Physics Notes for NEET Chapter 15 Transmission of Heat PDF

60 Transmission of Heat 695 Chapter E3 15 Transmission of Heat (8) In non-metallic solids and fluids the conduction takes place only due to vibrations of molecules, therefore they are poor conductors. ID Heat energy transfers from a body at higher temperature to a body at lower temperature. The tran...

60 Transmission of Heat 695 Chapter E3 15 Transmission of Heat (8) In non-metallic solids and fluids the conduction takes place only due to vibrations of molecules, therefore they are poor conductors. ID Heat energy 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 one of the following modes. Conduction, Convection and Radiation (9) In metallic solids free electrons carry the heat energy, therefore they are good conductor of heat. U Conduction in Metallic Rod Conduction D YG When one end of a metallic rod is heated, heat flows by conduction from the hot end to the cold end. U The process of transmission of heat energy in which the heat is transferred from one particle to other particle without dislocation of the particle from their equilibrium position is called conduction. ST (1) Heat flows from hot end to cold end. Particles of the medium simply oscillate but do not leave their place. (2) Medium conduction is necessary Radiation A Conduction Hot end 15.2 (1) Variable state : In this stateFig.Temperature of every part of the rod increases Heat received by each cross-section of the rod from hotter end used in three ways. (i) A part increases temperature of itself. (ii) Another part transferred to neighbouring cross-section. (iii) Remaining part radiates. 1 for   Changing 3 4 5 Metallic rod Cold Hot Fig. 15.3 Fig. 15.1 (5) Conduction is a process which is possible in all states of matter. (6) When liquid and gases are heated from the top, they conduct heat from top to bottom. (7) In solids only conduction takes place 2  1 > 2 > 3 > 4 > 5 (3) It is a slow process (4) The temperature of the medium increases through which heat flows Cold end (2) Steady state : After sometime, a state is reached when the temperature of every cross-section of the rod becomes constant. In this state, no heat is absorbed by the rod. The heat that reaches any crosssection is transmitted to the next except that a small part of heat is lost to surrounding from the sides by convection & radiation. This state of the rod is called steady state. (3) Isothermal surface : Any surface (within a conductor) having its all points at the same temperature, is called isothermal surface. The direction Plane isothermal surfaces Cylinderical isothermal surfaces 696 Transmission of Heat of flow of heat through a conductor at any point is perpendicular to the isothermal surface passing through that point. (6) More about K : It is the measure of the ability of a substance to conduct heat through it. (i) Units : Cal/cm-sec C (in C.G.S.), kcal/m-sec-K (in M.K.S.) and o W/m- K (in S.I.). Dimension : [MLT 3 1 ] (ii) The magnitude of K depends only on nature of the material. (iii) Substances in which heat flows quickly and easily are known as good conductor of heat. They possesses large thermal conductivity due to large number of free electrons e.g. Silver, brass etc. For perfect conductors, K . 60 (iv) Substances which do not permit easy flow of heat are called bad conductors. They possess low thermal conductivity due to very few free electrons e.g. Glass, wood etc. and for perfect insulators, K  0. Spherical isothermal surfaces S (C) Due to point source of heat Fig. 15.4 (4) Temperature gradient (T.G.) : The rate of change of temperature with distance between two isothermal surfaces is called temperature gradient. Hence  Heat (b) KSolid  KLiquid  KGas 2 x (c) KMetals  K Non metals U l Fig. 15.5 (i) Temperature gradient =   x D YG 1   2 l   x 1 (iv) Unit : K/m or °C/m (S.I.) and Dimensions [L  ] (5) Law of thermal conductivity : Consider a rod of length l and area of cross-section A whose faces are maintained at temperature  and  respectively. The curved surface of rod is kept insulated from surrounding to avoid leakage of heat 1 U 1 Q Q A 2 Q ST l (i) In steady state the amount of heat flowing from one face to the KA( 1   2 ) t other face in time t is given by Q  l where K is coefficient of thermal conductivity of material of rod. (ii) Rate of flow of heat i.e. heat current Thermal conductivity Substance Thermal conductivity (W/m-K) (W/m-K) Aluminium 240 Concrete 0.9 Copper 400 Water 0.6 Gold 300 Glass wool 0.04 Iron 80 Air 0.024 Lead 35 Helium 0.14 Glass 0.9 Hydrogen 0.17 Wood 0.1-0.2 Oxygen 0.024 2 Fig. 15.6 KA(1   2 ) Q H  l t (iii) In case of non-steady state or variable cross-section, a more general equation can be used to solve problems. dQ d   KA dt dx Table 15.1 : Thermal conductivity of some material Substance (ii) The negative sign show that temperature  decreases as the distance x increases in the direction of heat flow. (iii) For uniform temperature fall (vii) Decreasing order of conductivity : For some special cases it is as follows (a) K Ag  KCu  K Al  –  A 1 (vi) Human body is a bad conductor of heat (but it is a good conductor of electricity). ID Heat E3 (v) The thermal conductivity of pure metals decreases with rise in temperature but for alloys thermal conductivity increases with increase of temperature. (7) Relation between temperature gradient and thermal conductivity : dQ d In steady state, rate of flow of heat = – KA  (T.G.)    