Current Affairs JEE Main & Advanced

(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. (2) Sodium vapours, on heating, emit two bright yellow lines. These are called \[{{D}_{1}},\,{{D}_{2}}\] 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 \[{{D}_{1}}\] and \[{{D}_{2}}\] lines. Hence sodium vapours when cold, absorbs the same wavelength, as they emit while hot. This is in accordance with Kirchoff's law. (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. (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. (iv) During total solar eclipse these lines appear bright because the gases and vapour present in the chromosphere start emitting those radiation which they had absorbed.  

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 \[\frac{{{e}_{1}}}{{{a}_{1}}}=\frac{{{e}_{2}}}{{{a}_{2}}}=...\,\,{{\left( \frac{E}{A} \right)}_{\text{Perfectly black body}}}\] But for perfectly black body \[A=1\] i.e. \[\frac{e}{a}=E\] If emissive and absorptive powers are considered for a particular wavelength l, \[\left( \frac{{{e}_{\lambda }}}{{{a}_{\lambda }}} \right)={{({{E}_{\lambda }})}_{\text{black}}}\] Now since \[{{({{E}_{\lambda }})}_{black}}\] 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. This in turn implies that a good absorber is a good emitter (or radiator)  

(1) Every body emits heat radiations at all finite temperature (Except 0 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 \[(0\,\,K\,\,\text{or}\,\,-{{273}^{o}}C)\] this law is not applicable because at this temperature the heat exchange among various bodies ceases. (5) If \[{{Q}_{emission}}>{{Q}_{absorbed}}\to \] temperature of body decreases and consequently the body appears colder. If \[{{Q}_{emission}}

(1) A perfectly black body is that which absorbs completely the radiations of all wavelengths incident on it. (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 \[r=0\Rightarrow a=1\]. (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. (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. (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 (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% of the incident radiations.  

If temperature of a body is more than it's surrounding then body emits thermal radiation (1) Monochromatic Emittance or Spectral emissive power \[\mathbf{(}{{\mathbf{e}}_{\mathbf{\lambda }}}\mathbf{)}\] : 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 l i.e. lying between \[\left( \lambda -\frac{1}{2} \right)\] to \[\left( \lambda +\frac{1}{2} \right)\]. Spectral emissive power \[({{e}_{\lambda }})=\frac{\text{Energy}}{\text{Area}\times \text{time}\times \text{wavelength}}\]  Unit :  \[\frac{Joule}{{{m}^{2}}\times \sec }\] and     Dimension : \[[M{{L}^{-1}}{{T}^{-3}}]\] (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.    \[e=\int_{\,0}^{\,\infty }{\,\,{{e}_{\lambda }}d\lambda }\] Unit : \[\frac{Joule}{{{m}^{2}}\times \sec }\] or \[\frac{Watt}{{{m}^{2}}}\]   and  Dimension : \[[M{{T}^{-3}}]\] (3) Monochromatic absorptance or spectral absorptive power \[\mathbf{(}{{a}_{\mathbf{\lambda }}}\mathbf{)}\]: 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 \[{{a}_{\lambda }}\]. (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. \[a=\int_{\,0}^{\,\infty }{{{a}_{\lambda }}d\lambda }\] (5) Emissivity \[\mathbf{(\varepsilon )}\] : 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 emissive power of a perfect black body (E) at that temperature i.e. \[\varepsilon =\frac{e}{E}\]         (\[\varepsilon \to \]read as epsilon) (i) For perfectly black body \[\varepsilon =1\] (ii) For highly polished body \[\varepsilon =0\] (iii) But for practical bodies emissivity \[(\varepsilon )\]lies between zero and one \[(0

When thermal radiations (Q) fall on a body, they are partly reflected, partly absorbed and partly transmitted. (1) \[Q={{Q}_{a}}+{{Q}_{r}}+{{Q}_{t}}\] (2) \[\frac{{{Q}_{a}}}{Q}+\frac{{{Q}_{r}}}{Q}+\frac{{{Q}_{t}}}{Q}=a+r+t=1\] (3) \[a=\frac{{{Q}_{a}}}{Q}\]= Absorptance or absorbing power \[r=\frac{{{Q}_{r}}}{Q}\] = Reflectance or reflecting power \[t=\frac{{{Q}_{t}}}{Q}\]= Transmittance or transmitting power (4) r, a and t all are the pure ratios so they have no unit and dimension. (5) Different bodies (i) If \[a=t=0\] and \[r=1\to \] body is perfect reflector (ii) If \[r=t=0\] and \[a=1\to \] body is perfectly black body (iii) If, \[a=r=0\] and \[t=1\to \] body is perfect transmitter (iv) If \[t=0\Rightarrow r+a=1\,\,\text{or}\,\,a=1-r\]  i.e. good reflectors are bad absorbers.  

