Mode of Heat Transfer

Heat is a form of energy which transfers from a body at higher temperature to a body at lower temperature. The transfer of heat from one body to another may take place by any one of the following modes : 1. Conduction The process in which the material takes an active part by molecular action and energy is passed from one particle to another is called conduction. It is predominant in solids. 2. Convection The transfer of energy by actual motion of particle of medium from one place to another is called convection. It is predominant is fluids (liquids and gases). 3. Radiation Quickest way of transmission of heat is known as radiation. In this mode of energy transmission, heat is transferred from one place to another without effecting the intervening medium. Conduction Heat Transfer due to Temperaturedifference Due to free electron or vibration motion of molecules Heat transfer in solid body (in mercury also) Slo w process Irregular path Convection Heat transfer due to density difference Actual motion of particles Heat transfer in fluids (Liquid + gas) Slow process Irregular path Radiation Heat transfer with o ut any medium Electromagnetic radiation All Fast p rocess (3 × 108 m/sec) Straight line (like light) THERMAL CONDUCTION The process by which heat is transferred from hot part to cold part of a body through the transfer of energy from one particle to another particle of the body without the actual movement of the particles from their equilibrium positions is called conduction. The process of conduction only in solid body (except Hg) Heat transfer by conduction from one part of body to another continues till their temperatures become equal. The process of transmission of heat energy in which heat is transferred from one particle of the medium to the other, but each particle of the medium stays at its own position is called conduction, for example if you hold an iron rod with one of its end on a fire for some time, the handle will get hot. The heat is transferred from the fire to the handle by conduction along the length of iron rod. The vibrational amplitude of atoms and electrons of the iron rod at the hot end takes on relatively higher values due to the higher temperature of their environment. These increased vibrational amplitude are transferred along the rod, from atom to atom during collision between adjacent atoms. In this way a region of rising temperature extends itself along the rod to your hand. THERMAL PROPERTIES OF MATTER Consider a slab of face area A, Lateral thickness L, whose faces have temperatures T H and TC(TH > TC). Now consider two cross sections in the slab at positions A and B separated by a lateral distance of dx. Let temperature of face A be T and that of face B be T + T. Then experiments show that Q, the amount of heat crossing the area A of the slab at position x in time t is given by Q dT = –KA t dx Here K is a constant depending on the material of the slab and is named thermal conductivity of the material, and the  dT  quantity   is called temperature gradient. The (–) sign in equation (2.1) shows heat flows from high to low  dx  temperature (T is a –ve quantity) STEADY STATE If the temperature of a cross-section at any position x in the above slab remains constant with time (remember, it does vary with position x), the slab is said to be in steady state. Remember steady-state is distinct from thermal equilibrium for which temperature at any position (x) in the slab must be same. For a conductor in steady state there is no absorption or emission of heat at any cross-section. (as temperature at each point remains constant with time). The left and right face are maintained at constant temperatures T H and TC respectively, and all other faces must be covered with adiabatic walls so that no heat escapes through them and same amount of heat flows through each cross-section in a given Interval of time. Hence Q1 = Q = Q2. Consequently the temperature gradient is constant throughout the slab. Hence, dT T Tf  Ti TC  TH    dx L L L and  TH  TC  Q Q T   KA  = KA  L  t t L Here Q is the amount of heat flowing through a cross-section of slab at any position in a time interval of t. Ex. One face of an aluminium cube of edge 2 metre is maintained at 100 °C and the other end is maintained at 0 °C. All other surfaces are covered by adiabatic walls. Find the amount of heat flowing through the cube in 5 seconds. (thermal conductivity of aluminium is 209 W/m–°C) Sol. Heat will flow from the end at 100ºC to the end at 0°C. Area of cross-section perpendicular to direction of heat flow, A = 4m2 Q (20 9 W / mº C)(4 m 2 )(1 00 º C  0º C)(5 sec)  20 9 kJ 2m Thermal conductivity (K) : (i) It's depends on nature of material. Order of thermal conductivity Ag > Cu > Au > Al K For Ag maximum is (410 W/mK) For Freon minimum