Heat for iit jee by S.Dharmaraj

PHYSICS:HEATand THERMOYNAMICS

HEAT In summer, we instal fans, air-conditioners in our homes to keep ourselves comfortable. Our bodies are highly sensitive to our surrounding climate. So, we wear clothes depending upon seasons. Why ? Just think about these :: 1. 2. 3. 4. 5. 6. 7.

What is energy ? Where does it come from ? Why do we need this thing we call energy ? What happens to energy after it is used ? What is heat ? Where does it come from ? What is hot and cold ? How does heating and cooling work ? How do we measure heat and energy ?

To answer these questions let us first understand what the matter is and what it is made up of.

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At the atomic level, let us understand what an atom is made up of :

What is energy ? Where does it come from ? A quark is an elementary particle and a fundamental constituent of matter. As far as we know, quarks are indivisible; i.e., quarks are the smallest unit matter in the nucleus. Our current understanding is that the quark is a point-like particle with no spatial extent! Quarks have various intrinsic properties, including electric charge, color charge, spin, and mass. Quarks are the only elementary particles in the Standard Model of particle physics to experience all four fundamental interactions, also known as fundamental forces (electromagnetism, gravitation, strong interaction, and weak interaction), as well as the only known particles whose electric charges are not integer multiples of the elementary charge. Quarks combine to form composite particles called hadrons, the most stable of which are protons and neutrons, the components of atomic nuclei. There are six types of quarks, known as flavors: up, down, charm, strange, top, and bottom. Up and down quarks have the lowest masses of all quarks. The heavier quarks rapidly change into up and down quarks through a process of particle decay: the transformation from a higher mass state to a lower mass state. Because of this, up and down quarks are generally stable and the most common in the universe, whereas charm, strange, top, and bottom quarks can only be produced in high energy collisions (such as those involving cosmic rays and in particle accelerators).

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Hence these quarks are the carriers of an intrinsic mechanical property known as ‘ENERGY’ As quarks are the constituents of the nucleons, nucleons do possess this. So, an atom has an inherent energy stored in it. Molecules are made up of atoms and substances are made from molecules. Hence all substances in the universe have this inherent ‘Internal Energy’. This is called Potential Energy. Now atoms do contain electrons moving around the nucleus in some specified paths called Orbits. Hence the energy possessed by these electrons in motion is called Kinetic Energy. There are three types of kinetic energy. (i) (ii) (iii)

If the KE is associated with the movement of a particle from one location to another location, this is translational KE. If the KE possessed by a particle that oscillates or vibrates about a fixed point, it is called vibrational KE. If the KE possessed by a particle that rotates about an imaginary axis of rotation, it is called rotaional KE.

Thermal energy is the total energy of the translational, vibrational and rotational energies of all the individual molecules of which the object is made up of.

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What is heat ? Where does it come from ? As studied in the earlier topic, every substance has a certain internal energy. If some of that internal energy is transferred to the random motions of the particles in another object, then you can say that a transfer has heat has occured. Heat is a measure of how many atoms there are in a substance multiplied by how much energy each atom possess. When we say that a body is heated, then its (i) (ii) (iii) (iv)

thermal energy increases molecules begin to move violently, hence, thermal vibrational energy increases rate of collision between neighbouring molecules increases inter-atomic seperation increases, thus thermal expansion takes place.

Hence this measurable manifestation of the thermal energy due to the average K.E of all the particles within a sample of matter is called TEMPERATURE. It is related to how fast the atoms within a substance are moving. It is the degree of hotness ( atoms moving faster ) and coldness (atoms moving slower ) of a body. Definition of Heat :

Heat is energy in transit which is transferred from one body to the other, due to difference in temperature, without any mechanical work involved.

HEAT It is a form of thermal energy and measures the total energy of all the molecules in an object. SI Unit : joule CGS Unit : calorie Flow of heat doesnot depend

TEMPERATURE It is a measure of average KE of molecules in an object. It is the degree of hotness or coldness. Unit : degree Centigrade / Fahrenheit / Kelvin

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THERMAL EXPANSION

SOLID :

Expansion due to increase in temperature is called Thermal Expansion. At any temperature the atoms of solid are vibrating . Figure represents potential energy for two adjecent atoms in a solid as a function of their internuclear separation r. When temperature is increased vibrational energy of atoms increases. Because the curve is asymmetric, the average separation (r 1, r2) increases. U r T1 > T2 T1 r0

Hence thermal expansion is direct consequence of deviation from symmetry of the characteristic potential energy curve of solid.

2.(A)

TYPE

THERMAL

EXPANSION:

For temperature change ∆T (∆T is small) change in length ∆l = l − l 0 = l 0α∆T (1) Linear (where l0 is initial length of the rod, α is Co-efficient of linear expansion) (2) Superficial change in Area ∆A = A − A0 = A0β∆T 1  ∆A  β = lim  ÷ ∆T →0 A (where A0 is initial area of the rod, β is 0  ∆T  Co-efficient of superficial expansion) (3) Volume change in volume ∆V = V − V0 = V0 γ∆T 1  ∆V  γ = lim  ÷ ∆T→0 V (where V0 is initial volume of the rod, γ is 0  ∆T  Co-efficient of volume expansion) (B) For isotropic solid, α1 = α 2 = α3 = α (say). So β = 2α and γ = 3α Coefficient of expansion 1  ∆l  α = lim  ÷ ∆T →0 l 0  ∆T 

(C) For anisotropic solids, β = α1 + α 2 and γ = α1 + α 2 + α3 Here α1 , α2 and α3 are coefficient of linear expansion in X, Y and Z directions respectively. (D) For Variable α : (i)

variation of α with distance Let α = ax + b l

Total expansion = ∫ (expansion of length dx) = ∫ ( ax + b) dx∆T 0

(ii)

variation of α with temperature T2

∆l = ∫ l0 f (T )dT

Let α = f (T )

Caution : If α is in ºC then put T1 and T2 in ºC Similarly if α is in K, put T1 and T2 in K.

VARIATION T0 = 2π

TIME PERIOD

PENDULUM CLOCKS:

l0 g

If temperature is increased by ∆t , 2 nd

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T = 2π

l0 (1 + α∆t ) g

T = 2π

l0 α (1 + ∆t ) (by using Binomial expension) g 2

T = T0 (1 +

α ∆t ) 2

∆T 1 = (α∆t ) T0 2

VARIATION

⇒

T − T0 = T0

α ∆t 2

∆T = increase in time period

DENSITY:

With increase in temperature, volume increases, so density decreases and vice-versa. d0 d= (1 + γ∆T) For solid, values of γ are generally small so we can write d = d 0 (1 − γ∆T) (using binomial expansion)

THERMAL STRESS:

A rod of length l0 clamped between two fixed walls For ∆T change in temperature ∆l F stress = (area assumed to be constant) , strain = ; ∆l = l0α∆T l0 A F / A Fl0 F Y= = = so, F F ∆l / l0 A∆l Aα∆T lo F = YAα | ∆T | or, F = Y α | ∆T | or, A 1 Energy stored in rod E = × stress × strain × volume 2 For increase in temperature stress will be compressive and for decrease in temperatue stress will be tensile 1. Illustration : When the temperature of a body increases from T to T + ∆T, its moment of inertia increases from I to I + ∆I. The coefficient of linear expansion of the body is α. Find the ratio of

