Modes of Heat Transfer

MODES 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 : 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. 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). 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 inter-venning medium. 3.1 Thermal conduction The process by which heat is transferred from hot part to cold part of 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 position is called conduction. The process of conduction takes place only in solid body (except Hg). Heat transfer by conduction from one part of body to another continues till their temperatures become equal. For example if you hold an iron rod with one of its end on a fire for some time the handle will become 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 relatively higher value 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. Consider a slab of face area A, Lateral thickness L, whose faces have temperatures T H and TC(T H > T c)· 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 Q KA(TH  TC )   Ka  or t dx t L K  Thermal conductivity It is the measure of the ability of material to conduct the heat. Here K is a constant depending on nature of the material of the slab and is named thermal  dT  conductivity of the material, and the quantity   is called temperature gradiant. The (–)  dx  sign shows heat flows from high temperature to low 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 and temperature of rod is not same. 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 faces are maintained at constant temperatures TH 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. dT T Tf  Ti TC  t H Hence,    dx L L L Q T Q  T  TC  and  = KA  H   KA  t L t  L  Here Q is the amount of heat flowing through a cross-section of slab at any position in a time interval t. Thermal conductivity (K) : • It depends on nature of material. • Order of thermal conductivity Ag > Cu > Au > Al For Ag maximum is (410 W/mK) K For Freon minimum is 12 (0.008 W/mK) • SI Unit : J s–1 m–1 K–1 Dimensions : M1 L–3 – • 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 • 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. • Two thin blankets are warmer than a single blanket of double the thickness. • Birds often swell their feathers in winter . • 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 tiffin-box, 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 (T  TL ) In terms of R, the amount of heat flowing though a slab in steady-state (in time t) =  H t R T  TL Q If we name as thermal current iT then, iT = H R t This is mathematically equivalent to OHM's law, with temperature doing 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 i L remains same at each cross-section. This is analogous to Kirchhoff's current law in electricity, which can be very conveniently applied to thermal conduction. For a slab of cross-section A, Lateral thickness L and thermal conductivity K, ● R Equivalent conductivity for Heat flow through slabs in series Req = R1 + R2 L1  L2 L L  1  2 K eq A K1A K 2 A Equivalent thermal conductivity of the system is L  L2  Li Keq = 1  L1 L 2 L   i K1 K 2 Ki ● Equivalent thermal conductivity for Heat flow through slabs in parallel L R= KA 1 1 1   R eq R1 R 2 K eq K1A1 K 2 A 2  L L L Equivalent thermal conductivity (A1  A 2 )  Keq = K1A1  K 2 A 2  K i A i  A1  A 2  Ai 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 Heat escaping through ice in time dt is [0  ()] dt x Due to escape of this heat increasing extra thickness of ice = dx Mass of this extra thickness of ice is m = V = A.dx dQ = mL = ( A.dx) L  L  KA = ( A.dx) L  dt= x dx x K L x 1 L 2 So time taken by ice to grow a thickness x is t = x dx  x  0 K 2 K So time taken by ice to grow from thickness x1 to thickness x2is 1 L 2 t = t2 – t1 = t  (x 22  x12 ) (x 2  x12 ) and 2 KT Time taken to double and triple the thickness ratio t1 : t2 : t3 :: 12 : 22 : 32 So t1 : t2 : t3 :: 1 : 4 : 9 dQ = KA 3.2 (i) Convection 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. Phenomena Based on convection : 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 see-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 towards 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 ant-clockwise