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# Energy is being emitted from the surface a black body at  ${127}^{\circ }C$ temperature at $1×{10}^{6}{\text{Js}}^{\text{-1}}{\text{m}}^{\text{-2}}$, temperature of the black body at which the rate of emission is $16×{10}^{6}{\text{Js}}^{\text{-1}}{\text{m}}^{\text{-2}}$ will be

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a
254∘c
b
508∘c
c
527∘c
d
272∘c
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detailed solution

Correct option is C

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

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