is 12 (0.008 W/mK) then (T  TC ) Q  KA H t L (ii) (iii) (iv) (v) SI Unit : J s–1 m–1 K–1 Dimensions : M1 L1 T–3 –1 For an ideal or perfect conductor of heat the value of K =  For an ideal or perfect bad conductor or insulator the value of K = 0 For cooking the food, low specific heat and high conductivity utensils are most suitable. APPLICATION OF THERMAL CONDUCTION (i) (ii) (iii) (iv) In winter, the iron chairs appear to be colder than the wooden chairs. Cooking utensils are made of aluminium and brass whereas their handles are made of wood. Ice is covered in gunny bags to prevent melting of ice. We feel warm in woollen clothes and fur coat. (v) Two thin blankets are warmer than a single blanket of double the thickness. (vi) Birds often swell their feathers in winter. (vii) A new quilt is warmer than old one. THERMAL RESISTANCE TO CONDUCTION If you are interested in insulating your house from cold weather or for that matter keeping the meal hot in your tiffinbox, you are more interested in poor heat conductors, rather than good conductors. For this reason, the concept of thermal resistance R has been introduced. L KA Q (TH  TL )  In terms of R, the amount of heat flowing though a slab in steady-state (in time t) t R For a slab of cross-section A, Lateral thickness L and thermal conductivity K, R  Q as thermal current iT then, t TH  TL R This is mathematically equivalent to OHM’s law, with temperature donning the role of electric potential. Hence results derived from OHM’s law are also valid for thermal conduction. More over, for a slab in steady state we have seen earlier that the thermal current iL remains same at each crosssection. This is analogous to kirchoff’s current law in electricity, which can now be very conveniently applied to thermal conduction. If we name RAl = 50ºC 12ºC Aluminium er pp co Sol. Three identical rods of length 1m each, having cross-section area of 1cm2 each and made of Aluminium, copper and steel respectively are maintained at temperatures of 12ºC, 4ºC and 50ºC respectively at their separate ends. Find the temperature of their common junction. [ KCu=400 W/m-K , KAl = 200 W/m-K , Ksteel = 50 W/m-K ] ste el Ex. iT  L 1 10 4  4  KA 10  200 200 51ºC iS 104 104 Similarly Rsteel = and Rcopper = 50 400 Let temperature of common junction = T 12ºC Rs iAl RAl T iCu then from Kirchoff's current laws, iAl + isteel + iCu = 0 RCu 4ºC  T  12 T  51 T  4   0 R Al R steel R Cu   4(T – 12) + (T – 50) + 8 (T – 4) = 0  13T = 48 + 50 + 32 = 130  T = 10°C (T – 12) 200 + (T – 50) 50 + (T – 4) 400 4ºC THERMAL PROPERTIES OF MATTER SLABS IN PARALLEL AND SERIES Slabs in series (in steady state) Consider a composite slab consisting of two materials having different thickness L1 and L2 different cross-sectional areas A1 and A2 and different thermal conductivities K1 and K2. The temperature at the outer surface of the states are maintained at TH and TC, and all lateral surfaces are covered by an adiabatic coating. L2 L1 Q K2 Heat reservoir at T C K1 Heat reservoir at temperature T H adiabatic coating Let temperature at the junction be T, since steady state has been achieved thermal current through each slab will be equal. Then thermal current through the first slab. I= Q TH – T =  TH – T = IR1 t R1 and that through the second slab, i = I = Q T – TC =  T – TC = IR 2 t R2 TH  TC TH – TL = (R1 + R2)I I = R  R 1 2 Adding equation Thus these two slabs are equivalent to a single slab of thermal resistance R1 + R2. If more than two slabs are joined in series and are allowed to attain steady state, then equivalent thermal resistance is given by Ex. R = R1 + R2 + R3 + ....... The figure shows the cross-section of the outer wall of a house built in a hill-resort to keep the house insulated from the freezing temperature of outside. The wall consists of teak wood of thickness L1 and brick of thickness (L2 = 5L1), sand witching two layers of an unknown material with identical thermal conductivities and thickness. The thermal conductivity of teak wood is K1 and that of brick is (K2 = 5K). Heat conduction through the wall has reached a steady state with the temperature of three surfaces being known. T1 L1 (T1 = 25°C, T2 = 20°C and T5 = –20°C). Find the interface temperature T4 and T3. Sol. Let interface area be A. Then thermal resistance of wood,R1 = T2 T3 L T4 L T5 L2 L1 L2 5L1 = and that of brick wall R2 = = R1 K1 A K 2 A 5K1A Let thermal resistance of the each sand witch layer = R. Then the above wall can be visualised as a circuit iT 25ºC R1 R 20ºC Thermal current through each wall is same. R T3 R1 T4 iT –20ºC Hence T  20 2 5  20 20  T3 T3  T4    4 R1 R R R1 also, • 20 – T3 = T3 – T4  T3 =  25 – 20 = T4 + 20  T4 = –15°C 20  T4 = 2.5°C 2 Equivalent conductivity for Heat flow through slabs in series Req = R1 + R2 L1 L2 L1  L2 = K A K A K eq A 1 2 Equivalent