rotation of Earth the warm wind blowing from equator to north drifts towards east. The steady wind blowing from north-East 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) Ventillation : Ventillator 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 forced 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. ● ● ● ● ● ● ● ● Some important points : Natural convection takes place from bottom to top due to gravity 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. Natural convection is not possible in a gravity free region such as a freely falling lift or an orbiting satellite. Natural convection plays an important role in ventilation, in changing climate and weather and in forming land and sea breezes and trade winds. The forced convection of blood in our body by a pump (heart) helps in keeping the temperature of body constant. Convection takes place in fluids (gases and liquid) GOLDEN KEY POINTS For heat propagation via natural convection, temperature gradient exists in vertical direction and not in horizontal direction. Most of heat transfer that is taking place on Earth is by convection, the contribution due to conduction and radiation is very small. Illustrations Illustration 20. One face of an aluminium cube of edge 2 meter 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) Solution: Heat will flow from the end at 100°C to the end at 0°C. (T  T ) Q Area of cross-section perpendicular to direction of heat flow, A = 4m2 then  KA H C t L 2 (209W / mC)(4m )(100C  0C)(5sec) Q= = 209 kJ 2m Illustration 21. Three identical rods of length lm each, having crosssection area of l cm2 each and made of Aluminium, copper and steel respectively are maintained at temperatures of l2°C, 4°C and 50°C 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] Solution: L 1 104 RAl =   KA 200 10 4 200 104 10 4 Similarly Rsteel = and Rcopper = 50 400 Let temperature of common junction = T then from Kirchhoff’s current law, iAl + isteel + iCu = 0 T  12 T  50 T  4   0  R Al R steel R Cu  (T –12) 200 + (T – 50) 50 + (T – 4) 400 = 0  4(T – 12) + (T – 50) + 8 (T – 4) = 0  13T = 48 + 50 + 32 = 130  T = 10°C Illustration 22. The thermal conductivity of brick is 1.7 W m–l 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? Assuming their area to be same. Solution: KA(T1  T2 )t Since Q = . For same insulation by the brick and cement ; Q, A (T1 – T2) and t L do not change. K Hence, remain constant. If K1 and K2 be the thermal conductivities of breick and cement L K K 1.7 2.9  respectively and L1 and L2 be the required thickness then 1  2 or L1 L 2 20 L 2 2.9  L2 = × 20 = 34.12 cm. 1.7 Illustration 23. Two vessels of different materials are identical in size and wall-thickness. They are filled with equal quantities of ice at 0°C. If the ice melts completely, hi 10 and 25 minutes respectively then compare the coefficients of thermal conductivity of the materials of the vessels. Solution: 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 L in both the cases is same. K t 25 min 5 K A(1  2 )t1 K 2 A(1  2 )t 2   Now, Q = 1  1  2  K 2 t1 10 min 2 L L Illustration 24. 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. Solution: K A(   2 ) Q K1A(1  )  Rate of flow of heat I the plates is = 2 ….(1) t L1 L2 (i) Here 1 = –20°C, 2 = 20°C L1 = 2 cm = 0.02 m, L2 = 5 cm = 0.05 m and K1 = K2 = K KA(20  ) KA(  20)  0.02 0.05  5(–20 – ) = 2( – 20)  –100 – 5 = 2 – 40  7 = –60  = –8.6°C K1 2 2  or K1 = K2 K2 5 5  equation (i) becomes (ii)  from equation (i) 2 / 5K 2 A( 20  ) K 2 A(  20)  –20 –  =  – 20  = 0°C  0.02 0.05 Illustration 25. Two thin concentric shells made of copper with radius r1 and r2 (r2 > r1) have a material of thermal conductivity K filled between them. The inner and outer spheres are maintained at temperatures TH and TC respectively by keeping a heater of power P at the centre of the two spheres. Find the value of P. Solution : Heat flowing per second through each cross-section of the sphere = P = i. Thermal resistance of the spherical shell of radius x and thickness dx, dx dR = K .4x 2 r2 dx 1 1 1   R=     2 4x .K 4K  r1 r2  r1 Thermal current T  TC 4K(TH  TC )r1r2  i=P= H R (r2  r1 ) Illustration 26. 