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CHAPTER - 00

HEAT AND THERMODYNAMICS Thermometry Heat Heat is the form of energy which gives the sensation of hotness or coldness of a body. It is the physical cause of sensation of hotness or coldness. Temperature Degree of hotness or coldness of a body is temperature. It determines the direction of heat flow. Thermometry is the technique for the quantitative determination of thermodynamic temperature

Different temperature Scale Scale

LFP

UFP

n

Repre se ntation

Celsius

00 C

1000 C

100

10 C

Fahrenheit

320 F

212 0 F

180

10 F

Reaum er

00 R

800 R

180

10 R

273 K

373 K

100

1K

Kelvin

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Conversion of temperature Scales Let x be a facility / Reference Scale

x  x LFP C0 F  32 R  0 K  273    x UFP  x LFP 100  0 212  32 80  0 373 273 x  x LFP C F  32 R K  273    x UFP  x LFP 100 180 80 180 Relation between temperature scales Celsius and Fahrenheit

C F  32  100 180

F1 

9 C  32 5 1

C F  32  5 9

F2 

9 C  32 5 2

F

9 C 32 5

[F2  F1 ] 

9 [C2  C1 ] 5

0

F Slope =

32

9 C 5 5 C  F 9 F 

9 5

Common Re ading in celsius and

0

C

Celsius and Kelvin

Fahrenheit is  40 0

Fahrenheit and Kelvin

C K  273  100 100

F  32 K  273  180 100

C  K  273

F  32 K  273  9 5

K  C  273

No common Reading in Celsius and

Common Reading in Fahrenheit and kelvin

Kelvin

is 574.25

Since the no. of divisions in both the scales are same Change in temperature in Celsius is equivalent to change in temperature Fahrenheit 2

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Note :As the no. of divisions in a temperature scale increases “size of the degree” decreases. The zero value in Kelvin scale is known as “the absolute zero”. It is 0 Kelvin or –273.150C. In modern temperature scales absolute zero and triple point of water are taken as fixed reference points. Triple point of water 0.010C or 273.16 K @ 4.58 mm Hg pressure or at pressure 6.11 × 102 Pa or 6 × 10–3 atm Thermometers Devices used for the measurement of temperature Principle : Linear variation of thermometric property with temperature Eg : Pressure of a gas Volume of a liquid Resistance of metal Light Intensity Thermo emf, magnetic properties Liquid Thermometers Mercury and alcohol are thermometric liquids advantages of Hg High conductivity, high boiling point, low specific heat, high visibility, high sensitivity to heat, high angle of contact Gas Thermometers (Most Sensitive) Constant Volume Gas Thermometers P  T Based on Gaylussac’s law

Constant Pressure Gas Thermometers V  T Based on Charles law

Resistance Thermometers Eg : Platinum Resistance Thermometer Germanium Resistance Thermometer To find unknown temp Let  be a thermometric property x0 = thermometric property at 00C x100 = thermometric property at 1000C xt = thermometric property at t0C then unknown temp

t

xt  x0 100 x 100  x 0 3

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for pressure

t

Pt  P0 100 P100  P0

for resistance

t

Rt R0 100 R100  R 0 Thermal Expansion

Expansion of Solids Three types of Expansion is possible in solids 1) Linear Expansion (in length) 2) Areal Expansion (in area) 3) Volume Expansion (in volume) Linear

Areal

Cubical

Expansion

Expansion

Expansion

  linear

  superficial

  cubical

Expansivity

Expansivity

Expansivity

L Lt



A A t

V Vt

Co-efficient



Change in value

L  L t

 A  A t

 V  V t

Fractional change

L   t L

A   t A

V   t V

% change

L 100   t 100 L

A  100   t  100 A

V  100    t  100 V

Final Value

L '  L [1  t]

A '  A [1  t]

V '  V [1  t]

Application of Linear Expansion Time loss/gain of a pendulum clock we have time period T 2   / g T

2 g

1/2 T  k 1/2 4



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Fractional charge

T 1    T 2  Fractional change in T

T 1    T 2 Change in time period

T 1    T T 2

Time loss/gain in a day T = 1 day = 24 hrs = 86400 s T 

1    86400 2

In summer

Temperature  length  time period  Time loss clock will become slow

In winter

Temperature  length  time period  Time gain clock will become fast

Thermal Stress When a metal rod is rigidly fixed at its both ends so that it is prevented from expansion or contraction. On heating stress will be developed due to thermal strain it is called thermal stress

Thermal strain =

Y

l   t l

Thermal stress Thermal strain

Thermal stress  Y  Thermal strain  Y t Tension or force  Thermal stress Area  YA  t

Note : Thermal stress developed in a rod is depending on the material of rod and rise in temperature and is independent of length of the rod.

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Bimetallic Strip Two metal rods of same length and different  joined together to form a bimetallic strip. On heating bimetallic strip will bend in such a way that metal having larger value of  will come on the convex side and metal having smaller value of  will come on the concave side. On cooling the reverse happens.