thermal conductivity of the system is Keq = L1  L2 L i = L1 L2 L   i K1 K 2 Ki Slabs in Parallel (in steady state): Consider two slabs held between the same heat reservoirs, their thermal conductivities K1 and K2 and crosssectional areas A1 and A2 L Heat reservoir at temperature TH SLAB 1 K1 A1 Q1 SLAB 2 K2 A2 Q2 adiabatic coating Heat reservoir at temperature TC then L R1 = K A , 1 1 L R2 = K A 2 2 thermal current through slab 1 : I1  TH  TC R1 and that through slab 2 : I2  Net heat current from the hot to cold reservoir Comparing with I  TH  TC R2  1 1  I  I1  I2  (TH  TC )   R1 R 2  1 1 1 TH  TC  , we get, = R eq R1 R 2 R eq If more than two rods are joined in parallel, the equivalent thermal resistance is given by 1 1 1 1     ..... R eq R 1 R 2 R 3 THERMAL PROPERTIES OF MATTER • Equivalent thermal conductivity for Heat flow through slabs in parallel 1 1 1 K1 A1 K2 A 2 L K eq    ,R= ; (A1 + A2) = R eq R1 R 2 KA L L L A1 Q t A2 Q t K2 T1 Equivalent thermal conductivity K eq  K1 T2 L R1 equivaleent to K i A i K1 A1  K 2 A 2 = A i A1  A 2 R2 Ex. Three rods of material X and three rods of material Y are connected as shown in figure. All the rods are identical in length and cross–sectional area. If the end A is maintained at 600C and the junction E at at 100C, calculate the temp. of the junctions B,C,D. The thermal conductivity of X is 0.92 CGS units and that of Y is 0.46 CGS units. Sol. RX  1 KY 0 .4 6 1 R 1 , RY   X = = = KX KX 0 .9 2 2 RY KY Let RX = R  RY = 2R The total resistance R  R Y + effective resistance in the bridge C R  2R  2R  4 R 4 10 R &  Δθ = l × R = 2R + R= 2R  4R 3 3 A  A  B  60 0  B  60  B  2R  I For B and C B  C  For A and E  A   E  60  1 0  50  2 I  R 3 ....(ii) C  E  B Y Y D ...(i) 2 RI 3 10 ....(iii)  R  I  15   3 R  I  50 2   A  B  2  15  3 0 , B  60  30 = 300C, B  C     15  10 3  C  30  10  20 0 C Obviously, C  D  20 0 C GROWTH OF ICE ON LAKES In winter atmospheric temperature falls below 0°C and water in the lake start freezing. Let at time t thickness of ice on the surface of the lake = x and air temperature = –° C The temperature of water in contact with the lower surface of ice = 0°C Let area of the lake = A Due to escape of this heat increasing extra thickness of ice = dx Mass of this extra thickness of ice is m = V=  A.dx 10 C E 0 Y 2 1 Further IBCE (2R) = IBDE(4R) and IBCE + IBDE = I IBCE = I and IBDE = I 3 3 For A and B X X 600C dQ = mL = ( A.dx) L  KA dt = ( A.dx) L x L x dx K L x 1 L 2 x dx  x So time taken by ice to grow a thickness x is t = K 0 2 K So time taken by ice to grow from thickness x1 to thickness x2 is  t = t2 – t1 =  dt  z 1 L 2 (x – x12) and t  (x22 – x12) 2 KT 2 Time taken to double and triple the thickness ratio t1 : t2 : t3 :: 12 : 22 : 32 So t1 : t2 : t3 :: 1 : 4 : 9 Ex. Sol. One end of a brass rod 2m long and having 1 cm radius is maintained at 250°C. When a steady state is reached, the rate of heat flow across any cross–section is 0.5 cal s –1 . What is the temperature of the other end K = 0.26 cal s–1 cm–1 °C–1. Q = 0.5 cal s–1; r = 1 cm t  Area A = r2 = 3.142 × 1 cm2 = 3.142 cm2 L = Length of rod = 2m = 200 cm, T1 = 250°C, T2 = ? We know 0.5  2 0 0 Q Q KA (T1  T2 ) x = or (T1 – T2) = × = =122.4°C t t L kA 0.26 C 1  3 .14 2  T2 = 250°C – 122.4°C = 127.6°C Ex. Steam at 373 K is passed through a tube of radius 10 cm and length 2 m. The thickness of the tube is 5 mm and thermal conductivity of the material is 390 W m–1 K–1, calculate the heat lost per second. The outside temp. is 0°C. Sol. Using the relation Q = KA (T1  T2 )t L Here, heat is lost through the cylindrical surface of the tube. A = 2r (radius of the tube) (length of the tube) = 2 × 0.1 × 2 = 0.4 m2 K = 390 W m–1 K–1 T1 = 373 K,  Q= T2 = 0°C = 273 K, L = 5 mm = 0.005 m and t = 1 s 3 9 0  0.4   (3 7 3  2 7 3 )  1 3 9 0  0.4   1 0 0 = = 98 × 105 J. 0.0 0 5 0.0 0 5 Ex. The thermal conductivity of brick is 1.7 W m–1 K–1, and that of cement is 2.9 W m–1 K–1. What thickness of cement will have same insulation as the brick of thickness 20 cm. Sol. Since Q = KA (T1  T2 )t . For same insulation by the brick and cement Q, A (T1 – T2) and t do not change. Hence, L K remain constant. If K 1 and K 2 be the thermal conductivities of brick and cement respectively and L L1 and L2 be the required thickness then K1 1.7 K 2.9 = 2 or = 20 L1 L2 L2  L2 = 2.9 × 20 = 34.12 cm 1.7 THERMAL PROPERTIES OF MATTER Ex. Two vessels of different material are identical in size and wall–thickness. They are filled with equal quantities of ice at 0°C. If the ice melts completely, in 10 and 25 minutes respectively then compare the coefficients of thermal conductivity of the materials of the vessels. Sol. Let K1 and K2 be the coefficients of thermal conductivity of the materials, and t1 and t2 be the time in which ice melts in the two vessels. Since both the vessels are identical, so A and x in both the cases is same. Now, Q = b g K 2 A 1  2 t 2 K1 A (1   2 )t1 = L L K1 t 2 25 min 5 K2 = t 1 = 10 min = 2  Two plates of equal areas are placed in contact with each other. Their thickness are 2.0 cm and 5.0 cm respectively. The temperature of the external surface of the first plate is –20°C and that of the external surface of the second plate is 20°C. What will be the temperature of the contact surface if the plate (i) are of the same material, (ii) have thermal conductivities in the ratio 2 : 5. Sol. Rate of flow of heat in the plates is (i) Q K A (1  ) K 2 A (   2 ) = 1 = ...