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 ? Solution: On heating the liquid at A will become lighter and will rise up. This will push the liquid in the tube upwards and so the liquid in the tube will move clockwise i.e. from B to A. BEGINNER’S BOX - 3 1. Explain why : (a) A brass tumbler feels much colder than a wooden tray on a chilly day (b) Two layers of cloth of equal thickness provide warmer covering than a single layer of cloth of double thickness ? (c) Mud-houses are colder in summer and warmer in winter ? (d) In winter birds sit with their wings spread out ? (e) Woollen clothes are warmer than cotton clothes ? 2. Two metal cubes with 3 cm-edges of copper and aluminium are arranged as shown in figure. Find (a) (b) 3. The total thermal current from one reservoir to the other. The ratio of the thermal current carried by the copper cube to that carried by the aluminium cube. Thermal conductivity of copper is 60 W/m-K and that of aluminium is 40 W/m-K. For shown situation, calculate the temperature of the common interface. 4. The temperature at the ends of a uniform rod of length 100 cm are respectively 95°C and 5°C. What will be the temperature at a point 30 cm far from the hotter end ? Also calculate the temperature gradient. 5. Three conducting rods of same material and cross-section are shown in figure. Temperature of A, D and C are maintained at 20°C, 90°C and 0°C. Find the ratio of length BD and BC if there is no heat flow in AB. 3.3 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 radiation by which 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, its shadow is formed at the detector. (b) (c) (d) (e) (f) It is reflected and refracted or can be made to interfere. The reflection or refraction are exactly as in case of light. It can travel through vacuum. Intensity of radiation follows the law of inverse square. Thermal radiation can be polarised in the same way as light by transmission through a nicol prism. Radiation takes place in solid, liquid & gases. 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 thermal radiation is larger than that of visible light. ● ● ● ● ● ● ● When radiation passes through any medium then radiations slightly absorbed medium according to its absorptive power so temperature of medium slightly increases. In order to obtain a spectrum of radiation, a special prism is used like KC Rock salt prism Flourspar prism. Normal glass prism or Quartz prism can not be used (because it absorbs some radiation). Radiation intensity is measured with a specific device named as Bolometer. Heat radiation are always obtained in infra-red region of electromagnetic wave spectrum so they are called Infra-red rays. Thermal radiation when incident on a surface, then exert pressure on the surface which is known as Radiation Pressure. Basic Fundamental definitions 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 Q of absorption a = a . It is unitless and dimensionless. Q Qa  Spectral absorptive power (a) a   : also called monochromatic absorptive coefficient Q  At a given wavelength a =  a  d . For ideal black body a and a = 1, a and a are unitless 0 ● ● Emissive power (e) : The amount of heat radiation emitted by unit surface area in unit second at a particular temperature. SI UNIT: J/m2-s or watt/m2 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 2 ● SI UNIT : W/m –Å Emissivity (e) 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 ● ● Q GB e emitted radiation by gray body  GB  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 Relative emissivity (er) : er = Q r Qa Q t   =1 r+a+t=1 Q Q Q Q Q Reflective Coefficient r = r , Absorptive Coefficient a = a , Q Q Q Transmittive Coefficient t = t Q r = 1 and a = 0, t=0  Perfect reflector a = 1 and r = 0, t=0  Ideal absorber (ideal black body) t = 1 and a = 0, r=0  Perfect transmitter (diathermanous) Q  Q  Reflection power (r) =  r  100  %, Absorption power (a) =  a 100  % Q  Q  Q = Qr + Qa + Qt  Q  Transmission power (t) =  t  100  % Q  Ideal Black Body • A body surface which absorbs all incident thermal radiations at low temperature, irrespective of their wave length and emits out all these absorbed radiations at high temperature is assumed to be an ideal black body surface. • The identical parameters of. an ideal black body is given by 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 • • • • • The radiations emitted from surface of ideal black body are called as either full or white radiations. 