KA dt dx 1 dQ ( = constant) (T.G.)  dt K Temperature difference between the hot end and the cold end in steady state is inversely proportional to K, i.e. in case of good conductors temperature of the cold end will be very near to hot end. In ideal conductor where K = , temperature difference in steady state will be zero. (8) Thermal resistance (R) : The thermal resistance of a body is a measure of its opposition to the flow of heat through it. It is defined as the ratio of temperature difference to the heat current (= Rate of flow of heat) (i) Hence R  1   2 H  1   2 l  KA(1   2 ) / l KA Transmission of Heat 697 Dimension : walls prevents transmission of heat from the house to the cold surroundings. (9) Wiedmann-Franz law : At a given temperature T, the ratio of thermal conductivity to electrical conductivity is constant i.e., (K / T ) = constant, i.e., a substance which is a good conductor of heat (e.g., silver) is also a good conductor of electricity. Mica is an exception to above law. For exactly the same reason, two thin blankets are warmer than one blanket of their combined thickness. The layer of air enclosed in between the two blankets makes the difference. o  C s e/ c a l or K  sec / kcal and [M 1 L2 T 3 ] (10) Thermometric conductivity or diffusivity : It is a measure of rate of change of temperature (with time) when the body is not in steady state (i.e., in variable state) It is defined as the ratio of the coefficient of thermal conductivity to the thermal capacity per unit volume of the material. Thermal capacity per mc unit volume = = c V Davy's safety lamp has been Fig. 15.9 designed on this principle. The gases in the mines burn inside the gauze placed around the flame of the lamp. The temperature outside the gauze is not high, so the gases outside the gauze do not catch fire. K c E3 (  = density of substance)  Diffusivity (D) = Unit : m /sec and Dimension : [L2T 1 ] 2 Table 15.2 : Electrical Analogy for Thermal Conduction Thermal conduction Electric charge flows from higher potential to lower potential Heat flows from higher temperature to lower temperature The rate of flow of charge is called the electric current, The rate of flow of heat may be called as heat current i.e. H  dQ dt Similarly, the heat current may be related with the temperature where R is the electrical resistance of the conductor where R is the thermal resistance of the conductor The electrical resistance is defined as The thermal resistance may be l l R  A A D YG The relation between the electric current and the potential difference is given by Ohm's law, that is V  V2 I 1 R where K = Thermal conductivity of conductor dQ    2 KA (1   2 ) H 1  dt R l ST dq V  V2  A I 1  (V1  V2 ) dt R l 1 2 3 n – 1 K1 K2 Kn l1 l2 ln R l defined as R  KA = (1) Series combination : Let n slabs each of cross-sectional area A, lengths l1 , l2 , l3 ,...... ln and conductivities K1 , K2 , K3...... Kn respectively be connected in the series 1   2 U where  = Resistivity and  Electrical conductivity difference as H  Combination of Metallic Rods U dq dt (5) Birds often swell their feathers in winter. By doing so, they enclose more air between their bodies and the feathers. The air, being bad conductor of heat prevents the out flow of their body heat. Thus, birds feel warmer in winter by swelling their feathers. ID Electrical conduction i.e. I  (4) Wire gauze is placed over the flame of Bunsen burner while heating the flask or a beaker so that the flame does not go beyond the gauze and hence there is no direct contact between the flame and the flask. The wire gauze being a good conductor of heat, absorb the heat of the flame and transmit it to the flask. 60 (ii) Unit : Applications of Conductivity in Daily Life. (1) Cooking utensils are provided with wooden handles, because wood is a poor conductor of heat. The hot utensils can be easily handled from the wooden handles and our Wooden handle hands are saved from burning. Frying pan (i) Fig. 15.10 Heat current : Heat current is the same in all the conductors.i.e., Q  H1  H 2  H 3.........  H n t K1 A(1   2 ) K 2 A( 2   3 ) Kn A( n 1   n )   l1 l2 ln (ii) Equivalent thermal resistance : R  R1  R2 ..... Rn (iii) Equivalent thermal conductivity : It can be calculated as follows From RS  R1  R2  R3 ... l1  l2 ... ln l l l  1  2 ....  n Ks K1 A K 2 A Kn A  Ks  l1  l2 ...... ln l1 l l  2 ........ n K1 K 2 Kn (2) We feel warmer in a fur Fig. 15.7 coat. The air enclosed in the fur coat being bad conductor heat does not allow the body heat to flow outside. Hence we feel warmer in a fur coat. (a) For n slabs of equal length K s  (3) Eskimos make double walled houses of the blocks of ice. Air enclosed in between the double (b) For two slabs of equal length, K s  Fig. 15.8 n n 1 1 1 1   ..... K1 K 2 K 3 Kn 2 K1 K 2 K1  K 2 698 Transmission of Heat (iv) Temperature of interface of composite bar : Let the two bars are arranged in series as shown in the figure. 1  2 shown in the figure, then the ratio of thermal conductivities is K1 : K2 : K3  l12 : l22 : l32  Thermal conductivity (K)  (Melted length l ) 2 K1 K2 Searle's Experiment l2 l1 It is a method of determination of K of a metallic rod. Fig. 15.11 3 Then heat current is same in the two conductors. Q K1 A(1   ) K 2 A(   2 )   t l1 l2 1 2 (b) If K = K and l = l then   1 2 1 Water K2 2 l2 K2 l2 l K11  K 2 2 K1  K 2 (a) If l = l then   1   2 2 2 l 2 A1 K1 K2 by water is given by, Q  mc( 4   3 )......(ii) (3) Where, m is the mass of the water which has absorbed this heat and temperature is raised and c is the specific heat of the water U A2 A3 D YG K3 A3 Kn (i) Equivalent resistance : For two slabs Rs  Steam 15.14 (1) In this experiment a Fig. temperature difference (1   2 ) is maintained across a rod of length l and area of cross section A. If the thermal conductivity of the material of the rod is K, then the amount of heat transmitted by the rod from the hot end to the cold end in time t is KA(1   2 ) t......(i) given by, Q  l (2) In Searle's experiment, this heat reaching the other end is utilized to raise the temperature of certain amount of water flowing through pipes circulating around the other end of the rod. If temperature of the water at the inlet is  3 and at the outlet is  4 , then the amount of heat absorbed ID (2) Parallel Combination : Let n slabs each of length l, areas A1 , A2 , A3 ,..... An and thermal conductivities K1 , K2 , K3 ,..... Kn are connected in parallel then 1 2 60 K1 1  l By solving we get   1 K1  l1 1 E3 i.e., 4 Steam 1 Fig. 15.12 1 1 1 1    ..... Rs R1 R2 R3 Rn R1 R2 R1  R2 U (ii) Temperature gradient : Same across each slab. (iii) Heat current : in each slab will be different. Net heat current will be the sum of heat currents through individual slabs. i.e., H  H1  H 2  H 3 .... Hn K( A1  A2 .....  An ) (1   2 ) l K1 A1 (1   2 ) K 2 A2 (1   2 ) K A (   2 ) =  ...  n n 1 l l l K1 A1  K 2 A2  K3 A3 ..... Kn An  K A1  A2  A3 ..... An Equating (i) and (ii), K can be determined i.e., K  (4) In numericals we may have the situation where the amount of heat travelling to the other end may be required to do some other work e.g., it may be required to melt the given amount of ice. In that case equation (i) will have to be equated to mL. i.e. mL  KA(1   2 ) t l Growth of Ice on Lake (1) Water in a lake starts freezing if the atmospheric temperature ST drops below 0 o C. Let y be the thickness of ice layer in the lake at any instant t and atmospheric temperature is   o C. (2) The temperature of water in contact with lower surface of ice will be zero. K1  K 2  K3 ..... Kn n K1  K 2 (b) For two slabs of equal area K . 2 (3) If A is the area of lake, heat escaping through ice in time dt is KA[0  ( )] dt dQ1  y (a) For n slabs of equal area K  (4) Suppose the thickness of ice layer increases by dy in time dt, due to escaping of above heat. Then dQ 2  mL  (dy A) L Ingen-Hauz Experiment It is used to compare thermal conductivities of different materials. If l1 , l2 and l are the lengths Hot –  °C Air 3 water of wax melted on rods as mc ( 4   3 ) l A (1   2 ) t Ice y dy K1 l3 l2 l1 K2 K3 Fig. 15.13 0°C Water 4°C Fig. 15.15 Transmission of Heat 699 proportional to the surface area of body and excess temperature of body over its surroundings i.e. (5) As dQ1  dQ 2 , hence, rate of growth of ice will be (dy / dt)  (K / Ly ) Q Q  A(T  T0 )   h A(T  T0 ) t t where h = Constant of proportionality called convection coefficient, T = Temperature of body and T = Temperature of surrounding 0 Convection coefficient (h) depends on properties of fluid such as density, viscosity, specific heat and thermal conductivity. So, the time taken by ice to grow to a thickness y is L K  y 0 y dy  L 2 y 2 K (3) Natural convection takes place from bottom to top while forced convection in any direction. (6) If the thickness is increased from y 1 to y 2 then time taken t L K  y2 ydy  y1 L 2 (y 2  y12 ) 2 K 60 t (4) In case of natural convection, convection currents move warm air upwards and cool air downwards. That is why heating is done from base, while cooling from the top. (5) Natural convection plays an important role in ventilation, in changing climate and weather and in forming land and sea breezes and trade winds. (8) Ice is a poor conductor of heat, therefore the rate of increase of thickness of ice on ponds decreases with