(i) t L1 L2 Here 1 = –20°C, 2 = 20°C, L1 = 2 cm = 0.02 m, L2 = 5 cm = 0.05 m and K1 = K2 = K  equation (i) becomes b KA 2 0   0.0 2 g= KAb  2 0 g 2cm 5cm plate 1 Ex. plate 2 0 20C 0 20C 0.0 5  5(–20 –) = 2( – 20)  –100 – 5 = 2 – 40  7 = – 60   = – 8.6°C (ii) K1 2 2 = or K = K 1 K2 5 5 2  from equation (i) Ex. Sol. 0.0 2 g= K Ab  2 0 g–20 –  =  – 20 or –2 = 0 2  = 0°C 0.0 5 An ice box used for keeping eatables cold has a total wall area of 1 metre2 and a wall thickness of 5.0 cm. The thermal conductivity of the ice box is K = 0.01 joule/metre–°C. It is filled with ice at 0°C along with eatables on a day when the temperature is 30°C. The latent heat of fusion of ice is 334 × 103 joule/kg. Calculate the amount of ice melted in one day. dQ KA 0 .0 1  1  d   3 0  6 joule/sec dt L 0 .0 5 So dQ  86400  6  86400 dt Q 6  8 640 0   1.5 52 kg L 3 34  10 3 T A hollow spherical ball of inner radius a and outer radius 2a is made of a uniform a material of constant thermal conductivity K. The temperature within the ball is 2T maintained at 2T0 and outside the ball it is T0. Find, (a) the rate at which heat K flows out of the ball in the steady state, (b) the temperature at r = 3a/2, where r 2a is radial distance from the centre of shell. Assume steady state condition. In the steady state, the net outward thermal current is constant, and does not depend on the radial position. Q = mL (L – latent heat), Ex. b 2 5 K 2 A 20   m 0 0 Sol.  dQ  dT C 1 2 dT  1 2 Thermal current, C1 =     K.  4 r  dt  dr dr 4 K r At r = a, T = 2T0 and r =2a, T=T0 T = T= C1  C2 4 kr 2a dQ T0 (a) = 8 aKT0 (b) T  r  3a / 2   4 T0 / 3 r dt CONVECTIONS Convection requires a medium and is the process in which heat is transferred from one place to other by actual movement of heated substance (usually fluid).The type of convection which results from difference in densities is called natural convection (for example, a fluid in a container heated through its bottom). However, if a heated fluid is forced to move by a blower, fan or pump, the process is called forced convection. The rate of heat convection from an object is proportional to the temperature difference () between the object and convective fluid and the area of contact A, i.e.,  dQ  = hA  where, h represents a constant of proportionality called  dt    convection A Flame convection coefficient and depends on the properties of fluid such as density, viscosity, specific heat and thermal conductivity, etc. PHENOMENA BASED ON CONVECTION : (i) Land and sea breezes : The heat from the Sun is absorbed more rapidly by land than by sea–water. Moreover, the specific heat of land is low as compared to that of sea–water. Consequently, the rise in temperature of land is higher as compared to that of sea–water. To sum–up, land is hotter than the sea during day time. As a result of this, the colder air over the sea blows towards the land. This is called sea–breeze. At night, air blows from land towards sea. This is called land breeze. (ii) Formation of trade winds : The surface of Earth near the equator gets heated strongly. So, the air in contact with the surface of Earth at the expands and rises upwards. As a result of this, a low pressure is created at the equator. At the poles, the air in the upper atmosphere gets cooled and comes down. So, a high pressure is created at the poles. Due to difference of pressures at the poles and equator, the air at the poles moves towards the equator, rises up, moves the poles and so on. In this way, a wind is formed in the atmosphere. The rotation of the Earth also affects the motion of the wind. Due to anti–clockwise rotation of Earth the warm wind blowing from equator to north drifts towards east. The steady wind blowing from north–Earth to equator, near the surface of Earth, is called trade wind. (iii) Monsoons : In summer, the peninsular mass of central Asia becomes more strongly heated than the water of the Indian Ocean. This is due to the fact that the specific heat of water is much higher than that of the soil and rocks. Hot air from the heated land mass rises up and moves towards the Indian ocean. Air filled with moisture flows over the Indian ocean on the south towards heated land mass. When