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 is continuous then it must be ideal black body like kerosene lamp; oil lamp, heating filament etc. There are two experimentally ideal black body (a) Ferry's ideal black body (b) Wien's ideal black body. At low temperature, surface of ideal black body is a perfect absorber and at a high temperature it proves to be a perfect emitter. An ideal black body need not be of black colour (eg. Sun). Prevost's theory of heat energy exchange According to Prevost, at every possible temperature (except ze.ro kelvin temperature) there is a continuous heat energy exchange between 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. (Heating effect)· When a hot body placed in cooler surrounding: The body radiates more energy and absorbs less energy from the surroundings. Therefore temperature of body decrease (cooling effect) 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 and the body is in thermal equilibrium with surrounding. Hence no heating and cooling effects are seen. 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  e = E = constant Hence e µ a      = constant a a a   1    2 Good absorbers are good emitters and bad absorbers are bad emitters ● Applications of Kirchoff Law Fraunhoffer's lines Fratinhoffer 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) some of the radiations are 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. ● 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 ideal black body is directly proportional to the fourth power of its absolute temperature. Amount radiation emitted E T4 where T = temperature of ideal black body (in K) E =  T4 (This law is true for only ideal black body) 2 SI Unit : E = watt/m  Stefen's constant = 5.67 × 10–8 watt /m2 K4 (universal constant) 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 = e  A T4 t Rate of emission of radiation When temperature of surrounding T0 (Let T0 < T) Rate of emission of radiation from per unit area of ideal black body surface E1 =  T4 Rate of emission or absorption of radiation (per unit area) from surrounding E2 =  T04 Net rate of loss of radiation per unit area from ideal black body surface is E = E1 =  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 dT dQ Rate of loss of heat (IBB) RH = = A(T4 – T04 ) J/s dt It is also equal to emitted power or radiation emitted per second dT A 4 dQ dT   ms Rate of fall in temperature (Rate of cooling) RF = (T – T04 )  dt ms dt dt Where T and T0 absolute temperature of body and surrounding, M = mass of body, s = Specific heat,  = density, V = volume ● When a body cools by radiation, its cooling depends on : (i) Nature of radiating surface : greater the emissivity (er), faster will be the cooling. (ii) Area of radiating surface : greater the area of radiating surface faster will be the cooling. (iii) Mass of radiating body : greater the mass of radiating body slower will be the cooling. (iv) Specific heat of radiating body : greater the specific heat of radiating body slower will be the cooling. (v) Temperature of radiating body : greater the temperature of radiating. body faster will be the cooling. NEWTON’S LAW OF COOLING  dT  Rate of cooling   is directly proportional to excess of temperature of the body over that  dt  of surrounding dT [(when (T – T)  35°C]  (T – T0)  dt T = temperature of body [all temperatures in °C] T0 = temperature of surrounding, T – T0 = excess of temperature (T > T0) If the temperature of body is decreased by dT in time dt then rate of fall of temperature dT = K (T – T0)  dt Where negative sign indictates that the rate of cooling is decreasing with time. For Numerical Problems, Newton’s Law of cooling If the temperature of body decreases from T1 to T2 and temperature of surroundings is T0 then T  T  Average excess of temperature =  1 2  T0  2    T1  T2   T1  T2    t    K  2  T0  Limitations of Newton’s Law of Cooling • Temperature difference should not exceed 35° C, (T – T0)  35°C • Loss of heat should only be by readiation. • This law is an extended form of Stefan-Boltzman's