∆I I

I = Σmr2 I + ∆I = Σm[r(1 + α ∆T)]2 or I + ∆I = Σ[mr2(1 + 2α ∆T)] = I(1 + 2α ∆T) or ∆I/I = 2α ∆T. 2. Illustration : A pendulum clock with a pendulum made of Invar (α = 0.7 × 10−6 / C º ) has a period of 0.5 s and is accurate at 25º C. If the clock is used in a country where the temperature averages 35º C, what correction is necessary at the end of a month (30 days) to the time given by the clock? Solution :

In time interval t , the clock will become slow (or will lose time) by 1 ∆t = αt ∆θ 2 1 ∆t = × (7 × 10−7 ) × (30 × 86400) × (35 − 25) = 9.1s . So, 2 3.Illustration: Two metal rods of the same length and area of cross-section, are fixed end to end between rigid supports. The materials of the rods have Young modulii Y 1 and Y2, and coefficients of Solution:

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linear expansion α1 and α2. For what condition the junction between the rods does not shift if the rods are cooled. Solution : Tension must be the same in both the rod for their junction to be in equilibrium. Y1Aα1T = Y2Aα2T 4. Illustration : A copper and a tungsten plate having a thickness δ = 2 mm each are riveted together so that at 0ºC they form a flat bimetallic plate. Find the average radius of curvature of this plate at T = 200°C. The coefficients of linear expansion for copper and tungsten are α c = 1.7 × 10−5 K −1 and α t = 0.4 × 10 −5 K −1 . Solution : L0 (1 + α1T) = (R + δ/2) φ ….. (1) L0 (1 + α2T) = (R − δ/2) φ ….. (2)

From (1) and (2), we get δ RT (α1 − α2) = δ + T (α1 + α2) 2  α1 and α 2 are very small

RT( α1 − α 2 ) ~ −δ δ R= = 0.769 m . (α1 − α 2 )T

6.(A)

THERMAL EXPANSION

LIQUID :

Liquids do not have fixed shape and take the shape of container in which it is kept. Therefore, liquids undergo volume expansion only. For liquid two types of co-efficient of volume expansion is defined (I)

Co-efficient of apparent expansion ( γ a ) : γa =

Apparent change in volume (∆V) a = Initial volume × rise in temp V0 ∆T

In apparent expansion change in volume of container is not considered . (II)

Co-efficient of real exparsion ( γ r ) : γr =

( ∆V) r Re al change in volume = Initial volume × rise in temp V0 ∆T

For the determination of real change in volume of liquid expansion of container is also taken into account. V0

Suppose at temp T1 volume of liquid is V0 (Initial volume) At temp T2 volume of liquid w.r.t. container = V ∆T = T2 − T1 ∴ V = V0 [1 + γ a ∆T] → (i) If Vr is real volume of liquid at temp T2

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Vr = V0 [1 + γ r ∆T] → (ii) Vr = V[1 + γ c ∆T] → (iii) Where γ c is co-efficient of volume expansion of container. (i) & (iii) in (iii) ⇒ V0 [1 + γ r ∆T] = V0 [1 + γ a ∆T][1 + γ c ∆T] [1 + Vr ∆T] = 1 + Va ∆T + rc ∆T neglecting γ a γ c ∆T 2 ∴ γr = γa + γc

6.(B)

EXPANSION IN

ENCLOSED VOLUME:

Increase in height of liquid level in tube when bulb was initially completely filled :

apparent change in volume of liquid area of tube V0 ( γ L − γ g ) ∆T h= ; A0 (1 + 2α g ∆T) h=

γ apparent = γ L − γ g

Initially Finally Note: (i)

γ for liquids are in order of 10−3 .

(ii)

For water, density increases from 0 to 4ºC so 4ºC density is maximum.

VOLUME COEFFICIENT

AND

γ is –ve (0 to 4ºC) and for 4ºC to higher temperature γ is +ve. At

PRESSURE

COEFFICIENT OF

GASES:

Volume Coefficient : Consider a given mass of a gas is heated through 1º C at constant pressure. Now the ratio of the increase in volume to the original volume at 0º C is defined as volume coefficient. If VT and V 0 be the volumes of a given mass of gas T ºC and 0º C respectively, then αV =

VT −V 0 V 0 ×T

where αv is known as volume coefficient. Pressure coefficient : Consider a given mass of a gas is heated through 1º C at constant volume Now the ratio of the increase in pressure to the original pressure at 0º C is defined as the pressure coefficient. If PT and P0 be the pressures of a given mass of a gas at T ºC and 0ºC respectively, then α p =

PT − P0 . where α p is known as pressure P0 ×T

coefficient. 5. Illustration : A sphere of diameter 7cm and mass 266.5 gm floats in a bath of liquid. As the temperature is raised, the sphere just begins to sink at a temperature of 35ºC. If the density of the liquid at 0ºC is 1.527 gm / cm3 , find the co-efficient of cubical expansion of the liquid. Neglect the expansion of the sphere. Solution :

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The density of sphere, ρs =

ρ35 =

266.5 4  22   7  × ×  3  7  2

( density of sphere is constant)

ρ35 = 1.483 gm / cm3 ρ0 = ρ35 [1 + γ∆T ] Now, 1.527 = 1.483[1 + γ × 35] 1.029 = 1 + γ × 35 1.029 − 1 γ= = 0.00083/ °C . 35 6. Illustration : A one litre glass flask contains some mercury. It is found that at different temperature the volume of air inside the flask remains the same. What is the volume of mercury in this flask if coefficient of linear expansion of glass is 9 × 10−6 / °C while volume expansion of mercury is 1.8 × 10 −4 / C ° ? If V is the volume of flask, VL of mercury and VA of air in it, V = VL + VA Now as with change in temperature volume of air remains constant, the expansion of mercury will be equal to that of the whole flask i.e., ∆V = ∆VL V γ G ∆θ = VL γ L ∆θ or [as ∆V = V γ∆θ ]

Solution:

γ G = 3α G = 27 × 10−6 / °C

Here

V = 1 litre = 1000 cc

VL = (1000 × 27 × 10 −6 /1.8 × 10 −4 ) = 150 cc.

and

8.(A) Calorie: The amount of heat needed to increase the temperature of 1 gm of water from 14.5ºC to 15.5ºC at STP is known as 1 calorie 8.(B) Specific Heat: It is heat required to raise temperature by 1º C or 1 K for unit mass of the body. dQ = mc dT T2

Q = m ∫ cdT (be careful about unit of temperature, use units according to the given units of c) T1

8.(C) Molar Heat Capacity: If instead of unit mass we consider one mole of a substance, the heat required to change the temperature of one mole of a substance through 1 ºC (or K) is called molar heat capacity or molar specific heat and is represented by C. If the molecular weight of a substance is M : Q  Q m C = Mc = as c = m ∆T and n = M  n ∆T   Its SI units are (J/mol K) 8.(D) Thermal-capacity : If instead of unit mass we consider the whole body, (of mass m), the heat required to raise the temperature of a given body by 1 ºC is called its thermal capacity , i.e., Thermal capacity = mc = nC = (Q / ∆T ) Thermal capacity of a body depends on the mass and nature of body. It has units (J/K) or cal/ºC and dimensions [ ML2T −2 K −1 ] . 8.(E) Water-Equivalent :

