Question

Consider a spherical body `A` of radius `R` which is placed concentrically in a hollow enclosure `H` of radius `4R` as shown in the The temperature of the body `A` and `H` are `T_(A)` and `T_(H)` respectively Emissivity transmittivity and reflactivity of two bodies `A` and `H` are `(e_(a),e_(H)),?(t_(A),t_(H)),` and `(r_(A),r_(H))` respectively (Assume no absorption of the thermal energy by the space in between the body and enclosure as well as outside the enclousre and all radiations to be emitted and absorbed normal to the surface `[Tske sigma xx 4pi r^(2) xx 300^(4) = betaJ//s]`

In the previous question if the enclosure is considered as perfect black body and is maintaind at same temperature as that of temperature of body A then in the two cases .
A. the body A emits radiation at the same are
B. the body A emits radiation at different rates
C. the temperature of body A remains constant .
D. None of these

Answer 1

Correct Answer - B
The diagram shows the situation clearly The rate at which energy is emitted by `A` is `BetaJ//s` while crossing the enclousure the rate at which energy is transmitted out is `(beta)/(2)`
So rate at which `A` loses energy is `beta J//s` and the rate at which `P` and `Q` receive energy are `beta//2J//s` and `betaJ//s` respectively This energy is received on the area of sphere passing through `P` and `Q` Now in this case each of incidence, refleaction absorption take place The rate at which energy has been lost by `A` is
`P = - [P_(absorbed)-P_(emitted)]`
`=-[(beta)/(8)+(beta)/(32)+...]+[(beta)/(2)+(beta)/(8)+(beta)/(32)+....]=(beta)/(2)`
The rate a which energy is received by `P` is `P_(1) =0` The rate at which energy is receivedly `Q` is
`P_(2)=[(beta)/(2)+(beta)/(8)+...]+[(beta)/(2)+(beta)/(8)+(beta)/(32)+.....]=(beta)/(2)`
`=(beta)/(2)xx(4)/(3)+(beta)/(4)xx(4)/(3)=beta`
`If beta = sigma 4piR^(2) xx 300^(4)` then
`sigma xx 4pi (4R)^(2) xx 600^(4) = 256beta =gamma`
Let `a_(H) =e_(H) = 0.5` and for `A` in `2nd` case
`e_(A) = a_(A) = 0.5`
For `1st` case, `P_("emitted") = beta J//s P_("absorbed") = (gamma)/(2) + (beta)/(2)`
rate at which energy is lost
`P=(beta-(gamma)/(2)-(beta)/(2))J//s`
For `2nd` case `P_("emitted") = ((beta)/(2) + (beta)/(8) + (beta)/(32))J//s`
For `2nd` case
`P_("emitted")= ((beta)/(2)+(beta)/(8)+(beta)/(32)+....)+((gamma)/(4)+(gamma)/(16)+....)=(2beta)/(3)+(gamma)/(3)`
`P_("absorbed")=((beta)/(2)+(beta)/(32)+...)+((gamma)/(4)+(gamma)/(16)+....)=(beta)/(6)+(gamma)/(3)`
Rate at which heat is lost `P = (beta)/(2)` In these questions thermal equilibrium is not acheived and infinite no of reflection absorption can take place before thermal equilibrium has been actived and it is because of very large speed of radiation