time. (6) Natural convection is not possible in a gravity free region such as a free falling lift or an orbiting satellite. (9) It follows from the above equation that time taken to double and triple the thickness, will be in the ratio of (7) The force of blood in our body by heart helps in keeping the temperature of body constant. ID t1 : t 2 : t 3 :: 1 2 : 2 2 : 3 2 , i.e., t1 : t 2 : t 3 :: 1 : 4 : 9 Convection D YG  t1 : t 2 : t 3 :: 1 : 3 : 5 U Mode of transfer of heat by means of migration of material particles of medium is called convection. It is of two types. ST (1) Natural convection : This arise due to difference of densities at two places and is a consequence of gravity because on account of gravity the hot light particles rise up and cold heavy particles Convection current try setting down. It mostly occurs on heating a liquid/fluid. (2) Forced convection : If a fluid is forced to move to take up heat from a hot body then the convection process is called forced convection. In this case Newton's law of cooling holds good. According to which rate of loss of heat from a hot body due to moving fluid is directly (8) If liquids and gases are heated from the top (so that convection is not possible) they transfer heat (from top to bottom) by conduction. (9) Mercury though a liquid is heated by conduction and not by convection. U (10) The time intervals to change the thickness from 0 to y, from y to 2y and so on will be in the ratio t1 : t 2 : t 3 :: (1 2  0 2 ) : (2 2  1 2 ) : (3 2 : 2 2 ) E3 (7) Take care and do not apply a negative sign for putting values of temperature in formula and also do not convert it to absolute scale. Radiation (1) The process of the transfer of heat from one place to another place without heating the intervening medium is called radiation. (2) Precisely it is Fig. 15.18 electromagnetic energy transfer in the form of electromagnetic wave through any medium. It is possible even in vacuum e.g. the heat from the sun reaches the earth through radiation. (3) The wavelength of thermal radiations ranges from 7.8  10 7 m to 4  10 4 m. They belong to infra-red region of the electromagnetic spectrum. That is why thermal radiations are also called infra-red radiations. (4) Medium is not required for the propagation of these radiations. (5) They produce sensation of warmth in us but we can’t see them. (6) Every body whose temperature is above zero Kelvin emits thermal radiation. (7) Their speed is equal to that of light i.e. ( 3  10 8 m / s). (8) Their intensity is inversely proportional to the square of distance Fig. 15.16 of point of observation from the source (i.e. I  1 / d 2 ). (9) Just as light waves, they follow laws of reflection, refraction, interference, diffraction and polarisation. (10) When these radiations fall on a surface then exert pressure on that surface which is known as radiation pressure. Fig. 15.17 700 Transmission of Heat (13) Diathermanous Medium : A medium which allows heat radiations to pass through it without absorbing them is called diathermanous medium. Thus the temperature of a diathermanous medium does not increase irrespective of the amount of the thermal radiations passing through it e.g., dry air, SO 2 , rock salt (NaCl). (i) Dry air does not get heated in summers by absorbing heat radiations from sun. It gets heated through convection by receiving heat from the surface of earth. (ii) In winters heat from sun is directly absorbed by human flesh while the surrounding air being diathermanous is still cool. This is the reason that sun’s warmth in winter season appears very satisfying to us. (i) If a = t = 0 and r = 1 body is perfect reflector (ii) If r = t = 0 and a = 1 body is perfectly black body (iii) If, a = r = 0 and t = 1 body is perfect transmitter (iv) If t = 0  r  a  1 or a  1  r i.e. good reflectors are bad absorbers. Emissive Power, Absorptive Power and Emissivity If temperature of a body is more than it's surrounding then body emits thermal radiation (1) Monochromatic Emittance or Spectral emissive power (e ) : For a given surface it is defined as the radiant energy emitted per sec per unit area of the surface with in a unit wavelength around  i.e. lying between 1 1       to    . 2 2   Spectral emissive power (e  )  Colour of Heated Object Joule and Dimension : [ML1 T 3 ] m 2  sec  Å (2) Total emittance or total emissive power (e) : It is defined as the total amount of thermal energy emitted per unit time, per unit area of the body for all possible wavelengths. Unit : U When a body is heated, all radiations having wavelengths from zero to infinity are emitted. (1) Radiations of longer wavelengths are predominant at lower temperature. D YG (2) The wavelength corresponding to maximum emission of radiations shifts from longer wavelength to shorter wavelength as the temperature increases. Due to this the colour of a body appears to be changing. (3) A blue flame is at a higher temperature than a yellow flame Table 15.3 : Variation of colour of a body on heating Temperature 525°C 900°C 1100°C Colour Dull red Cherry red Orange red U 1200°C 1600°C Yellow White Interaction of Radiation with Matter ST Qa = Absorptance or absorbing power Q Q r  r = Reflectance or reflecting power Q t Qt = Transmittance or transmitting power Q   0 Joule m 2  sec e  d or W att m2 Q Qr (4) Total absorptance or total absorpting power (a): It is defined as the total amount of thermal energy absorbed per unit time, per unit area of the body for all possible wavelengths.   