Energy is being emitted from the surface of a black body at `127^(@)C` temperature at the rate of `1.0xx10^(6)J//sec-m^(2)`. Temperature of the black at which the rate of energy emission is `16.0xx10^(6)J//sec-m^(2)` will be

Options

A. `254^(@)C`
B. `381^(@)C`
C. `527^(@)C`
D. `80^(@)C`

Answer

C

Explanation

`T_(1)=127+273=400K.`
`R_(2)/R_(1)=16=(T_(2)/T_(1))^(4)=(2)^(4)`
`T_(2)/T_(1)=2/1" "thereforeT_(2)=2T_(1)=800K`
`thereforeT_(2)=800-273=527^(@)C`