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If thermal capacity of a body is expressed in terms of mass of water it is called water-equivalent of the body, i.e., water-equivalent of a body is the mass of water which when given same amount of heat as to the body, changes the temperature of water through same range as that of the body, i.e., W = (m × c) gram [where : W = mass of water ; m = mass of substance ] The unit of water equivalent W is g while its dimension [M]. 9. Latent Heat: The amount of heat required to change the phase of 1 gm of a substance to another phase at melting or boiling point. Q = mL L = latent heat of substance in cal/gm or in Kcal/kg. Lice = 80 cal / gm – Specific latent heat of fusion of ice Lsteam = 540 cal/gm – Specific latent heat of vaporization of water 10.(A) Principle of Calorimetry : When two bodies (one being solid and other liquid or both being liquid) at different temperature 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 at lower temperature absorbs it, so that : Heat lost = Heat gained, i.e. principle of calorimetry represents the law of conservation of heat energy. 10.(B) DETERMINATION OF SPECIFIC HEAT OF LIQUID USING A CALORIMETER: The principle of calorimetry states that the total heat given by the hot objects equals the total heat received by the cold objects. Objects at different temperatures are made to come in contact with each other in the calorimeter. As a result heat is exchanged between the objects as well as with the calorimeter. Neglecting any heat exchange with the surroundings, The method is similar to that for determining the specific heat of the solid. In this case, the calorimeter is half filled with a liquid of unknown specific heat and a solid of known specific heat, which is insoluble in the liquid is steam heated and then put inside the calorimeter. Let the mass of the solid = M Mass of the calorimeter and stirrer = m1 Mass of the liquid = m2 Specific heat capacity of the solid = s1 Specific heat capacity of the material of calorimeter and stirrer = s 2 Specific heat capacity of water = s Initial temperature of the solid = θ1 Initial temperature of the calorimeter, stirrer and water = θ2 Final temperature of the mixture = θ From the principle of calorimetry Ms1 ( θ1 − θ ) = m1s2 ( θ − θ2 ) + m2 s ( θ − θ2 )

Heat gained = Heat lost s =

Ms1 ( θ1 − θ ) m2 ( θ − θ 2 )

−

m1 s2 m2

7. Illuatration : The temperature of equal massesof three different liquids A, B and C are 12ºC, 19º C and 28º C respectively. The temperature when A and B are mixed is 16º C and when B and C are mixed is 23º C. What would be the temperature when A and C are mixed? Solution:

In accordance with principle of calorimetry : When A and B are mixed mcA (16 − 12) = mcB (19 − 16) ⇒ c A = (3 / 4)cB VYAHRUTI EDUCATIONALINSTITUTIONS: 2 nd Floor, Samsung Plaza, Eluru Rd, Near Benz Circle, VIJAYAWADA

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and when B and C are mixed mcB (23 − 19) = mcC (28 − 23) ⇒ cC = (4 / 5)cB Now when A and C are mixed if T is the common temperature of mixture: mc A (T − 12) = mcC (28 − T ) Substituting c A and cC from above, (3/ 4)(T − 12) = (4 / 5)(28 − T ) which on solving gives, T = 20.25º C . 8. Illustration : A solid material is supplied heat at a constant rate. The temperature of the material is changing with the heat input as shown in figure. Study the graph carefully and answer the following questions : What do the horizontal regions AB and CD represent? If CD = 2 BA , what do you infer? What does slope DE represent? The slope of OA > the slope of BC . What does this indicate?

Y E T e m p e ra tu re

(i) (ii) (iii) (iv)

Solution :

C D

A B

H e a t in p u t

The horizontal portions AB and CD of the graph represent the change of phase. The portion AB represents the change of phase from solid to liquid at constant temperature and the portion CD represents the change of phase from liquid to vapour at constant temperature or the portion CD represents the latent heat of vaporization. CD = 2 AB , (ii) i.e., latent heat of vaporization is twice the latent heat of fusion. (iii) The slope DE is equal to dT / dQ for vapour, i.e., this gives the rate of increase of temperature of vapour with heat input. (i)

1 Specific heat of the vapour 1 or, Specific heat of vapour µ Slope of DE (iv) Slope OA > slope BC The slope OA is related with specific heat as 1 Specific heat of solid µ Slope of OA Now slope OA > slope BC , represents that specific heat of the liquid is more than that of the ∴ Slope of DE µ

solid.

11. HEAT–TRANSFER: (A)

Conduction:

Heat energy is transferred (usually through solids) from one part of the material medium to other without transferring the material particles. (i)

Steady State : In this state heat absorption stops and temperature gradient throughout the rod becomes dT = constant constant i.e. dx

(ii)

Before steady State : Temparture of rod at any point changes 11.(B) LAW FOR THERMAL CONDUCTION IN

2 nd

STEADY STATE:

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PHYSICS:HEATandTHERMOYNAMICS L In steady state heat passing through a bar of length L and cross-section Q T2 T1 A A in time t when its ends are at temperatures T1 and T2 (< T1 ) is given x = L x = 0 by: (T − T ) Q = KA 1 2 t … (i) L So rate of flow of heat will be dQ dT = − KA … (ii) dt dx The quantity (dT / dx) is called temperature gradient (minus sign indicates that with increase in x , temperature θ decreases) and the constant K depends on the nature of metal and is called coefficient

of thermal conductivity or simply thermal conductivity and is a measure of the ability of a substance to conduct heat through it. The dimensions of coefficient of thermal conductivity are [ MLT −3 K −1 ] while its SI units are W/mK. Let the two ends of rod of length l is maintained at temperature T1 and T2 dQ T1 − T2 = Thermal current dt RTh l Where thermal resistance RTh = KA R (a) Two rods joined A dQ T1 − T2 = T1 dt R1 + R2 (b)

Three rods joined to a common point T − 100 T − 20 T − 5 + + =0 l1 l2 l3 K1 A K2 A K3 A

R B

( T 1> T 2) T

2 0 ºC l2

K 1 0 0 ºC

l3 5 ºC

11.(C) STATE:

SERIES

AND

PARALLEL CONNECTION

RODS

STEADY

R 2, l B K2

Series Connection K1 = thermal conductivity of A K 2 = thermal conductivity of B T1 > T2 length l and cross section area A of both rods are same l l R1 = R2 = K1 A K2 A

R 1, l A K1

∆Q T1 − T T − T2 T1 − T2 = = = ∆t R1 R2 R1 + R2 (T − T2 ) R1 = (T1 − T ) R2 TR1 − T2 R1 = T1 R2 − TR2 T ( R1 + R2 ) = T1 R2 + T2 R1 T R +T R T= 1 2 2 1 R1 + R2

Thermal current i =

⇒ ⇒

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T1 − T2 T1 − T2 = R R1 + R2 ⇒ R = R1 + R2 Two rods together is equivalent to a single rod of thermal resistance R1 + R2

⇒

Parallel Connection ∆Q1 T1 − T2 ∆Q2 T1 − T2 i1 = = = ; i2 = ∆t R1 ∆t R2

R T

 1 T −T 1  i = i1 + i2 ; 1 2 = (T1 − T2 )  + ÷ R  R1 R2 

The system of the two rods is equivalent to a single rod of thermal resistance R is given by

11.(D)

B C

1 1 1 = + R R1 R2

CONDUCTION BEFORE STEADY STATE:

Differential form : dQ dT = KA dt dx

dT = temperature gradient dx

T - dT

CONDUCTION

IN A

SECTION dQ

MEDIUM BEFORE STEADY STATE: dQ

(T 1 > T 2 )

dQ = dQ1 − dQ2 (In steady state dQ = 0 , i.e. dQ1 = dQ2 ) dT dT ∴ mS dT = KA 1 dt − KA 2 dt dx dx dT KA  dT1 dT2  = −   dt mS  dx dx  dT = increase in temperature of the section in time dt.

