Solution - Career Point

Unit of λ : Different units are used,e.g., cal cm s ºC –1 , cal m –1 s –1 ºC –1 , jm´1 s –1 ºC –1 .Convection :It is a process by which heat is conveyed by theactual movement of particles. Particles closest to thesource receive heat by conduction through the wall ofthe vessel. They rise up-wards and are replaced bycolder particles from the sides. Thus, a circulation ofparticles is set up – hot particles constitute theupward current and cold particles, the side anddownward current.The transfer of heat by convection occurs only influids, and is the main mode of heat transfer in them.Most fluids are very poor conductors.Radiation :Thermal Radiation : Thermal radiations areelectromagnetic waves of long wavelengths.Black Body : Bodies which absorb the whole of theincident radiation and emit radiations of allwavelengths are called black bodies.It is difficult to realize a perfect black body inpractice. However, a cavity whose interior walls aredull black does behave like a black body.Absorption : Every surface absorbs a part or all ofthe radiation falling on it. The degree of absorptiondepends on the nature and colour of the surface. Dull,black surfaces are the best absorbers. Polished, whitesurfaces absorb the least. The coefficient ofabsorption for a surface isradiation absorbeda λ =radiation incidentThe suffix λ denotes the wavelength of the radiationbeing considered, Clearly, a λ = 1 for a black body, forall values of λ.Emission : Each surface emits radiation (radiates)continuously. The emissive power (e λ ) is defined asthe radiation emitted normally per second per unitsolid angle per unit area, in the wave-length range λand λ + dλ. Clearly, the emissive power of a blackbody (denoted by E λ ) is the maximum.Kirchhoff's Law : According to this law, for thesame conditions of temperature and wavelength, theratio e λ /a λ is the same for all surfaces and is equal toE λ . This simply means that good absorbers are goodemitters. Hence, a black body is the best emitter, anda polished white body, the poorest emitter.Prevost's Theory of Exchanges : All bodies emitradiations irrespective of their temperatures. Theyemit radiations to their environments and receiveradiations from their environments simultaneously. Inthe equilibrium state the exchange between a bodyand the environment of energy continues in equalamounts.Stefan-Boltzmann Law : If a black body at anabsolute temperature T be surrounded by anotherblack body at an absolute temperature T 0 , the rate ofloss of radiant energy per unit area isE = σ(T 4 – T 4 0 )where σ is a constant called Stefan constant and itsvalue is 5.6697 × 10 –8 W m –2 K –4The total energy radiated by a black body at anabsolute temperature T is given byE = σT 4 × surface area × timeNote : Remember that rate of generation of heat byelectricity is given by H = I 2 V 2R or or VI Js –1 or W.RSolved Examples1. An earthenware vessel loses 1 g of water per seconddue to evaporation. The water equivalent of thevessel is 0.5 kg and the vessel contains 9.5 kg ofwater. Find the time required for the water in thevessel to cool to 28ºC from 30ºC. Neglect radiationlosses. Latent heat of vaporization of water in thisrange of temperature is 540 cal g –1 .Sol. Here water at the surface is evaporated at the cost ofthe water in the vessel losing heat.Heat lost by the water in the vessel= (9.5 + 0.5) × 1000 × (30 – 20) = 10 5 calLet t be the required time in seconds.Heat gained by the water at the surface= (t × 10 –3 ) × 540 × 10 3(Q L = 540 cal g –1 = 540 × 10 3 cal kg –1 )∴ 10 5 = 540t or t = 185 s = 3 min 5s2. 15 gm of nitrogen is enclosed in a vessel attemperature T = 300 K. Find the amount of heatrequired to double the root mean square velocity ofthese molecules.Sol. The kinetic energy of each molecule with mass m isgiven by1 m2 3v rms = kT ...(1)2 2If we want to increase the r.m.s. speed to η times,then the temperature has to be raised to T´. Then,1 2 3 1mv rms = kT´ or mη2 2 3v rms = kT´2 2 2 2 ...(2)From eqs. (1) and (2), T´ = η 2 T ...(3)XtraEdge for IIT-JEE 25 APRIL 2010