 A > B A B

Radius of bimetallic arc R

d [ A   B ]  t

if  A   B

Expansion of Liquids For heating a liquid, it has to be kept in a container. On heating the container will also expand.

Real Expansion of = Apparent Expansion of + Expa nsion of Liquid Liquid the container

 Real   apparent   container  Real   apparent  3 vessel Anomalous Expansion Water Volume of given amount of water decreases with increase in temperature from 0 to 40C. But beyond 40C water will normally expand. Water has the least volume and maximum density at 40C

Volume

P max 0 0C

40C

Temp

It plays an important role in the survival of aquaticf life in cold winter season in polar regions. There is expansion of water above and below 40C

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Condition for two metal rods of different length and  to have same difference in length at all temperature For difference in length independent of temperature  = constant  1    2  1 1 t   2  2 t

11   2  2

1  2  2  1

The correct the reading of a metallic scale Case 1 : When scale is expanding

True reading Scale reading(1   t)  t  temp diff (temp at which measurement is taken and temperature at which scale is calibrated

Case 2 : When object (only) is expanding Measurement value = True value (1  0 t) [ 1     0  t]  0   of object material

Case 3 : When both scale and object are expanding Measured value

1   [1 (0  s )] t

 0  of the object  s  of the scale Variation of density with temperature

density 



mass volume

m v

as temp increases volume increases and density decreases

 '

m v' 7

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 '

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m V [1  t]

' 

 1  t

 '   (1   t) expanding binomially Variation of upthrust with temperature Apparent weight = Actual weight – Upthrust At normal temperature Upthrust FB  VL g When temperature increases by t Upthrust FB'  V ' 'L g

F'B  V [1   st] 

L g [1  L t]

 1  s t  F'B  VLg    1  L t  1  s t  F'B FB   1   Lt  FB' 1  s t  FB 1 L  t In other way

FB' FB [1 ( s  L ) t Special Cases 1.

When a metallic disc with an inner hole i heated the diameter of the hole will also increases.

2.

When a metallic sphere with a cavity is heated size and volume of cavity increases.

3.

When a broken ring with gap between ends is heated the gap will also increases.

4.

For an anisotropic solids (shows different properties in different directions)   x  y   z

Calorimetry Calorimetry deals with two types of heats Specific heat and latent heat

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Joules mechanical equivalent of heat (J) Joule found that when mechanical work is converted into heat (Q) the ratio of W and Q is always a constant W  J or W  JQ Q J  joules mechanical equivalent of heat (It is a conversion factor)

J = 4.186 Joule/ Calorie Calorie It is the amount of heat energy required to rise the temperature of 1 g water by 10C (14.50C to 15.50C) 1 Calorie = 4.2 J Principle of Calorimetry Law of mixtures Heat lost by hot body = heat gained by cod body [when no heat is lost to the surroundings] Specific heat capacity (C) It is the amount of heat energy required to rise the temperature of unit mass (1 g on 1 kg) of any substance by 10C or 1 K for a given mass m heat required is H = m H is called heat capacity or thermal capacity for a given mass m and t rise in temperature Amount of heat required

dQ = mcdt t2

Q   mcdt

Q mc t

t

Unit of specific heat capacity

C

Q Joule  m t kg K

C

Q Calorie  m t g  0C

SI unit : J Kg–1K–1

CGS unit : Calg–1 0C–1

for water C = 4200 J Kg–1K–1

C = 1 Calg–1 0C–1

for ice C = 2100 J Kg–1K–1

C = 0.5 Calg–1 0C–1

Latent Heat It is the amount of heat energy exclusively utilized for phase transition at certain fixed temperatures like melting point or boiling point etc 9

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Latent heat of fusion (Lf) For solid  Liquid Transition @ melting point L Eg : 1 g ice @ 00 C   1 g water @ 00C f

Lf of ice = 80 cal/g For a given mass m Q = mLf Latent heat of Vapourisation (Lv) For liquid  Gas transition @ boiling point L Eg : 1 g water @ 1000C  1 g steam @ 1000C V

LV of water = 540 Cal/g For a given mass m Q = mLV Unit of Latent Heat Capacity Q = mL

S.I. Unit : Joule/kilogram

L = Q/m

CGS Unit : Calorie/gram

Water Equivalent It is the amount of water which has got the same heat capacity as that of a given substance H = mC

mC m C 

for water

H  m C

m 

mC C

Water equivalent is numerically equal to heat capacity of the substance in CGS system Temperature - Time Graph

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Graph plotted with time along x axis and temperature along y axis Phase Change

dQ  m L

p

dQ m L  dt dt

Increase in temperature

dQ  m cdT p

dQ m cdT  dt dt

P  L    dt m 

dT  P  1   dt  m  C

L = a constant × dt

dT 1  a constant  dt C

L  dt

L  time interval

Slope 

L  change in x value

C

1 C

1 Slo pe

P-T diagram A graph between the temperature T and pressure P of the substance is called phase diagram or P-T diagram. Phase Diagram