11.(E)

Growth Of Ice:

Consider a layer of ice of thickness x. The air temperature is − θº C and water temperature below the ice is 0º C. Considering unit cross-section area of ice, if a layer of thickness dx grows in time dt .

Temperature of air −θº C

dx Then heat given by this layer = 1 × dx ×ρ × L × = mass latent heat T e m p e ra tu re o f w a te r 0 ºC ρ = density of ice L = latent heat of fusion of ice. If this quantity of heat is conducted upwards through the ice layer of area A in time dt . { 0 − (−θ)} ∴ Adx ×ρ ×L = K dt A x x ρL 2 ρL x ×dx = ( x22 − x12 ) time taken t = ∫ K θ x1 2 K ×θ  dx K ×θ  Rate of increase of thickness of the ice layer  = ÷.  dt ρLx 

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9. Illustration : Three cylindrical rods A, B and C of equal lengths and equal diameters are joined in series as shown in figure. Their thermal conductivities are 2 K , K and 0.5K respectively. In steady state, if the free ends of rods A and C are at 100º C and 0º C respectively, calculate the temperature at the two junction points. Assume negligible loss through the curved surface. What will be the equivalent thermal conductivity? θBC

θAB

L A

1 0 0 ºC

Solution:

0 ºC

0 .5 K

As the rods are in series, Req = RA + RB + RC with R = ( L / KA) L L L 7L Req = + + = i.e., … (i) 2 KA KA 0.5 KA 2 KA dQ ∆θ (100 − 0) 200 KA H= = = = And hence, dt R (7 L / 2 KA) 7L Now in series, rate of flow of heat remains same, i.e., H = H A = H B = H C .  dQ   dQ  So for rod A ,   =    dt  A  dt  or,

i.e.,

(100 − θ AB )2 KA 200 KA = L 7L

θ AB = 100 − (100 / 7) = (600 / 7) = 85.7º C

(θ BC − 0) × 0.5KA 200 KA = L 7L

 dQ   dQ  And for rod C ,   =   i.e.,  dt  C  dt  θBC = (400 / 7) = 57.1º C or,

Furthermore if K eq is equivalent thermal conducitivity, L + L + L 7L Req = = [from equation (i)] K eq A 2 KA i.e.,

K eq = (6 / 7) K .

10. Illustration : A 2m long wire of resistance 4 ohm and diameter 0.64 mm is coated with plastic insulation of thickness 0.06 mm. When a current of 5 ampere flows through the wire, find the temperature difference across the insulation in steady state if [ K = 0.16 × 10−2 cal / cm − °C s] Solution :

θ2 dr r

Considering a concentric cylindrical shell of radius r and thickness dr as shown in figure. The radial rate of flow of heat through this shell in steady state will be dQ dθ H= = − KA Negative sign is used as with increase in r, θ decreases dt dr Now as for cylindrical shell A = 2πr L θ b dr −2πLK 2 dθ = − dθ H = −2πrLK or ∫a r H θ∫1 dr which on integration and simplification gives dQ 2πLK (θ1 − θ2 ) H= = …(i) dt ln b a I 2 R (5) 2 × 4 cal L = 2m = 200 cm Here, H = = = 24 4.2 4.2 s r1 = (0.64 / 2) mm= 0.032 cm and R2 = r1 + d = 0.032 + 0.006 = 0.038 cm

( )

θ1

(θ1 − θ2 ) =

(

24 × ln 38

)

24 × 2.3026[log10 38 − log10 32] 32 o = − 2 C % 3.14 × 0.64 2 × 3.14 × 200 × .16 × 10 −2

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(θ1 − θ 2 ) =

Solution:

55 × [1.57 − 1.50] = 2 ºC. 2

11. Illustration : Consider a lake that is getting frozen at an atmospheric temperature of -10 ° C. Assuming that most of the heat that is lost comes from the latent heat of fusion released when the water freezes. Find the rate at which the thickness of ice increases as a function of time. Take the conductivity of ice as K and the density of ice ~ − density of water = ρ The water just beneath the ice is almost at 0°C. Assume that the thickness of ice at time t is x ( t ) , that the area of the lake is A0 and that the density of ice is ρ . If the latent heat of ice is L, then

10KA0 dQ LA0 dxρ KA0 = = 0 − ( −10 )  = dt dt x x or,

dx 10 K = dt xLρ

or,

∫

x 2 10 K = t + constant 2 Lρ

xdx =

10K Lρ

∫

At t = 0 , we assume that x = 0 : i.e. initially the lake is not frozen. 2 Therefore, x =

where 12.

20K 20 K t or x ( t ) = t = C t, Lρ Lρ

20K is a constant. Lρ

Convection:

Heat energy is transferred (usually through liquids and gases) by mass movement of molecules from one point to another. (Due to gravity & buoyant force). 13.

Radiation:

Heat energy is transferred by electromagnetic waves even in absence of medium. Absorptive Power a Absorptive power of a body is defined as the fraction of the incident radiation that is absorbed by the body. Energy absorbed Absorptive power a = Energyincident Emissive Power ‘e’ The emissive power denotes the energy radiated per unit area per unit time per unit solid angle along the normal to the area. Emissivity ε Emissivity of a surface is the ratio of the emissive power of the surface to the emissive power of black body at the same temperature.

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Emissivity,

ε=

(Emissive power of the surface) e = (Emissive power of black body at the same temperature) E

Black body A perfectly black body is one which absorbs completely all the radiation, of whatever wave-length, incident on it. (a = 1) KIRCHHOFF’S LAW It states that the ratio of the emissive power to the absorptive power for radiation of a given wave length is the same for all bodies at the same temperature, and is equal to the emissive power of a perfectly black body at that e temperature. = E a 14. STEFAN’S LAW OF RADIATION : The total radiant energy emitted E per unit time by a black body of surface area A is proportional to the fourth power of its absolute temperature and surface area A of the black body E ∝ AT 4 or, ( σ = Stefan’s constant = 5.67 × 10−8 kg sec−2 k−4) E = σAT 4 For a body which is not a black body where ε is emissivity of the body E = εσAT 4 Using Krichoff’s law E( body ) =a E(black body ) εσAT 4 ε=a or, =a σAT 4 Emissivity and absorptive power have the same value. So, good absorbers are good radiators and bad absorbers are bad radiators. or,

Net Loss Of Thermal Energy: If a body of surface area A is kept at absolute temperature T in a surrounding of temperature T0 (T0 < T ) . Then energy emitted by the body per unit time E = εσAT 4 And energy absorbed per unit time by the body E0 = εσAT04 Net, loss of thermal energy per unit time. ∆E = E − E0 = εσA(T 4 − T04 ) . 15.

