Q.

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

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a

254c

b

508c

c

527c

d

272c

answer is C.

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Detailed Solution

  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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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