obstructed by mountains, the moist air rushes upwards to great height. In the process, it gets cooled. Consequently, the moisture condenses and falls as rain. (iv) Ventilation : Ventilator of exhaust fan in a room help of remove impure and warm air from a room. The fresh air from outside blows into the room. This is all due to the convection current set up in the room. (v) To regulate temperature in the human body : Heat transfer in the human body involves a combination of mechanisms. These together maintain a remarkably uniform temperature in the human body inspite of large changes in environmental conditions. The chief internal mechanism is forced convection. The heart serves as the pump and the blood as the circulating fluid. THERMAL PROPERTIES OF MATTER ETOOS KEY POINTS  (i)  (ii)     Natural convection takes place from bottom to top while forced convection in any direction. In case of natural convection, convection currents move warm air upwards and cool air downwards. This is why heating is done from base, while cooling from the top. (iii) Natural convection is not possible in a gravity free region such as a freely falling lift or an orbiting satellite. (iv) Natural convection plays an important role in ventilation, in changing climate and weather and in forming land and sea breezes and trade winds. (v) The forced convection of blood in our body by a pump (heart) helps in keeping the temperature of body constant. (vi) For heat propagation via convection, temperature gradient exists in vertical direction and not in horizontal direction. (vii) Most of heat transfer that is taking place on Earth is by convection, the contribution due to conduction and radiation is very small. Ex. Water in a closed tube is heated with one arm vertically placed above the lamp. In what direction water will begin the circulate along the tube ? Sol. On heating the liquid at A will become lighter and will rise up. This will push the liquid A B in the tube upwards and so the liquid in the tube will move clockwise i.e. form B to A. THERMAL RADIATION The process of the transfer of heat from one place to another place without heating the intervening medium is called radiation. When a body is heated and placed in vacuum, it loses heat even when there is no medium surrounding it. The heat can not go out from the body by the process of conduction or convection since both of these process require the presence of a material medium between source and surrounding objects. The process by which heat is lost in this case is called radiation. This does not require the presence of any material medium. It is by radiation that the heat from the Sun reaches the Earth. Radiation has the following properties: (a) Radiant energy travels in straight lines and when some object is placed in the path, it's shadow is formed at the detector. (b) It is reflected and refracted or can be made to interfere. The reflection or refraction are exactly as in case of light. (c) It can travel through vacuum. (d) Intensity of radiation follows the law of inverse square. (e) Thermal radiation can be polarised in the same way as light by transmission through a nicol prism. All these and many other properties establish that heat radiation has nearly all the properties possessed by light and these are also electromagnetic waves with the only difference of wavelength or frequency. The wavelength of heat radiation is larger than that of visible light. Types of thermal Radiation :– Two types of thermal radiation. Plane Radiation Diffuse Radiation Radiations which are incident on a surface at certain angle Incident on the surface at all angles (i) When radiation passes through any medium then radiations slightly absorbed by medium according to its absorptive power so temperature of medium slightly increases. (ii) In order to obtain a spectrum of radiation, a special prism used like KC prism, Rock salt prism Flourspar prism. Normal glass prism or Quartz prism can not be used (because it absorbed some radiation). (iii) Radiation intensity measured with a specific device named as Bolometer. (iv) Heat radiation are always obtained in infra–red region of electromagnetic wave spectrum so they are called Infra red rays. (v) Thermal radiations is incident on a surface, it exerts pressure on the surface, which is known as Radiation Pressure. BASIC FUNDAMENTAL DEFINITIONS 1. Energy Density (u) The radiation energy of whole wavelength (0 to ) present in unit volume at any point in space is defined as energy density. S I UNIT : J/m3 2. 3.  