NEWTON’S LAW OF COOLING:

For a small temperature difference between a body and its surrounding, the rate of cooling of the body is directly proportional to the temperature difference. If a body of surface area A is kept at absolute temperature T in a surrounding of temperature T0 (T0 < T ) . Then net loss of thermal energy per unit time. dQ = εσA(T 4 − T04 ) dt If the temperature difference is small ∴ T = T0 + ∆T

  ∆T  4  4 4 = εσ A = εσA{(T0 + ∆T ) − T } T0  1 + ÷ − T0  T0     4

4 0

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 ∆T ∆T  = εσAT04 1 + 4 + higher powers of − 1 T0 T0  

= 4εσAT03 ∆T

… (i)

Now, rate of loss of heat at temperature T dQ dT = − ms … (ii) dt dt From equation (1) and (2), we get dT dT −4εσAT03 ∴ ms = −4εσAT03 (T − T0 ) ; = (T − T0 ) dt dt ms dT = − k (T − T0 ) ... (iii) where k = 4εσAT03 dt dT µ (T − T0 ) . i.e., dt dT = −Kdt From equation (iii), T − T0 t dT T − T0 = − ∫ Kdt ; = e −Kt ⇒ T = T0 + ( T1 − T0 ) e −Kt T − T T − T T1 0 0 1 0 T

By integrating this equation ∫

16.

Wien’s black body radiation :

At every temperature (>0K) a body radiates energy in the form of electromagnetic wave of all wavelengths. T3 According to Wien’s displacement law if the wavelength corresponding I T1 < T2 < T3 to maximum energy is λ m T2 Then λ mT = b where b is a constant (Wien’s constant) T1 T = temperature of body Intensity at a specific temp. T I m αT 5 This is wien’s fifth power law. λm λm λm λ 3

12. Illustration:One end of a rod of length 20 cm is inserted in a furnace at 800 K. the sides of the rod are covered with an insulating material and the other end emits radiation like a blackbody. The temperature of this end is 750 K in the steady state. The temperature of the surrounding air is 300 K. Assuming radiation to be the only important mode of energy transfer between the surrounding and the open end of the rod, find the thermal conductivity of the rod. Stefan constant σ = 6.0 × 10−8 W / m 2 − K 4 Solution :

Quantity of heat flowing through the rod in steady state dQ K . A.d θ = …(i) dt x Quantity of heat radiated from the end of the rod in steady state dQ = Aσ T 4 − T04 …(ii) dt From (i) and (ii) F u rn a c e K .d θ 800K = σ T 4 − T04 x K × 50 = 6.0 × 10−8 [(7.5) 4 − (3)4 ] × 108 0.2 Or K = 74 W/m – K.

(

2 nd

)

750K

20cm

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13. Illustration: A body initially at 80°C cools to 64°C in 5 minutes and to 52°C in 10 minutes. What will be the temperature of the body after 15 minutes and what is the temperature of the surroundings ? Solution : If T0 is the temperature of the surroundings, then We have  80 − T0   = 5K loge  ….. (1)  64 − T0   64 − T0   = 5K loge  ….. (2)  52 − T0  Equating Equations (1) and (2), we get 80 − T0 64 − T0 = 64 − T0 52 − T0 which gives T0 = 16°C If T temperature after 15 minutes, then  52 − T0   52 − 16   = 5K or loge  loge   = 5K  T − 16   T − T0  and

….. (3) From equations (1) and (3), we get 80 − 16 52 − 16 = 64 − 16 T − 16 which gives T = 43°C 14. Illustration : The intensity of solar radiation, just outside the earth’s atmosphere, is measured to be 1.4 kW/m 2 . If the radius of the sun 7 × 108 m, while the earth-sun distance is 150 × 106 km , then find (i) the intensity of solar radiation at the surface of the sun, (ii) the temperature at the surface of the sun assuming it to be a black body, (iii) the most probable wavelength in solar radiation, Solution:

(i)

The radiation emitted from the solar surface per unit time is spread over the surface of a sphere having a radius equal to earth-sun distance where it is received on the earth (just outside the atmosphere)

∴

W × 4πRS2 = I 0 × 4πDSe2 where DSe is the distance between the sun and the earth, and I 0 is the intensity outside the earth’s atmosphere. 2

 R  I0 = W ×  S ÷  DSe 

 7 × 108  49 ∴ 1.4 × 10 = W ×  =W × × 10 −4 or W = 6.4 × 107 W/m 2 9 ÷ 225  150 × 10  3

(ii) Assuming the sun to be a blackbody, W = σT04 6.4 × 107 = σT04 = ( 5.67 × 10 −8 ) T04 ∴ T04 = 18

6.4 × 1015 or 5.67

T0 ≈ 0.58 × 10 4 K = 5800K

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(iii) Using Wien’s displacement law, λ mpT0 = 0.29cm-K = 2.9 × 10 −3 m-K

λ mp =

2.0 × 10 −3 = 5 × 10−7 m = 5000 A° 5800

[Note : λ mp is also referred to as λ max ]

17. IDEAL GAS Any gas which obey gas law is called ideal gas, where R = universal gas constant and its value is equal to 8.31 J/mol K. An ideal gas is a simple theoretical model of a gas. No real gas is truly ideal. Real gas approaches the ideal gas behavior for low pressures and hight temperatures.

18. KINETIC THEORY OF GASES (A)

Gases are made-up of tiny particles, consisting of molecules, atoms or even ions (sometimes) which retain the chemical properties of the sample of which they are composed. The kinetic theory of gases develops a model of the molecular behavior which should result in the observed behavior of an ideal gas.

Assumptions (i)

Molecules (Whose size is extermely small in comparision to the separation between them) are moving randomly in all direction.

(ii)

Molecules exert no appreciable force on one another or on the walls of the container except during collision.

(iii)

All collisions between the molecules of with the wall of the container are perfectly elastic.

(iv)

The duration of a collision is negligible in comparsion to the time spend by a molecue between collisions.

(v)

The molecules obey Newton’s Laws of motion.

The assumptions of kinetic theory are close to the situation at low densities. (B)

Pressure of an ideal gas

Consider an ideal gas enclosed in a cubical vessel of edge L. Considering a molecule moving with velocity Y Face 1

Face 1 Vz

r r r r r V = Vx + Vy + Vz ; | V |= Vx2 + Vy2 + 2z The change in momentum of the molecule on collision with face 1 ∆P = (− mv x ) − (mv x ) = 2mv x Thus, momentum imparted to the wall = 2mvx Time between two successive collisions of face 1, ∆t =

2L vx

2 Rate at which momentum is imparted to the wall , ∆F = ∆P / ∆t = mv x / L

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Total force on the face 1 due to all the molecules F = ∑ mv 2x / L = (m / L) ∑ mv 2x As

...........(1)

∑ v 2x = ∑ v 2x = ∑ v 2x 1 1 ∑ Vx2 = ∑ ( Vx2 + Vy2 + z z2 ) = ∑ V 2 ……………… (2) 3 3 F=

1m 1 mN ∑ v 2 ∑ V2 = , 3L 3 L N

where N = tottal number of molecules in the sample. Now, Pressure =

force 1 mN ∑ V 2 = area 3 L N

mN 1 ∑ V2 P= ρ , where ρ = 3 L 3 N   1 ∑ V2 P = ρV 2 ,  V 2 = is the mean square speed. 3 N  

RMS Speed It is defined as Vrms

 v 2 dN  ∫  =  ∫ dN   

1/ 2

 N1V12 + N 2 V22 + .....  =   N1 + N 2 + ...... 