a d (5) Emissivity () : Emissivity of a body at a given temperature is defined as the ratio of the total emissive power of the body ( e) to the total e emissive power of a perfect black body (E) at that temperature i.e.   E ( read as epsilon) Qa Qt Fig. 15.19 and Dimension : [MT 3 ] (3) Monochromatic absorptance or spectral absorptive power (a ) : It is defined as the ratio of the amount of the energy absorbed in a certain time to the total heat energy incident upon it in the same time, both in the unit wavelength interval. It is dimensionless and unit less quantity. It is represented by a. 0 (1) Q  Qa  Qr  Qt Q Q Q (2) a  r  t  a  r  t  1 Q Q Q Unit : e a When thermal radiations (Q) fall on a body, they are partly reflected, partly absorbed and partly transmitted. (3) a  Energy Area  time  wavelength ID (14) Athermanous medium : A medium which partly absorbs heat rays is called a thermous medium As a result temperature of an athermanous medium increases when heat radiations pass through it e.g., wood, metal, moist air, simple glass, human flesh etc. (5) Different bodies 60 (12) Spectrum of these radiations can not be obtained with the help of glass prism because it absorbs heat radiations. It is obtained by quartz or rock salt prism because these materials do not have free electrons and interatomic vibrational frequency is greater than the radiation frequency, hence they do not absorb heat radiations. (4) r, a and t all are the pure ratios so they have no unit and dimension. E3 (11) While travelling these radiations travel just like photons of other electromagnetic waves. They manifest themselves as heat only when they are absorbed by a substance. (i) For perfectly black body  = 1 (ii) For highly polished body  = 0 (iii) But for practical bodies emissivity () lies between zero and one (0 <  < 1). Perfectly Black Body Transmission of Heat 701 (1) A perfectly black body is that which absorbs completely the radiations of all wavelengths incident on it. If Q = Qabsorbed temperature of body remains constant (thermal equilibrium) (2) As a perfectly black body neither reflects nor transmits any radiation, therefore the absorptance of a perfectly black body is unity i.e. t = 0 and r = 0  a = 1. Kirchoff's Law But for perfectly black body A = 1 i.e. e E a If emissive and absorptive powers are considered for a particular e wavelength ,    a    (E )black   Now since (E ) is constant at a given temperature, according to this law if a surface is a good absorber of a particular wavelength it is also a good emitter of that wavelength. black This in turn implies that a good absorber is a good emitter (or radiator) Applications of Kirchoff's Law ID (5) Ferry’s black body : A perfectly black body can’t be realised in practice. The nearest example of an ideal black body is the Ferry’s black body. It is a doubled walled evacuated spherical cavity whose inner wall is blackened. The space between the wall is evacuated to prevent the loss of heat by conduction and radiation. There is a fine hole in it. All the radiations incident upon this hole are absorbed by this black body. If this black body is heated to high temperature then it emits radiations of all wavelengths. It is the hole which is to be regarded as a black body and not the total enclosure e1 e 2 E  ...   a1 a 2  A  Perfectly black body 60 (4) When perfectly black body is heated to a suitable high temperature, it emits radiation of all possible wavelengths. For example, temperature of the sun is very high (6000 K approx.) it emits all possible radiation so it is an example of black body. According to this law the ratio of emissive power to absorptive power is same for all surfaces at the same temperature and is equal to the emissive power of a perfectly black body at that temperature. Hence E3 (3) We know that the colour of an opaque body is the colour (wavelength) of radiation reflected by it. As a black body reflects no wavelength so, it appears black, whatever be the colour of radiations incident on it. emission P U O D YG (6) A perfectly black body can not be realised in practice but materials like Platinum black or Lamp black come close to being ideal black bodies. Such materials absorbs 96% to 85%Fig.of15.20 the incident radiations. U Prevost Theory of Heat Exchange (1) Every body emits heat radiations at all finite temperature (Except 0 ST