law. Derivation of Newton’s law from Stefan’s Boltzman law T  T0  T  dT A 4  (T  T04 )   dt ms T  T0  T  dT A  [(T0 + T)4 – T04 ] if x <<< 1 then (1 + x)n = 1 nx (Binomial theorem) dt ms  A 4  dT A  4 T 4 T  T 4  A 4   )  T04  = T0 (1  )  1 = T0 1  4  1 T0 (1  dt ms  T0 ms  T0 ms  T0    4AT03 dT dT  A 3   4 T0 T  = KT here constant K = dt  ms  dt ms dT Newton’s law of cooling  T (for small temperature difference) dt Spectral Energy distribution curve of Black Body radiations Edubull Results from these Graphs 1 (i) m  Wien’s displacement law T m T = b m T1 =  m 2 T2 (ii) E  m  T5 (iii) i.e. Area  E  d = E T4 Area between curve and  axis gives the emissive power of body 0 A T  Hence 1   1  A 2  T2  4 Wein’s Displacement Law The wavelength corresponding to maximum emission of radiation decrease with increasing 1  temperature   m   . This is known as Wein’s displacement law. T  m T = b where b is Wein’s constant = 2.89 × 10–3 mK 0 1 0 1 Dimensions of b : M L T  c Relation between frequency and temperature vm = T [c =  × v] b ● 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 watt/m 2 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) So, The energy falling per unit time per unit area of Earth 4R 2  E ER 2 =  2 4d 2 d 8 R = 7 × 10 m, d = 1.5 × 1011 m,  = 5.7 × 10–8 W m–2 K–4 ER 2 Solar constant S= 2 d By Stefan’s Law E = T4 1 1  S  d 2  4  1340  (1.5 1011 ) 2  4 T R S=  T =    R 2    5.7  10 8  (7 108 ) 2  = 5732 K d2     4 ● 4 GOLDEN KEY POINT At absolute zero temperature (zero kelvin) all atoms of a given substance remains in ground state, so, at this temperature emission and absorption of radiation from any substance is impossible, so Prevost's heat energy exchange theory does not applied at this temperature, so it is called limiting temperature of prevost's 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 Stefan Boltzman law, Newton's law of cooling.) ● For a constant temperature the spectral emmisive power of an ideal black body is a constant parameter ● 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 Good absorber Good emitter ƒ Bad absorber Bad emitter ƒ (at Low temperature) (at high temperature) ● If all of T, T0, m, s, V, , are same for different shape body then RF and RH will be maximum for the flat surface. ● If a solid and hollow sphere are taken with all the parameters same then hollow will cool down at fast rate. 1 dT Rate of temperature fall, RF   dt s so dt  s. If condition of specific heat is s1 > s2 > s3 and if all cooled same temperature i.e. temperature fall is also identical for all then required time t  s  t1 > t2 > t3 ● ● Spectral energy distribution curves are continuous. At any temperature in all possible wavelength radiation between (0 – ) are emitted but quantity of radiation are different for different wavelength. ● As the wave length increases the amount of radiation emitted first increase, becomes maximum and then decreases. ● At a particular temperature the area enclosed between the spectral energy curve and wavelength axis shows the emissive power of the body.  Area = E  d = E = T4 0 Illustrations Illustration 27. If m for the moon is 14.5 micron, then find its temperature. Solution: Wien’s displacement law m T = b 3 b 2.89 10  T= = 199.3K  m 14.5 106 Illustration 28. Total radiation incident on body is 400 J, If 20% of incident radiation reflected back and 120 J is absorbed by body. Then find out transmittive power in percentage. Solution: Q = Qt + Qr + Qa  400 = 80 + 120 + Qt Qt = 200 Q So transmittive power is t × 100% = 50% Q Illustration 29. 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 s = 5.67 × 10–8 Wm–2 K–4 Solution:  Rate of emission = wattage of the lamp W 25   W = AeT4  = 0.918 × 10–4 m2 4 8 4 eT 0.3  5.67 10  (2000) Illustration 30. Draw a graph between log E and log T Solution: E =  T4 (taking log) logE = log (T4) logE = 4logT + log This is equivalent to y = mx – C ( < 1 so its log is negative quantity) Illustration 31. If temperature of ideal black body is increased by 50%, what will be percentage increase in quantity of radiations emitted from its surface. Solution: 4 4 81 4 15  3 E'  (1.5)4T4    T4    T4  T 16 10  2  81 4  T  T4   E ' E × 100% =  16 4  × 100% = 406 %  400 % E T     E  T4 and  Illustration 32. If temperature of ideal black body is decreased from T to T then find out percentage loss in 2 emissive rate Solution: 4 T4 T  E'    16 2 15 1  E  E '   E  × 100% = 1  16  × 100% = 16 × 100%  94% E  T4, Remaining is 6% (Approx.) Illustration 33. Calculate the temperature at which a perfect black body radiates at the rate of 5.67 W cm –2. Stefan’s constant is 5.67 × 10–8 J s–1 m–2 K–4. Solution: Given E = 5.67 W cm–2 = 5.67 × 10+4 W m–2,  = 5.67 × 10–8 J s–1 m–2 K–4 1 Using, E = T4; E T4 =  1 4  E  4  5.67 10  4 or T =     = (1012)1/4 = 1000 K 8   5.67  10     Illustration 34. 