Here, N1 molecules have speed V1, N2 molecules have speed V2, and so on. Since P =

1m 2 v rms 3V 1 3

If we take n mole gas, PV = n.Mv 2rms 1 ∴ nRT = n.Mv 2rms , where R = universal gas constant = 8.314 J/mol-K 3

or v rms =

3RT M

Kinetic interpretation of temperature Kinetic energy of radom motionis the intermal energy, given by 1 1 3RT 2 K.E.int = m.Vrms = n.M 2 2 2 K.E.int = µ T This means temperature of a body depends on its internal KE. At absolute zero temperature, K.E. int becomes zero. (C)

Degrees of Freedom

The degree of freedom of a particle is the number of independent motions which the particle can undergo. Law of Equipartition of Energy

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For a system in equilibrium at absolute temperature T, the average energy per particle associated with each 1 degree of freedom is kT, where k is Boltzmann’s constant. 2 The internal energy of an ideal gas is entirely the kinetic energy of its molecules. Thus, internal energy ofone mole of an ideal gas, having ‘f’ degrees of freedom 1 1 U = N × f × kT = fRT , N = Avogadro’s Number, R = kN 2 2 The specific heat of the gas at constant volume , CV =

dU fR = dT 2

f f  ∴ C P = CV + R = R + R =  + 1÷R 2 2  f  +1 + R C P  2 ÷ 2  = =1+ Thus, ratio of the two specific heats, γ = f CV f R 2 A molecule of a monatomic gas has only 3 (translational) degrees of freedom, i.e. f = 3.



f 3 R= R 2 2 C P = (f / 2 + 1)R = 5 / 2R; γ = 1 + 2 / f = 5 / 3 = 1.66

∴ CV =

 A molecule of a diatomic gas has 5 degrees of freedom (3-translational and 2-rotational) at ordinary atmospheric temperatures (Because the vibrational modes are not excited). The moment of inertia about the line joining the two atoms is negligibly small. Hence, the roational energy about that axis is zero. F=5 f 5 CV = R = R 2 2 CP = (f / 2 + 1)R = 7 / 2R γ = 1 + 2 / f = 7 / 5 = 1.40 Note : At very hight temperatures, if vibrational mode is also considered, then f = 7 A molecule of a triatomic or polyatomic gas has 6 degrees of freedom (3 translational and 3 rotational ) f=6 f C v = R = 3R 2 f  Cp =  + 1÷R = 4R 2  2 4 γ = 1 + = = 1.33 f 3 However, if the atoms of the molecule are arranged in one line (like the molecule of CO 2), then the degrees of freedom are only 5 and its Cv, Cp and γ values will be similar to those of the diatomic gases.

18. Illustration : An air bubbles starts rising from the bottom of a lake. Its diameter is 3.6 mm at the bottom and 4 mm at the surface. The dpeth of the lake is 250 cm and the temperature at the surface is 40°C. What is the temperature at the bottom of the lake ? Given atmospheric pressure = 76 cm of Hg and g = 980 cm/sec2. (Specific gravity of mercury = 13.6) Solution:

At the bottom of the lake, volume of the bubble

4 4 V1 = πr13 = π(0.18)3 cm3 3 3 2 nd

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Pressure on the bubble P1 = Atmospheric pressure + Pressure due to a column of 250 cm of water. = 76 × 13.6 × 980 + 250 × 1 × 980 = (76 × 13.6 + 250) 980 dyne / cm2 T1 = Temperature at the bottom At the surface of the lake, Volume of the bubble 4 4 V2 = πr23 = π(0.2)3 cm 3 3 3

Pressure on the bubble P2 = atmospheric pressure = 76 × 13.6 × 980 dyne/cm2 T2 = 273 + 40°C = 313 K Now,

P1V1 P2 V2 = T1 T2

(76 × 13.6 + 250)980 × (4 / 3) π(0.18) 3 (76 × 13.6) × 980(4 / 3)π(0.2)3 = T1 313

1283.6 × (0.18)3 1033.6(0.2)3 = T1 313 T1 =

∴

1283.6 × (0.18) 3 × 313 = 287.37K 1033.6(0.2)3

T1 = 283.37 – 273 = 10.37°C

19. Illustration : Given : Avogardo’s Number N = 6.02 × 10 23 and Boltzmann’s constant k = .38 × 10–23 J/K. (i) 100°C. (ii)

Calculate the average kinetic energy of translation of the molecules of an ideal gas at 0°C and at Also calculate the crossesponding energies per mole of the gas.

Solution: (i) According to the kinetic theory, the average kinetic energyo f translationper molecule of an ideal gas at Kelvin temperature T is (3/2) kT, where kis Boltzmann’s constant. At 0°C (T = 273 K) The kinetic energy of translation = 3/2 kT = (3/2) × (1.38 × 10–23) × 273 = 5.65 × 10–23 J/molecule At 100°C (T = 373 K), (ii)

1 mole of gas contains N (=6.02 × 1023 ) molecules. Therefore, at 0°C, the kinetic energy of translation of 1 mole of the gas = (5.65 × 10–21) (6.02 × 1023) ≈ 3401 J/mole And at 100°C, the kinetic energy of translation of 1 mole of gas = (7.72 × 10–21) (6.02×1023) ≈ 4647 J/mole.

20. Illustration : How many degrees of freedom does the gas molecule have if undr standard conditions the gas density is 1.3 kg/m3 and the velocity of sound in it is v = 330 m/s. Solution:

As v =

γP ρv 2 ⇒γ= ρ P

If f be the number of degree of freedom, then 22

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2 2 2 γ − 1 ρv / P  − 1

P = 1.013 × 105 N/m2 = 1.3 kg/m3 , v = 330 m/s. f = 5

19.(A) INTERNAL ENERGY OF AN IDEAL GAS In case of an ideal gas, no internal force of interaction exists. Hence, the internal potential energy is zero. The interrnal energy of an ideal gas comprises of molecular kinetic energy. f U = nRT 2

19.(B) WORK DONE Work done by the gas (or system) over the surroundings can be calculated as P p

dV V

dW = P dV vf

∴ W = ∫ P dV vi

Here, dW is elemental work done by pressure P, of the system during elemental change in voluem eV. Work done in the process AB is equalt o the area under the curve AB and V-axis. 20.

Mechanical Equivalent of Heat:

Whenever mechanical work is transformed into heat or heat into mechanical work, there is a constant ratio between the work and the amount of heat. This ratio is called “mechanical equivalent of heat” and is denoted by J . Thus, if W be the amount of work done and Q the amount of heat prouduced, we have W =J , W = JQ Q If Q = 1unit then J =W . Therefore, J is numerically equal to the mechanical work required to produce one

unit of heat. 21.

First law of Thermodynamics :

It is the consequence of conservation of energy for gaseous system. Heat supplied to the gas = Increase in internal energy + work done by the gas. Q = ∆U + W Q = +ve ⇒ heat is supplied to the gas Q = –ve ⇒ heat is taken out from the gas in differential form dQ = dU + dW dQ = nCdT C = molar specific heat and C = CP (constant pressure); C = CV =(constant volume) 22.

Indicator Diagram:

This is graph between pressure and volume of a system under going operation,

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(1)

Every point of Indicator Diagram represents a unique state (P, V, T) of gases.

(2)

Every curve on Indicator Diagram represents a unique process.