K) as well as it absorbs radiations from the surroundings. (2) Exchange of energy along various bodies takes place via radiation. (3) The process of heat exchange among various bodies is a continuous phenomenon. (4) At absolute zero temperature (0 K or – 273°C) this law is not applicable because at this temperature the heat exchange among various bodies ceases. (5) If Q > Qabsorbed temperature of body decreases and consequently the body appears colder. emission If Q hotter. emission (1) Sand is rough black, so it is a good absorber and hence in deserts, days (when radiation from the sun is incident on sand) will be very hot. Now in accordance with Kirchoff’s law, good absorber is a good emitter so nights (when sand emits radiation) will be cold. This is why days are hot and nights are cold in desert. < Qabsorbed temperature of body increases and it appears (2) Sodium vapours, on heating, emit two bright yellow lines. These are called D , D lines of sodium. When continuos white light from an arc lamp is made to pass through sodium vapours at low temperature, the continuous spectrum is intercepted by two dark lines exactly in the same places as D and D lines. Hence sodium vapours when cold, absorbs the same wavelength, as they emit while hot. This is in accordance with Kirchoff's law. 1 1 2 2 (3) When a shining metal ball having some black spots on its surface is heated to a high temperature and is seen in dark, the black spots shine brightly and the shining ball becomes dull or invisible. The reason is that the black spots on heating absorb radiation and so emit these in dark while the polished shining part reflects radiations and absorb nothing and so does not emit radiations and becomes invisible in the dark. (4) When a green glass is heated in furnace and taken out, it is found to glow with red light. This is because red and green are complimentary colours. At ordinary temperatures, a green glass appears green, because it transmits green colour and absorb red colour strongly. According to Kirchoff's law, this green glass, on heating must emit the red colour, which is absorbed strongly. Similarly when a red glass is heated to a high temperature it will glow with green light. (5) A person with black skin experiences more heat and more cold as compared to a person of white skin because when the outside temperature is greater, the person with black skin absorbs more heat and when the outside temperature is less the person with black skin radiates more energy. (6) Kirchoff' law also explains the Fraunhoffer lines : (i) Sun's inner most part (photosphere) emits radiation of all wavelength at high temperature. 702 Transmission of Heat (ii) When these radiation enters in outer part (chromosphere) of sun, few wavelength are absorbed by some terrestrial elements (present in vapour form at lower temperature) (iii) These absorbed wavelengths, which are missing appear as dark lines in the spectrum of the sun called Fraunhoffer lines. Chromosphere proportional to the fourth power of its absolute temperature, i.e. E  T 4  E = T 4 where  is a constant called Stefan’s constant having dimension 60 (ii) Area of radiating surface, i.e. greater the area of radiating surface, faster will be the cooling. (iii) Mass of radiating body i.e. greater the mass of radiating body slower will be the cooling. (iv) Specific heat of radiating body i.e. greater the specific heat of radiating body slower will be cooling. (v) Temperature of radiating body i.e. greater the temperature of body faster will be cooling. (vi) Temperature of surrounding i.e. greater the temperature of surrounding slower will be cooling. (i) For ordinary body : e = E   T 4 D YG (ii) Radiant energy : If Q is the total energy radiated by the ordinary Q   T 4  Q  A  T 4 t At (iii) Radiant power (P) : It is defined as energy radiated per unit area Q i.e. P   A T 4. t (iv) If an ordinary body at temperature T is surrounded by a body at temperature T , then Stefan's law may be put as Table 15.4 : Comparison of rate of heat loss (R ) and rate of cooling (R ) for different bodies Body H Condition U (1) Rate of loss of heat (or initial rate of loss of heat) : If an ordinary body at temperature T is placed in an environment of temperature T (T < T) then heat loss by radiation is given by 0 Q  Qemission  Qabsorption  A  (T (2) Rate of loss of heat (R H )  4 c Rate of heat loss dQ RH  dt Two solid sphere T, T0 , c,  are same RH  A  r 2  Two solid sphere of diff. material T, T0 – same 2 1 2 2 (R H )1 r  (R H )2 r RH  A  r 2 dQ  A  (T 4  T04 ) dt If two bodies are made of same material, have same surface finish  dQ    A dQ  dt 1  1 and are at the same initial temperature then  A A2 dt  dQ     dt  2 Rc  Rc   dT or dt Different shape bodies like cube, sphere plate T, T0, c,  - same Bodies of different materials T, T0, m, A are same but c diff. RH  A A max Plate A  V r2 1  r r3 Rc   0  T04 ) Rate of cooling d dt  T04 ) Rate of Loss of Heat (RH) and Rate of Cooling (RC) ST (i) Nature of radiating surface i.e. greater the emissivity, faster will be the cooling. U [MT 3 4 ] and value 5.67  10 8 W / m 2 K 4. (i) A  4 (T  T04 ) ; where m = density ()  volume (V) V c ID According to it the radiant energy emitted by a perfectly black body per unit area per sec (i.e. emissive power of black body) is directly e    (T d  (dQ / dt) A  4  (T  T04 )  dt mc mc E3 Stefan's Law 4 (i) Rate of cooling (Rc )  (4) Dependence of rate of cooling : When a body cools by radiation the rate of cooling depends on 15.21lines appear bright because the (iv) During total solar eclipse Fig. these gases and vapour present in the chromosphere start emitting those radiation which they had absorbed. 0 ( Q  mc T and dT  d ) (ii) for two bodies of the same material under identical environments, (Rc )1 A V the ratio of their rate of cooling is  1. 2 (R c )2 A 2 V1 Radiations having all wavelengths except wavelengths absorbed by various elements in chromosphere body then e  dQ dT d  mc.  mc dt dt dt  Radiations of wavelengths ranging 0 to  Photo sphere (3) Initial rate of fall in temperature (Rate of cooling): If m is the body and c is the specific heat then A V c 1 r c Rc  A V Rc  1 c Amin sphere RH same for all. bodies Newton's Law of Cooling Transmission of Heat 703 When the temperature difference between the body and its surrounding is not very large i.e. T – T = T then T 0 may be 4 T03 T By Stefan’s law, Hence  T04 This is a straight line intercept R-axis at  K 0 (4) Curve between rate of cooling (R) and temperature difference between dT A  4  [T  T04 ] dt mc body () and surrounding ( ) dT dT A  d  4 T03 T   T or   0 dt dt mc dt R  (   0 ). This is a straight line i.e., if the temperature of body is not very different from surrounding, rate of cooling is proportional to temperature difference between the body and its surrounding. This law is called Newton’s law of cooling. (1) Greater the temperature difference between body and its surrounding greater will be the rate of cooling. d  0 i.e. a body can never be cooled to a dt temperature lesser than its surrounding by radiation. If volume, radiating surface area, nature of surface, initial temperature and surrounding of water and given liquid are equal and they are allowed to cool down (by radiation) then rate of loss of heat and fall in temperature of both will be same. D YG (ii) Adding milk in hot tea reduces the rate of cooling. (1) Curve between log( –  ) and time loge ( – 0) 0 loge (   0 )  Kt  loge A O t U ST As loge (   0 )  Kt  loge A  loge   0 1 t 2 2 Water circulation  dQ   dQ  Fig. 15.26     dt dt   water  liquid (mW cW  W ) (1   2 ) (   2 )  (m lcl  W ) 1 t1 t2  mW cW  W   m lcl  W     t1 t2     or W = m c = Water equivalent of calorimeter, where m and c are mass and specific heat of calorimeter. c c c c If density of water and liquid is  and  respectively then mW  VW and ml  V l Specific heat of liquid cl  1 ml   tl  (mW cW  W )  W  t  W Distribution of Energy in the Spectrum of Black Body A Langley and later on Lummer and Pringsheim investigated the distribution of energy amongst the different wavelengths in the thermal spectrum of a black body radiation. The results obtained are shown in figure. From these curves it is clear that   Kt     0  Ae kt increasing time.   2 A perfectly black body emits radiation of all possible wavelength. (2) Curve between temperature of body and time decreases exponentially with 1 Fig. 15.22 This is a straight line with negative slope which indicates temperature   U (i) Hot water loses heat in smaller duration as compared to moderate warm water. Water out Water in i.e. (4) Practical examples T Liquid 1 This form of law helps in solving numericals. Integrating loge (   0 )  Kt  C Water t    2   1   2  becomes  1   K  2  0 . t     d d  (   0 )    Kdt As dt (   0 ) T ID  2 d  1   2 and    av  1. The Newton’s law of cooling  dt t 2 ( – 0) Fig. 15.25 Determination of Specific Heat of Liquid E3 (3) If a body cools by radiation from  1o C to  2o C in time t, then O passing through origin. (2) If    0 , Cooling Curves R 0 60 approximated as 4 (1) At a given temperature energy is not uniformly distributed among different wavelengths.  (2) At a given temperature intensity of heat radiation increases with wavelength, reaches a maximum at a particular wavelength and with further increase in wavelength it decreases. 