Spectral energy density (u) : Energy density per unit spectral region. u   u  d Absorptive power or absorptive coefficient 'a' : The ratio of amount of radiation absorbed by a surface (Qa) to the amount of radiation incident (Q) upon it is defined as the coefficient of absorption a  4. SI UNIT : J/m3 Å 0 Spectral absorptive power (a) : z  At a given wavelength a  a= Qa .It is unitless Q Qa  : Also called monochromatic absorptive coefficient Q a  d . For ideal black body a and a = 1, a and a are unitless 0 5. Emissive power (e) : The amount of heat radiation emitted by unit area of the surface in one second at a particular temperature. SI UNIT : J/m2s 6. Spectral Emmisive power (e) : The amount of heat radiation emitted by unit area of the body in one second in  unit spectral region at a given wavelength. Emissive power or total emissive power e   e  d 0 SI UNIT : W/m2 Å EMISSIVITY (e) (i) Absolute emissivity or emissivity : Radiation energy given out by a unit surface area of a body in unit time corresponding to unit temperature difference w.r.t. the surroundings is called Emissivity. S I unit : W/m2 °K THERMAL PROPERTIES OF MATTER (ii) Relative emissivity (er) : er = Q GB e GB emitted radiation by gray body = = Q IBB E IBB emitted radiation by ideal black body GB = gray or general body, IBB = Ideal black body (i) No unit (ii) For ideal black body er = 1 (iii) range 0 < er < 1 Spectral, Emissive, Absorptive and Transmittive Power of a given body surface Due to incident radiations on the surface of a body following phenomena occur by which the radiation is divided into three parts. (a) Reflection (b) Absorption (c) Transmission • From Energy Conservation Q Q Q Q = Qr + Qa + Qt  r  a  t  1  r + a + t = 1 Q Q Q Qr Qa Reflective Coefficient r = , Absorptive Coefficient a = , Q Q amount of incident radiation Q amount of absorbed radiation Qa amount of transmitted radiation Q t Qt Transmittive Coefficient t = Q r = 1 and a = 0 , t=0 a = 1 and r = 0, t = 1 and a = 0, r=0 t=0 amount of reflected radiation Qr  Perfect reflector  Ideal absorber (ideal black body)  Perfect transmitter (diathermanous) Qr  Q   1 0 0  % , Absorption power (a) =  a  10 0  % Reflection power (r) =  Q  Q  Qt   10 0  % Transmission power (t) =  Q  Ex. Total radiations incident on body = 400 J, 20% radiation reflected and 120 J absorbs. Then find out % of transmittive power Sol. Q = Qt + Qr + Qa  400 = 80 + 120 + Qt  Qt = 200 . So % of transmittive poweris 50% IDEAL BLACK BODY (i) For a body surface which absorbs all incident thermal radiations at low temperature irrespective of their wave length and emitted out all these absorbed radiations at high temperature assumed to be an ideal black body surface. (ii) The identical parameters of an ideal black body is given by (iii) a = a = 1 and r = 0 = t, er = 1 The nature of emitted radiations from surface of ideal black body only depends on its temperature (iv) The radiations emitted from surface of ideal black body called as either full or white radiations. (v) At any temperature the spectral energy distribution curve for surface of an ideal black body is always continuous and according to this concept if the spectrum of a heat source obtained to be continuous then it must be placed in group of ideal black body like kerosene lamp; oil lamp Heating filament etc. (vi) There are two experimentally ideal black body (a) Ferry's ideal black body (b) Wien's ideal black body. Ferry's ideal black body (vii) At low temperature surface of ideal black body is a perfect absorber and at a high temperature it proves to be a good emitter. (viii) An ideal black body need not be black colour (eg. Sun) PREVOST'S THEORY OF HEAT ENERGY EXCHANGE According to Prevost at every possible temperature (Not absolute temperature) there is a continuous heat energy exchange between a body and its surrounding and this exchange carry on for infinite time. The relation between temperature difference of body with its surrounding decides whether the body experience cooling effect or heating effect. When a cold body is placed in the hot surrounding : The body radiates less energy and absorbs more energy from the surrounding, therefore the temperature of body increases. When a hot body placed in cooler surrounding : The body radiates more energy and absorb less energy from the surroundings. Therefore temperature of body decreases. When the temperature of a body is equal to the temperature of the surrounding The energy radiated per unit time by the body is equal to the energy absorbed per unit time by the body, therefore its temperature remains constant. ETOOS KEY POINTS  (i)  (ii) At absolute zero temperature (0 kelvin) all atoms of a given substance remains in ground state, so, at this temperature emission of radiation from any substance is impossible, so Prevost's heat energy exchange theory does not applied at this temperature, so it is called limited temperature of prevosts theory. With the help of Prevost's theory rate of cooling of any body w.r.t. its surroundings can be worked out (applied to Stefen Boltzman law, Newton's law of