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.  (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   Variation of colour of a body on heating

Temperature	Colour
\[{{525}^{o}}C\]	Dull red
\[{{900}^{o}}C\]	Cherry red
\[{{1100}^{o}}C\]	Orange red
\[{{1200}^{o}}C\]	Yellow
\[{{1600}^{o}}C\]	White

 

(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 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 \[7.8\times {{10}^{-7}}\,m\] to \[4\times {{10}^{-4}}\,m\]. 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. \[(=3\times {{10}^{8}}\,m/s)\]. (8) Their intensity is inversely proportional to the square of distance of point of observation from the source (i.e. \[I\propto 1/{{d}^{2}}\]). (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. (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. (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. (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, \[S{{O}_{2}}\], 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. (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.  

Mode of transfer of heat by means of migration of material particles of medium is called convection. It is of two types. (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 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 proportional to the surface area of body and excess temperature of body over its surroundings i.e.  \[\frac{Q}{t}\propto A(T-{{T}_{0}})\]\[\Rightarrow \] \[\frac{Q}{t}=h\,A(T-{{T}_{0}})\] where h = Constant of proportionality called convection coefficient, T = Temperature of body and \[{{T}_{0}}=\] Temperature of surrounding Convection coefficient (h) depends on properties of fluid such as density, viscosity, specific heat and thermal conductivity. (3) Natural convection takes place from bottom to top while forced convection in any direction. (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. (6) Natural convection is not possible in a gravity free region such as a free falling lift or an orbiting satellite. (7) The force of blood in our body by heart helps in keeping the temperature of body constant. (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.  

(1) Water in a lake starts freezing if the atmospheric temperature drops below \[{{0}^{o}}C\]. Let y be the thickness of ice layer in the lake at any instant t and atmospheric temperature is \[-{{\theta }^{o}}C\]. (2) The temperature of water in contact with lower surface of ice will be zero. (3) If A is the area of lake, heat escaping through ice in time dt is \[d{{Q}_{1}}=\frac{KA[0-(-\theta )]\,dt}{y}\] (4) Suppose the thickness of ice layer increases by dy in time dt, due to escaping of above heat. Then \[d{{Q}_{2}}=mL=\rho (dy\,A)\,L\] (5) As \[d{{Q}_{1}}=d{{Q}_{2}}\], hence, rate of growth of ice will be \[(dy/dt)=(K\theta /\rho Ly)\] So, the time taken by ice to grow to a thickness y is \[t=\frac{\rho L}{K\theta }\int_{\,0}^{\,y}{\,y\,dy}=\frac{\rho L}{2K\theta }{{y}^{2}}\] (6) If the thickness is increased from \[{{y}_{1}}\] to \[{{y}_{2}}\] then time taken \[t=\frac{\rho L}{K\theta }\int_{\,{{y}_{1}}}^{\,{{y}_{2}}}{\,ydy}=\frac{\rho L}{2K\theta }(y_{2}^{2}-y_{1}^{2})\] (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. (8) Ice is a poor conductor of heat, therefore the rate of increase of thickness of ice on ponds decreases with time. (9) It follows from the above equation that time taken to double and triple the thickness, will be in the ratio of \[{{t}_{1}}:{{t}_{2}}:{{t}_{3}}::{{1}^{2}}:{{2}^{2}}:{{3}^{2}},\] i.e., \[{{t}_{1}}:{{t}_{2}}:{{t}_{3}}::1:4:9\] (10) The time intervals to change the thickness from 0 to y, from y to 2y and so on will be in the ratio \[\Delta {{t}_{1}}:\Delta {{t}_{2}}:\Delta {{t}_{3}}::({{1}^{2}}-{{0}^{2}}):({{2}^{2}}-{{1}^{2}}):({{3}^{2}}:{{2}^{2}})\] \[\Rightarrow \] \[\Delta {{t}_{1}}:\Delta {{t}_{2}}:\Delta {{t}_{3}}::1:3:5\]