or only phase changes: A very common example for this category is
From W = J(Q +Q )
Temperature change Phase change
ice-water mixing.
1
 mv 2  J (mc   mL ) ; L = Latent heat of bullet Suppose water at temperature  °C is mixed with ice at 0 °C, first ice
2 W i

will melt and then it's temperature rises to attain thermal equilibrium.
 v2  Hence; Heat given = Heat taken
   L

 2 J  m W CW (W   mix )  m i Li  m iCW ( mix  0 )
 Rise in temperature      C

 c  m i Li
  mW W 
  CW
  mix 
(6) If m kg ice-block falls down through some height (h) and melts mW  m i
partially (m' kg) then its potential energy gets converted into heat
of melting. Li
W 
CW
m '  JL  (i) If m W  m i then  mix 
From W = JQ  mgh  J m ' L  h    2
m  g 
JL (ii) By using this formulae if  mix   i then take  mix  0 C
If ice-block melts completely then m' = m  h  meter
g
Heating Curve
Principle of Calorimetry
If to a given mass (m) of a solid, heat is supplied at constant rate P
Calorimetry means 'measuring heat'. and a graph is plotted between temperature and time, the graph is as
When two bodies (one being solid and other liquid or both being shown in figure and is called heating curve. From this curve it is clear that
liquid) at different temperatures are mixed, heat will be transferred from
body at higher temperature to a body at lower temperature till both acquire
same temperature. The body at higher temperature releases heat while body E
at lower temperature absorbs it, so that C
T2 Boiling
Boiling D point
Temp.

T1 B Melting
A Melting point

O Time
t1 t2 t3 t4

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 564 Thermometry, Thermal Expansion and calorimetry

the atmosphere to melt down.

So, in the mountains, when
snow falls, one does not feel
too cold, but when ice melts, he feels too cold.

(1) In the region OA temperature of solid is changing with time so,  There is more shivering effect of ice-cream on teeth as compared to
that of water
Q  mc S T  P t  mc S T [as Q = Pt]
(obtained from ice).
But as (T/t) is the slope of temperature-time curve This is because, when ice-cream
1 melts down, it absorbs large
cs 
Slope of line OA amount of heat from teeth.
i.e. specific heat (or thermal capacity) is inversely proportional to the  Branch of physics dealing with production and measurement of
slope of temperature-time curve. temperatures close to 0K is known as cryogenics while that dealing with
the measurement of very high temperature is called as pyrometry.
(2) In the region AB temperature is constant, so it represents change
of state, i.e., melting of solid with melting point T . At A melting starts and
1
 It is more painful to get burnt by steam rather than by boiling
at B all solid is converted into liquid. So between A and B substance is water at same temperature. This is so because when steam at 100°C gets
partly solid and partly liquid. If L is the latent heat of fusion. Q  mL F or
F converted to water at 100°C, then it gives out 536 calories of heat. So, it
P(t 2  t1 ) is clear that steam at 100°C has more heat than water at 100°C (i.e.,
LF  [as Q  P(t 2  t1 ) ] boiling of water).
m

or L  length of line AB  A solid and hollow sphere of same radius and material, heated to
the same temperature then expansion of both will be equal because
F

i.e. Latent heat of fusion is proportional to the length of line of zero thermal expansion of isotropic solids is similar to true photographic
1 enlargement. It means the expansion of cavity is same as if it has been a
slope. [In this region specific heat  ] solid body of the same material. But if same heat is given to the two
tan 0
spheres, due to lesser mass, rise in temperature of hollow sphere will be
(3) In the region BC temperature of liquid increases so specific heat
  Q  
(or thermal capacity) of liquid will be inversely proportional to the slope of more  As     .
line BC   mc  

1 Hence its expansion will be more.

i.e., c L 
Slope of line B C  Specific heat of a substance can also be negative. Negative specific
heat means that in order to raise the temperature, a certain quantity of
(4) In the region CD temperature is constant, so it represents the
heat is to be withdrawn from the body.
change of state, i.e., boiling with boiling point T . At C all substance is in
2

liquid state while at D in vapour state and between C and D partly liquid e.g. Specific heat of saturated vapours.
and partly gas. The length of line CD is proportional to latent heat of
 Specific heat for hydrogen is maximum (3 .5 cal / gm o C ) and it
vaporisation
is minimum for radon and actinium ~ 0 . 022 cal / gm  C  .
i.e., L  Length of line CD [In this region specific heat 
V

1  The minimum possible temperature is 0 K.

]
tan 0  Amount of steam at 100°C required to just melt m gm of ice at 0°C
(5) The line DE represents gaseous state of substance with its is m/8 gm.
temperature increasing linearly with time. The reciprocal of slope of line will  If we put the beaker containing water in melting ice, the water in
be proportional to specific heat or thermal capacity of substance in vapour the beaker will cool to 0°C but will never freeze.
state.

Water

Melting Ice

 A pressure in excess of 25 atm is required to make helium solidfy.

 After snow falls, the temperature of the atmosphere becomes very At 1 atm pressure, helium remains a liquid down to absolute zero.
low. This is because  Boiling temperature of water, if pressure is different from normal
the snow absorbs the heat from pressure is t = [100°C – (760 – P in mm)  0.037]°C
Boili ng

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 Thermometry, Thermal Expansionand and Calorimetry 565

5. The resistance of a resistance thermometer has values 2.71 and 3.70

 Confusing S.I. and C.G.S. units ohm at 10°C and 100°C. The temperature at which the resistance is
It is advised to do questions on calorimetry in C.G.S. as 3.26 ohm is [CPMT 1994]
calculations becomes simple. If the final answer is in joules, then convert (a) 40°C (b) 50°C
cal into joules. (c) 60°C (d) 70°C
 Invar and quartz have very small values of co-efficient of linear 6. No other thermometer is as suitable as a platinum resistance
expansion. thermometer to measure temperature in the entire range of
[MNR 1993]
 In S.I. nomenclature " degree" is not used with the kelvin scale; e.g. (a) 0°C to 100°C (b) 100°C to 1500°C
273°K is wrong while 273 K is correct to write.
(c) – 50°C to +350°C (d) – 200°C to 600°C
 Magnetic thermometer is recommended for measuring very low
temperature (2K).
 The most sensitive thermometer is gas thermometer.
 Dew formation is more probable on a cloudiness calm night.
 In winters, generally fog disappear before noon. Because, the
atmosphere warms up and tends to be unsaturated. The condensed
vapours reevaporates and the fog disappears.
 Standardisation of thermometer is obtained with gas thermometer.
Because coefficient of expansion of gas is very large.
 Dogs hang their tongues in order to expose a surface to the air for
evaporation and hence, cooling. They do not sweat.

Thermometry
1. On the Celsius scale the absolute zero of temperature is at
[CBSE PMT 1994]
(a) 0°C (b) – 32°C
(c) 100°C (d) – 273.15°C
2. Oxygen boils at – 183°C. This temperature is approximately
[CPMT 1992]
(a) 215°F (b) – 297°F
(c) 329°F (d) 361°F
3. Recently, the phenomenon of superconductivity has been observed
at 95 K. This temperature is nearly equal to
[CPMT 1990]
(a) – 288°F (b) – 146°F
(c) – 368°F (d) +178°F
4. The temperature of a substance increases by 27°C. On the Kelvin
scale this increase is equal to [CPMT 1993]
(a) 300 K (b) 2.46 K
(c) 27 K (d) 7 K

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