The ratio of energy of emitted radiation of a black body at `27^(@)C` and `927^(@)C` is

Options

A. `1:4`
B. `1:8`
C. `1:16`
D. `1:256`

Answer

D

Explanation

According to Stefan's law `EpropT^(4)" or " E=sigmaT^(4)`
where, `sigma` is Stefan's constant It's value is `5.67xx10^(-8)Wm^(-2)K^(-4)`
Here `T_(1)=27+273=300K`
`T_(2)=927+273=1200K`
`therefore" " (E_(1))/(E_(2))=((300)/(1200))^(4)=1.256 `