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. Solution: by using m T = b, 4000 (2000 + 273) = 2000(T)  T = 4546K. The temperature of furnace = 4546 – 273 = 4273 °C Illustration 35. Two bodies A and B have thermal emissivities of 0.01 and 0.81 respectively. The outer surface areas of the two bodies are same the two bodies emit total radiant power at the same rate. The wavelength B corresponding to maximum spectral radiancy of B is shifted from the wavelength corresponding to maximum spectral radiancy in the radiation of A by 1.0 m. If the temperature of A is 5802K, Calculate :(a) The temperqtt.ire of B (b) Wavelength B Solution: (a) As both bodies A and B having same radiant poweer  PA = PB  eAAA TA4 = eBAB TB4  (0.01)A TA4 = (0.81)A TB4 /4 (b) TA 5802  0.01  TB =  = 1934 K   TA  3 3  0.81  According to wein’s displacement law  5802  ATA = BTB    A = 3A  1934   2 B As B – A = 1 m  B – B = 1 m  = 1 m B = 1.5 m. 3 3 Illustration 36. When a metallic body is heated in a furnace, then what colour will appear as temperature increases. Solution: 1 T m As Temperature increases colour of body will appear from red, yellow, green, blue and the white. Illustration 37. Define (i) Steady state and (ii) Temperature gradient in conduction of heat through a conducting rod. Solution: (i) When one end of a rod is heated, the temperature of various points of the rod changes continuously but after some time a state is reached, when the temperature of each cross-section becomes steady which is called steady state. In this state the heat received by any section will be totally transferred to the next section so no heat is absorbed by any cross section . (ii) Temperature gradient is defined as the rate of change of temperature with distance in the direction of flow of heat. Illustration 38. Assuming Newton’s law of cooling to be valid. The temperature of body changes from 60°C to 40°C in 7 minutes. Temperature of surroundings being 10°C, find its temperature after next 7 minutes. Solution: T t T T  According to Newton’s low of cooling 1 2  K  1 2  T0  t  2  Since the temperature decreases from 60°C to 40°C in 7 minutes 60  40 1  60  40   K  10   K (50 – 10)  K = 7 14  2  40  T2 1  40  T2     10  If the temperature of object becomes T2 in next 7 minutes then 7 14  2  1  40 – T2 = (40 + T2 – 20)  160 – 4T2 = 20 + T2 4  5T2 = 140  T2 = 28°C 1. BEGINNER’S BOX - 4 Explain why : (a) a body with large reflectivity is a poor emitter. (b) an optical pyrometer (for measuring high temperatures) calibrated for an ideal black body radiation gives too low value for the temperature of a red hot iron piece in the open, but gives a correct value for the temperature when the same piece is in the furnace (c) the earth without its atmosphere would be inhospitably cold (d) Heat is generated continuously in an electric heater but its temperature remains constant after some time. (e) The bottom of a cooking vessel is made black ? (f) On winter nigh you feel warmer when clouds cover the sky than when the sky is clear. (g) A thermos or vacuum flask can keep hot things hot and cold things cold for a long time, how? 2. Two spherical ideal black bodies of radii r1 and r2 are having surface temperature T1 and T2 T respectively, If both radiate the same power. Then calculate the ratio of 1 . T2 3. If a liquid takes 30 sec. in cooling from 80°C to 70°C and 70 sec in cooling from 60°C to 50°C, then find the room temperature. 4. A body cools in 10 minutes from 60°C to 40°C. What will be its temperature after next 10 minutes? The temperature of the surrounding is 10°C. 5. Calculate the temperature of the black body from given graph.