Isochoric Process (V = constant)

dV = 0 ⇒ dW = 0 By First Law of Thermodynamic dQ = dU = nCV dT

P 2 V

Is o c h o r ic

Q = ∫ nCV dT = nCV (T2 − T1 ) T1

Isobaric Process (P = constant) dP = 0 By First Law of Thermodynamics

dQ = dU + dW

* *

f  nC p (T2 − T1 ) =   nR (T2 − T1 ) + nR (T2 − T1 ) 2 W = nR (T2 − T1 ) where d Be careful if ∆V = 0 then not necessarily an Isochoric Process. If ∆P = 0 then not necessarily an Isobaric Process.

Is o b a ric

Isothermal Process (T = constant): dU = 0 (Q dT = 0) PV = K By First Law of Thermodynamics

∫ dQ = ∫ dW ⇒

T V

∫ dQ = ∫ PdV

2 1

T 2> T

Is o th e rm a l P ro c e s s V2

Q = W = (nRT ) ∫ dV / V V1

V P W = nRT ln 2 = nRT ln 1 . V1 P2 Adiabatic Process : dQ = 0 but if ∆Q = 0 , it is not necessarily adiabatic. dW = − dU By First Law of Thermodynamics T2

− PV nRdT nR(T1 − T2 ) PV = = 1 1 2 2 γ −1 γ −1 T1 γ − 1

W = −∫

How to get the process Equation for adiabatic 24

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(i)

(ii)

First Law of Thermodynamics with process condition nRdT dU = −dW = …(i) γ −1 Differential form of gas law d ( PV ) = d (nRT ) PdV + VdP = nRdT

−nRdT γ −1 PdV + VdP = −( γ − 1) PdV VdP = −( γPdV ) dP dV = −γ P V ln P = −γ ln V + ln C

But dW = PdV = So

…(ii)

PV γ = Const. TV γ−1 = Const. T γ P1−γ = Const. For Adiabatic Process PV γ = constant dP dP =γ dV adiabatic dV isothermal Slope of adiabatic curve is more in magnitude in comparison to the slope of the isothermal curve.

Is o th e rm a l A d ia b a tic

∆P ∆P = −V (∆V )/ V ∆V dP  ∂P  = −V  Isothermal bulk modulus of Elasticity ET = −  dV / V  ∂V  Isothermal

Bulk Modulus of Gases: β = −

Adiabatic bulk modulus of Elasticity E adia = −

E adia dP  ∂P  = −V  =γ. ÷ ; dV / V  ∂V adia E T

CYCLIC PROCESS : It is the combination of various processes such that initial and final state is same. Therefore initial and final temperature is equal i.e., ∆T = 0 ⇒ ∆U = 0 ⇒ ∆Q = ∆W

Efficiency of a Cyclic Process So

∆U = 0 ⇒ no rise in internal energy ∆Q = ∆W

work done by gas heat input Q W η= = 1 − out Qin Qin

Efficiency η =

Q P

2 Q

out

23.(A) Molar Specific Heat For Polytropic Process PV n = K

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molar heat capacity of polytropic process C =

R R R R = − + γ −1 n −1 γ −1 1 − n

So C is constant for polytropic process

23.(B) Work done on gas in some process :

∆V = 0 but dV ≠ 0 work done = + ve

∆W can be zero but dW ≠ 0

For clockwise ∆W = + ve For anticlockwise ∆W = − ve

Work done is least for monoatomic gas in expansion : P Iso th e rm a l P o ly a to m ic D ia to m ic M o n a to m ic V

A d ia b a tic P ro c ess

21. Illustration : Three moles of an ideal gas being initially at a temperature T0 = 273 K were isothermally expanded η = 5.0 time its initial volume and then isochorically heated so that the pressure in the final state became equal to that in the initial state. The total amount of heat transferred to the gas during the process equals Q = 80 KJ. Find the ratio γ = C P / CV for this gas. Solution : In isothermal process, the heat transferred to the gas is given by Q1 = nRT0 ln(V2 / V1 ) = nRT0 lnη …(i) [∴ η = (V2 / V1 ) = ( P1 / P2 )] In isochroric process, Q2 = ∆U (W = 0) ∴ Q2 = nCV ∆T = n{R /( γ − 1)}∆T …(ii)

P T = T0  1  = ηT0  P2  ∴ ∆T = ηT0 − T0 = (η − 1)T0 …(iii) substituting the value of ∆T from equation (iii) in equation (ii), we get  R  Q2 = n   (η − 1)T0  γ −1 Now

∴

P2 T0 = P1 T

 R  Q = nRT0 lnη + n   (η − 1)T0  γ −1   η −1 Q − ln η =   nRT0  γ −1  η −1 γ −1 = Q − lnη nRT0

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η −1 Q − lnη nRT0 Substituting given values, we get (5 − 1) γ =1+ 80 × 103 − ln5 3 × 8.3 × 273 Solving we get γ = 1.4 ∴

γ =1+

22. Illustration : If 2 moles of an ideal monoatomic gas at temperature T 0 is mixed with 4 moles of another ideal monoatomic gas at temperature 2T 0, then find the temperature of the mixture ? Solution : Let T be the temperature of the mixture. Then U =U1 + U2 or

f f f (n1 + n 2 )RT = (n1 )RT0 + (n 2 )(R)(2T0 ) 2 2 2

or (2 + 4) T = 2T0 + 8T0 (n1 = 2, n2 = 4) or T =

5 T0 3

23. Illustration : In a given gas during a process one third of heat supplied is used to raise internal energy of gas. Find molar specific heat of the gas and their process. Solution:

Heat supplied = nCdT

C = molar specific heat

nCdT f 3 fR 3R  dQ  = nRdT ⇒ C = = = dU  . Q 3 2 2 γ −1  3  24. Illustration:An ideal gas is taken through a process in which the pressure and the volume are changed according to the equation P = KV. Show that the molar heat capacity of the gas for the process is given by C = CV +R/2. Solution :

PV = nRT P = KV From (i) and (ii), KV2 = nRT Differentiating 2 KVdV = nRdT dQ = dU + PdV C = CV + R/2

2 nd

. . . . . (i) . . . . . (ii)

PdV = nRdT/2 nCdT = nCVdT + nRdT/2

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25. Illustration : One mole of monoatomic ideal gas is taken A B through the cycle shown in figure. A → B Adiabatic expansion P B → C Cooling at constant volume C D C → D Adiabatic compression V D → A Heating at constant volume The pressure and temperature at A, B etc., are denoted by PA , TA ; PB , TB etc/ respectively. Given TA = 1000 K , PB = (2 / 3) PA and PC = (1/ 3) PA . Calculate (a) The work done by the gas in the process A → B (b) The heat lost by the gas in the process B → C and (c) Temperature TD given (2 / 3) 2 / 5 = 0.85 and R = 8.31 J/mol K. Solution :

(a)

As for adiabatic change Tγ = constant P γ−1

1−

γ−1

where γ =

5 3

2/5

 2 γ  2 TB = TA   = 1000   = 850 K  3  3 µR[Ti − T f ] 1 × 8.31[1000 − 850] WAB = = [ γ − 1] [(5 / 3) − 1] WAB = (3/ 2) × 8.31 ×150 = 1869.75 J

i.e. so i.e. (b)

T   P  so  B  =  B   TA   PA 

For B → C , V = constant so ∆W = 0 so from first law of thermodynamics ∆Q = ∆U + ∆W = µCV ∆T + 0

3 3  or ∆Q = 1×  R  (TC − 850) as Cv = R 2 2  Now along path BC, V = constant; P ∝ T PC TC (1/ 3) PA T 850 = , TC = × TB = B = = 425 K …(ii) i.e. PB TB (2 / 3) PA 2 2 3 ∆Q = 1 × × 8.31(425 − 850) = −5297.625 J So 2 [Negative heat means, heat is lost by the system] (c)

D → A process is isochoric PD TD T = , PD = PA D i.e. PA TA TA But C and D are on the same adiabatic γ

 TD   PD    =   TC   PC  or ( TD )

i.e.

