A plane surface A is at a constant temperature `T_(1)`` = 1000 K`. Another surface B parallel to A, is at a constant lower temperature `T_(2) = 300 K`. There is no medium in the space between two surfaces. The rate of energy transfer from A to B is equal to `r_(1)(J/s)`. In order to reduce rate of heat flow due to radiation, a heat shield consisting of two thin plates C and D, thermally insulated from each other, is placed between A and B in parallel. Now the rate of heat transfer (in steady state) reduces to `r_(2)`. Neglect any effect due to finite size of the surfaces, assume all surfaces to be black bodies and take Stefan’s constant `sigma = 6 xx 10^(- )8 Wm^(- 2)K^(-4)`. Area of all surfaces `A = 1m^(2)`.
(i) Find `r^(1)`
(ii) Find the ratio `(r^(2))/(r^(1))`
(iii)Find the ratio `(r^E(2))/(r_(1))` if temperature of A and B were 2000 K 600 K respectively.

Answer

(i) `r_(1)=59513W` (ii) `(r_(2))/(r_(1))=(1)/(3)` (iii) `(r_(2))/(r_(1))=(1)/(3)`