cooling.) KIRCHHOFF'S LAW At a given temperature for all bodies the ratio of their spectral emissive power (e) to spectral absorptive power (a) is constant and this constant is equal to spectral emissive power (E) of the ideal black body at same temperature e  e       = constant e  a  a  1  a  2 e  E  = constant a Good absorbers are good emitters and bad absorbers are bad emitters ETOOS KEY POINTS (i) For a constant temperature the spectral emmisive power of an ideal black body is a constant parameter (ii) The practical confirmation of Kirchhoff's law carried out by Rishi apparatus and the main base of this apparatus is a Lessilie container. The main conclusion predicted from Kirchhof's law can be expressed as (iii) Good absorber ƒ Good emitter Bad absorber ƒ Bad emitter (at Low temperature) (at high temperature) THERMAL PROPERTIES OF MATTER APPLICATIONS OF KIRCHOFF LAW (i) Fraunhoffer's lines Fraunhoffer lines are dark lines in the spectrum of the Sun. When white light emitted from the central core of the Sun (Photosphere) passes. Through its atmosphere (Chromosphere) radiations of those wavelengths will be absorbed by the gases present, resulting in dark lines in the spectrum of Sun. At the time of total solar eclipse direct light rays emitted from photosphere cannot reach on the Earth and only rays from chromosphere are able to reach on the Earth surface. At that time we observe bright fraunhoffer lines. (ii) K Ca Na vapour state chromosphere SUN photo sphere 10 K 7 6000K In deserts days are hot and nights cold Sand is rough and black, so it is a good absorber and hence in deserts, days (When radiation from Sun is incident on sand) will be very hot. Now in accordance with Kirchhoff's Law, good absorber is a good emitter. So nights (when send emits radiation) will be cold. STEFAN'S LAW The amount of radiation emitted per second per unit area by a black body is directly proportional to the fourth power of its absolute temperature. Amount of radiation emitted E where T = temperature of ideal black body (in K) E  This law is true for only ideal black body SI Unit : E = watt/m2 Stefen's constant = 5.67 x10–8 watt /m2 K4 Dimensions of : M1 L0 T–3 –4 Total radiation energy emitted out by surface of area A in time t : Ideal black body QIBB = A T4 t and for any other bodyQGB = erA T4 t Rate of emission of radiation When Temperature of surrounding T0 (Let T0 < T) Rate of emission of radiation from ideal black body surface E1 = T4 Rate of emission of radiation from surrounding E2 = T04 Net rate of loss of radiation from ideal black body surface is E = E1 – E2 = T4– T04 = ( T4 – T04 ) Net loss of radiation energy from entire surface area in time t is QIBB = A( T4 – T04 ) t For any other body QGB = er A( T4 – T04 ) t If in time dt the net heat energy loss for ideal black body is dQ and because of this its temperature falls by d Rate of loss of heat RH = dQ   A(T 4  T04 ) dt It is also equal to emitted power or radiation emitted per second Rate of fall in temperature (Rate of cooling) RF = d  d A 4  dQ  msJ   (T  T04 ) Q dt dt  dt ms J  Note : (i) If all of T, T0, m, s, V, , are same for different shape body then RF and RH will be maximum in the flat surface. (ii) If a solid and hollow sphere are taken with all the parameters same then hollow will cool down at fast rate. (iii) Rate of temperature fall , R F  1 d so dt  s. If condition in specific heat is  s1 > s2 > s3  s dt If all cooled same temperature i.e. temperature fall is also identical for all then required time t s  t1 > t2 > t3 When a body cools by radiation the cooling depends on : (i) (ii) (iii) (iv) (v) Ex. Nature of radiating surface : greater the emissivity (er), faster will be the cooling. Area of radiating surface : greater the area of radiating surface, faster will be the cooling. Mass of radiating body : greater the mass of radiating body slower will be the cooling. Specific heat of radiating body : greater the specific heat of radiating body slower will be the cooling. Temperature of radiating body : greater the temperature of radiating body faster will be the cooling. The operating temperature of a tungesten filament in an incandescent lamp is 2000 K and its emissivity is 0.3. Find the surface area of the filament of a 25 watt lamp. Stefan's constant  = 5.67 × 10–8 Wm–2 K–4 Sol.  Rate of emission = wattage of the lamp  W = AeT4  A = W eT 4 = 25 = 0.918 m2 0.3  5.67  10 8  (200 )4 NEWTON'S LAW OF COOLING Rate of loss of heat dQ I F G Hdt JKis directly proportional to excess of temperature of the body over that of surrounding. dQ dQ d (–0)  = ms dt dt dt = temperature of body [ in °C], = temperature of surrounding,–= excess of temperature  [(when (–0)  35°C] If the temperature of body decrease d in time dt then rate of fall of temperature – d  (  0 ) dt Where negative sign indictates that the rate of cooling is decreasing with time. Excess of temperature If the temperature of body decreases from 1 to 2 and temperature of surroundings is 0 then average excess  1  2   0   of temperature =   2  Ex. If a liquid takes 30 seconds in cooling of 80°C to 70°C and 70 seconds in cooling 60°C to 50°C, then find the room temperature. Sol.  1  2   1  2   t  = – K  2  0      1 –  2 t =K In first case,   F G H2 1 2 – 0 IJ K 80 – 70  80  70  – 0  =K    30 2 1 = K (75 – 0) 3 ...