1−

1/ γ

 P  = TC  A   PC TA 

1 γ

γ −1

PT  = A D   PC TA  3/ 5 C

, i.e. T

γ−1

 PA T  =  B    2   (1/ 3) PA1000 

2/5

 1  2 2 / 3   3 2/ 5 TD3/ 5 =    × 1000    i.e. TD = 500 K  2  3   1000 

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PHYSICS:HEATand THERMOYNAMICS

26. Illustration : An ideal gas is taken through a cyclic thermodynamical process through four steps. The amount of heat involved in these steps are Q1 = 5960 J ; Q2 = −5585 J Q3 = −2980 J ; Q4 = 3645 J respectively. and The corresponding works involved are W1 = 2200 J ; W2 = −825 J ; W3 = −1100 J and W4 respectively. (a) Find the value of W 4 (b) What is the efficiency of the cycle? Solution :

(a)

According to the given problem ∆Q = Q1 + Q2 + Q3 + Q4 = 5960 − 5585 − 2980 + 3645 ∆Q = 9605 − 8565 = 1040 J ∆W = W1 + W2 + W3 + W4 = 2200 − 825 − 1100 + W4 = 275 + W4 ∆U = U F − U I = 0 and as for cyclic process U F = U I , So from first law of thermodynamics, i.e., ∆Q = ∆U + ∆W , we have 1040 = (275 + W4 ) + 0, i.e. W4 = 765 J

(b)

Efficiency of a cycle is defined as Network ∆W ∆Q η= = = Input heat (Q1 + Q4 ) (Q1 + Q4 ) 1040 η= = 0.1082 = 10.82% 9605

KEY CONCEPTS : 1.

For small changes in temperature: L = L0(1 + α∆T); A = A0 (A + β∆T); V = V0(1 + γ∆T) α : β : γ = 1 : 2 : 3 (for isotropic substances only)

For liquid: ga =gR - gc gR coefficient of real expansion of liquid 2 nd

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PHYSICS:HEATandTHERMOYNAMICS

ga coefficient of apparent expansion of the liquid gc coefficient of cubical expansion of container 3.

r0 Variation of density with temperature, r = 1 +gDT

Thermal stress =

Thermal capacity of a body is amount of heat required to raise its temperature though one degree, DQ =mc.

If temperature of a body of mass m rises by ∆T, then ∆Q = mc ∆T.

F =aYDq (where Dq is the increases in temperature) A

c = specific heat of the material. 7.

When the sate of body of mass m changes, then ∆Q = mL L → latent heat

According to principle of calorimetry Heat lost = Heat gained

TC £ T £ TH Where

TC ® temperature of cold object, T ® temperature of equilibrium, and TH ® temperature of hot object

10.

Thermal Conduction in steady state DQ KA(T1 - T2 ) = Dt x Where K is the co-efficient of thermal conductivity of the material of the slab. Also R = thermal resistance = In parallel connection,

x KA

1 1 1 1 = + +... R eq R1 R 2 Rn

In series connection, R eq =R1 +R 2 +... +R n 11.

In steady state, temperature of each pint is constant w.r.t. time but different at different points. Actually, it decreases along the direction of heat flow.

12.

Absorptive power of a body is defined as the fraction of the incident radiation that is absorbed by the body Absorptive power, a =

Energy absorbed Energy incident

13.

The emissive power (E) denotes the energy radiated per unit time per unit area of the surface.

14.

Emissivity (ε) of a surface is the ratio of the emissive per of the surface to the emissive power of block body at the same temperature.

15.

A perfectly black body is one which absorbs completely all the radiation, of whatever wave length, incident on it

16.

Kirchoff’s Law 30

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It states that the ratio of the emissive power to the absorptive power for radiation of a given wavelength is el =E (cons tan t) the same for all bodies at the same temperature, and is equal to the emissive al 17.

Stefan’s Law of Radiation Energy emitted per unit time by a black body of surface area A is given by dq =s AT 4 [Stefan’s constant σ = 5.67 × 10−8 Wm−2Kelvin−4] dt Energy emitted per unit time by a body other than a black body, dq =s AeT 4 [ e=emissivity] dt

18.

Net loss of thermal energy, DE =se A(T 4 - T04 )

19.

Newton’s Law of Cooling For a small temperature difference between a body and its surrounding the rate of cooling of the body is directly proportional to the temperature difference. dT =- k(T - T0 ) (rate of fall of temperature) dt T = temperature of the body; T0 = temperature of the surroundings.

20.

Wien’s displacement law: It states that the wavelength (l m ) corresponding to which energy emitted by a perfectly black body is maximum is inversely proportional to the temperature (T) of the black body

21.

Ideal gas equation: PV = nRT

22.

Pressure exerted by a gas: It is due to continuous. Collision of gas molecules with the wall of container. 1 MC2 1 2 P= = rC 3 V 3 C → rms velocity of gas molecules

23.

Average K.E. molecule of a gas 1 3 = mc 2 = kT 2 2

24.

Internal energy of an ideal gas is wholly kinetic energy of molecules and is a function of temperature.

25.

If ∆Q is heat given to a system, and ∆W is work done by the system, the ∆U (the change in its

internal energy) can be written as ∆Q = ∆U + ∆W ,where ∆U = nC v ∆T vf

∆W = ∫ PdV = Area enclosed under P – V v ∆Q = nC∆T , where C = specific heat capacity of a gs, for the process. Note: For a cylclic process, work done = area enclosed under P-V graph, and is positive if the cycle is clockwise. 26.

Important Cases of the First Law of Thermodynamics (a)

Isobric process (Pressure = Constant)

2 nd

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∆Q = ∆U + ∆W nC p ∆T = nCv ∆T + nR∆T

Cp – Cv = R (Mayer’s equation) (b)

Isochoric process ( ∆W = 0 , volume = constant) ∆Q − ∆U = nC v ∆T

(c)

Isothermal process ( ∆U = 0, temperature = constant) v2

∆Q = ∆W ∫ P dV = v1

(d)

nRT V  P  dV = nRT ln  2  = nRT ln  2   V1   P1  v1 V

∫

Adiabatic process ( ∆Q = 0) TV γ−1 = constant

(e)

Cyclic process ∆U = 0, since process returns to the same initial state

∆U = ∆Q − ∆W = 0 ∴∆Q = ∆W (f)

Free expansion

If a gas expands in such a way that no heat enters or leaves the system (adiabatic process) and also no work is done by or on the system, then the expansion is called the free expansion. Uf – U1 = Q – W, Now, Q = 0, W = 0 ∴ 27.

Uf – U1

P  dp  = For an isothermal process,  ÷  dV isothermal V P  dp  = −γ For an adiabatic process,  ÷ V  dV adiabatic Efficiency of a process, η =

∆W × 100 ∆Q

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