0 0 Time t Fig. 15.23 E (3) Curve between the rate of cooling (R) and body temperature (). R R  K(  0 )  K  K0  K0 m Fig. 15.27 Fig. 15.24 704 Transmission of Heat (3) For all wavelengths an increase in temperature causes an increase in intensity. (4) The area under the curve will represent the total intensity of radiation at a particular temperature i.e. Area = E  E hc   (6) For radiations of long wavelength     Planck's law KT   reduces to Rayleigh-Jeans energy distribution law E  d    d From Stefan's law E = T  Area under E -  curve (A)  T 4 8KT 4 d Temperature of the Sun and Solar Constant 4 (5) The energy (E ) emitted corresponding to the wavelength of maximum emission ( ) increases with fifth power of the absolute max m temperature of the black body i.e., E max  T 5 If R is the radius of the sun and T its temperature, then the energy emitted by the sun per sec through radiation in accordance with Stefan’s law will be given by According to Wien's law the product of wavelength corresponding to maximum intensity of radiation and temperature of body (in Kelvin) is constant, i.e. m T  b  constant 4R 2 T 4 4r 2  r  2 S    R     r 1/4 R i.e. T   2 Earth Sun 3    1.4  10  5.67  10 8  1/4  Fig. 15.29 ~ 5800 K    As r  1.5  10 8 km, R  7  10 5 km, U T1 m  m 2 m 1 3  ID m 3 < m 2 < m 1 T2 4r 2  1.5  10 8    7  10 5  T3 > T2 > T1 T3 P E3 As the temperature of the body increases, the wavelength at which the spectral intensity (E ) is maximum shifts towards left. Therefore it is also called Wien's displacement law. E In reaching earth this energy will spread over a sphere of radius r (= average distance between sun and earth); so the intensity of solar radiation at the surface of earth (called solar constant S) will be given by S  where b is Wien's constant and has value 2.89  10 3 m - K. 60 P  AT 4  4R 2T 4 Wien's Displacement Law  S 2 15.28 This law is of greatFig.importance in ‘Astrophysics’ as through the D YG analysis of radiations coming from a distant star, by finding m the temperature of the star T ( b / m ) is determined. cal kW W  1.4 2 and   5.67  10 8 2 4 cm 2min m m K This result is in good agreement with the experimental value of temperature of sun, i.e., 6000 K. Law of Distribution of Energy (Plank's Hypothesis) (1) The theoretical explanation of black body radiation was done by Planck. (2) According to Plank's atoms of the walls of a uniform temperature enclosure behave as oscillators, each with a characteristic frequency of oscillation. ST U (3) These oscillations emits electromagnetic radiations in the form of photons (The radiation coming out from a small hole in the enclosure are called black body radiation). The energy of each photon is h. Where  is the frequency of oscillator and h is the Plank's constant. Thus emitted energies may be h, 2h, 3h... nh but not in between. According to Planck's law E  d   8hc 1 5 [e hc / KT  1] d where c = speed of light and k = Boltzmann's constant. This equation is known as Plank's radiation law. It is correct and complete law of radiation (4) This law is valid for radiations of all wavelengths ranging from zero to infinite. hc   (5) For radiations of short wavelength     Planck's law KT   A reduces to Wien's energy distribution law E  d   5 e  B / T d    Glass and water vapours transmit shorter wavelengths through them but reflects longer wavelengths. This concept is utilised in Green house effect. Glass transmits those waves which are emitted by a source at a temperature greater than 100°C. So, heat rays emitted from sun are able to enter through glass enclosure but heat emitted by small plants growing in the nursery gets trapped inside the enclosure.  Suppose two metallic rods are first connected in series then in parallel. 1 1 2 1 Series 2 2 1 Parallel (ii) (i) 2 If Q s heat flows in time t s in series combination and Q p heat flows in time t p in parallel combine, then Qp Qs  tp ts  Rs Rp Transmission of Heat 705 If Rods are identical then RS  R 2 t   4  p  Qs  ts  Qp and R p  2 R   If temperature of a body becomes  to  in t time and it becomes 1 2  to  in next time then use 2 3  2  0 3  0 ( = temperature of enviroment)  1   0  2   0 0  Newton's law of cooling can be used to compare the specific heat of 60 the two liquids.  Radiations from sun take 8 min and 20 sec to reach earth.  Suppose temperature of a body decreases  °C to  °C in time t and 1 2 1  °C to  °C in time t in the same invirment 2 3 2 If ( –  )  ( –  ) then t > t 1 2 2 3 2 1 ID  Green glass is a good absorber of red light and a good reflector of E3 If equal masses of two liquids having same surface are and finish cools from same initial temperature to same final temperature with same t K C surrounding then 1  2  1 t2 K1 C2 Hence Also Green Yellow Red Blue U green light. Consequently at lower temperature it is a good emitter of red light. equation D YG  While solving the problems of heat flow, remember the following e.g. If we are interested in finding the mass of ice which transfoms into water in unit time. For this we will take T.D. dm  Lf. R dt dm T.D.   dt (L f )(R)  Confusion 100°C L, K, A ice (0°C) U The rate of cooling has been used in many books, with double dQ meanings. At some places. Rate of cooling  and at other places, dt d. Our suggestion is that look for the units, if the dt ST rate of cooling  rate of cooling is in cal/m in or J/sec etc., then it is cooling is in °C/min it means d. dt dQ. But if rate of dt