(i) THERMAL PROPERTIES OF MATTER In second case , 60 – 50 =K 70 60 50 F I1 G H 2 – J K7 0 ... (ii) = K (55 – 0) (7 5 –  0 ) 7 1 60 = (5 5 –  )  385 – 70 = 225 – 30 0 = = 40°C 3 4 0 Equation (i) divide by equation (ii) Limitations of Newton's Law (i) Temperature difference should not exceed 35° C, ( – 0)  35° C (ii) Loss of heat should only be by radiation. (iii) This law is an extended form of Stefan–Boltzman's law. For Heating, Newton's law of heating 1  2   2   where H heating constant.   H 0  1 t 2   Derivation of Newton's law from Steafen's Boltzman law d A 4  (T 4  T0 ) dt m s J  T  T0  T     T  T0  T  d A (T0  T)4  T04   dt m s J  d A  dt m s J If x <<< 1 then (1 + x)n = 1 + nx  4  A 4   A 4   T 4 T 4 T )  T04  = T0  (1  ) 1 = T0 1  4 1  T0 (1  T0 T0 T0   msJ   msJ   d   A 3  d  4 T0  T   K T dt  m s J dt  d Newton's law of cooling dt  T constant K  4  A T03 msJ (for small temperature difference) APPLICATION OF NEWTON'S LAW OF COOLING • To find out specific heat of a given liquid If for the two given liquids their volume, radiating surface area, nature of surface, initial temperature are allowed to cool down in a common environments then rate of loss of heat of these liquids are equal . stirrer air water ms  1 – 2 t1 air stirrer liquid m' s' 1 –  2 t2 air water out water in water and t for2 liquid fall of temperature from 1 2 in time t for 1  dQ   dQ      dt   Water  dt  Liquid  1  2        (m s+w)    (m 's ' w)    t1   t2   m 's' w msw = t2 t1 where w = water equivalent of calorimeter. Cooling curve : Ex. When a calorimeter contains 40g of water at 50°C, then the temperature falls to 45°C in 10 minutes. The same calorimeter contains 100g of water at 50°C, it takes 20 minutes for the temperature to become 45°C. Find the water equivalent of the calorimeter. Sol. m 1 s1  W m s W = 2 2 where W is the water equivalent t1 t2  40  1  W 10 0  1  W = 80 + 2W = 100 + W  W = 20 g 10 20 SPECTRAL ENERGY DISTRIBUTION CURVE OF BLACK BODY RADIATIONS Practically given by : Lumers and Pringshem Mathematically given by : Plank 1 T (i) m  (ii) E m  T 5  (iii) Area E 0  d  E  T 4 A1  T1    A 2  T2  4 THERMAL PROPERTIES OF MATTER ETOOS KEY POINTS (i) (ii) Spectral energy distribution curves are continuous. At any temperature in between possible wavelength (0 – ) radiation emitted but for different wavelength quantity of radiations are different. As the wave length increases, the amount of radiation emitted first increase, becomes maximum and then decreases. (iii) At a particular temperature the area enclosed between the spectral energy curve shows the spectral emissive power  of the body. Area   E  d  E  T 4 0 WEIN'S DISPLACEMENT LAW 1  The wavelength corresponding to maximum emission of radiation decrease with increasing temperature  m   . T  –3 This is known as Wein's displacement law. mT = b where b Wein's constant = = 2.89 x 10 mK. Dimensions of b : = M0 L1 T0  Relation between frequency and temperature  m  Ex. c T b The temperature of furnace is 2000°C, in its spectrum the maximum intensity is obtained at about 4000Å, If the maximum intensity is at 2000Å calculate the temperature of the furnace in °C. Sol. By using mT = b, 4000 (2000+273) = 2000(T)  T = 4546K The temperature of furnace = 4546 – 273 = 4273 °C SOLAR CONSTANT 'S' The Sun emits radiant energy continuously in space of which an in significant part reaches the Earth. The solar radiant energy received per unit area per unit time by a black surface held at right angles to the Sun's rays and placed at the mean distance of the Earth (in the absence of atmosphere) is called solar constant. The solar constant S is taken to be 1340 watts/m2 or 1.937 Cal/cm2–minute • Temperature of the Sun Let R be the radius of the Sun and 'd' be the radius of Earth's orbit around the Sun. Let E be the energy emitted by the Sun per second per unit area. The total energy emitted by the Sun in one second = E.A = E × 4R2. (This energy is falling on a sphere of radius equal to the radius of the Earth's orbit around the Sun i.e., on a sphere of surface area 4d2) d So, The energy falling per unit area of Earth = R = 7× 108m , By Stefan's Law E = T4 d2 E R2 d2 R E R2 S= S= 4 d 2 = d = 1.5 × 1011 m, s = 5.7 × 10–8 W m–2 K–4 Solar constant  T 4 R2 4 R 2  E T d2 1 A 1  S  d 2  4  134 0  (1.5  1 0 11 )2  4  T=  = 5732 K 2  8 8 2    R   5 .7  10  (7  10 )  A'