Note : for water melting point decreases with increase in pressure 11

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Heating Curve

12

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Regelation Melting of ice at lower temperature due to increase in pressure and refreezing when pressure is with drawn is regelation Heat Transfer Transport of heat energy from one point to another can be done in 3 days Conduction Convection Radiation Conduction Particle to particle heat transfer without actual transport of matter. Its common to solids and mercury Gravity has no effect in conduction Metals are good conductors of heat Heat Current (H) Rate of flow of heat energy through a conductor

H

Q t

unit : Watt

Thermal Conductivity The ability of a conductor to allow the passage of heat energy through it Expression for Thermal conductivity

Consider a metallic rod with length  , uniform cross sectional area A. Its both ends are maintained at two different temperature T1 and T2 (T1 > T2). Steady State is a condition in which heat current through every cross section of the conduction become a constant. There is no further absorption of heat energy by molecules. Only effective method of heat transfer is conduction is steady state. At Steady State Rate of heat flow

H

Q  area of cross section t

Q  temperature gradient t 13

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Q Q T  T2 A  1 t t  Temperature gradient  

Q (T  T2 ) KA 1  t

Temp diff length T1  T2 

Q K A T   t

Unit of thermal conductivity

H

K

K A T  H Watt  m  2 A T m  Kelvin

Unit of K  Watt m–1K–1 Thermal Resistance (RT) It is the ability to oppose the flow of heat energy through a conduction

Heat Current 

H

T RT

We have H 

H

Temperature difference Thermal Resistance

(1)

K A T 

T [  / KA]

(2)

Thermal Resistance RT   / KA Unit of Thermal Resistance

H

T RT

RT 

T Kelvin  H Watt 14

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Combination of Thermal Conductors Series

Consider two rods of length  1 and 2 connected in series

RT 

 KA

RT1 

1 K 1A

RT2 

2 K2 A

RT '

RT  RT1  RT2 '

1  2    1  2 K S A K1A K 2A KS 

1

1   2 2 K1  K 2

If 1   2 Parallel

1  2    1 2 K SA K1 K 2

KS

KS 

K1K2 [ 1  2 ]  1K 2   2K 1

2K 1K 2 K1  K2

Consider two rods connected in parallel

RT   KA

RT1 

 K1 A1

RT2 

 K 2A 2

R'T 

1 1 1   R 'T R T1 R T2

     K P[A 1  A 2] K 1A 1 K 1A 2 15

 K P [A 1  A 2]

1  2 K s' A

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KP 

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K1 A1  K2 A2 A1  A 2

if A1 = A2 = A KP 

K1  K 2 2

Convection It is the transfer of heat energy along with actual transport of matter. Gravity plays an important role in natural convection. Natural Convection

Forced Convection

With the help of gravity.

With some mechanical support

In upward direction only

Possible in all directions

Eg : Sea breeze

Eg : Human circulatory system

Land breeze

Automobile cooling system

Trade winds

House hold heating system

Boiling water

furnaces

Growth of ice ponds

Ice starts forming in a pond at sub zero temperature 0 C in winter season. To find time taken for growth of ice upto a thickness y, we will consider a small thickness dy The heat energy released when ice of thickness dy is formed is dQ. Heat energy is transferred to the environment

dQ 

KA [0   ] dt y

For melting dQ  mL   A  dy  L

KA  dt   A dy L y 16

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dt 

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L y dy K

for total time taken to grow from o to y y

t  0

L y dy K y

t

L y dy K  0 y

L  y 2  t   K   2 0 t

  density of ice

1 L 2 y 2 K

L  Latent heat of fusion   Atmospheric temperature K  Thermalconductivity of ice y  Thickness of ice

Time taken by ice to grow up to thickness y, 2y and 3y from O is given by t1 : t2 ; t 3  1 : 4 : 9

t y 2

t 1, t 2, t 3 are time taken by

0 y

ice to grow upto y,2y and

2 0  2y t 2 (2y)

3y respectively

0  3y t 3 (3y) 2

t 1 y

2

t 1  y2 2 t 2  4y

t 3  9y 2

Time taken by ice ice to grow from o to y y to 2y and 2y to 3y is 0y

t1  t 1  0  y 2  0  y 2

y  2y

t 2  t 2  t 1  4y 2  y 2  3y 2

2y  3y

t 3  t 3  t 2  9y 2  4y 2  5y 2

t 1: t 2 : t 3 1 : 3 :5

Radiation Radiation is the fastest mode of heat transfer 17

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Properties of heat radiations It travels in straight lines It is universal, invisible All bodies above zero kelvin will emit radiations It belongs to IR Region It shows reflection, refraction, interference, diffraction and polarisation They are em waves with speed 3 × 108 m/s Let Q be total incident energy, then part of it may absorbed. Some part will be reflected and the rest is transmitted

Q A  R  T  ing by Q

A  absorptance Q R  reflectance r Q I t = = transmittance Q

a

Q A R T    Q Q Q Q a  r  t 1

Absorptive Power (absorptance) [a]

a

...