Energy is being emitted from the surface a black body at  127∘C temperature at 1×106Js-1m-2, temperature of the black body at which the rate of emission is 16×106Js-1m-2 will be

given temperature of black body is T1=127+273=400K      T2=?  energy emitted by black body is E1=1×106 Js-1m-2       E2=16×106 Js-1m-2  from stefan's law,   E =σ T4 ⇒E α T4 ⇒E1E2=T1T24 ⇒E1E2=400T24 ⇒(E1E2)14=400T   ⇒10616×10614=400T⇒  12 =   400T ⇒   T =   800  K =   800  -  273 =   527  °  C