A black body is heated from `27^(@)C` to `127^(@)C`. The ratio of their energies of radiation emitted will be

Options

A. `81 : 256`
B. `27 : 64`
C. `9 : 16`
D. `3 : 4`

Answer

A

Explanation

From Stefan's law, the total radiant energy emitted per second per unit surfaxe area of a black body is proportional to the fourth power of the absolute temperature of the body.
That is `E = sigma T^(4)`
where `sigma` is Stefan's constant.
Given
initial temperature `T_(1) = 273+27=300K`
final temperature `T_(2)=127+273=400K`
Hence, we have
`(E_1)/(E_2) = (T_(1)^(4))/(T_(2)^(4)) = ((300)^(4))/((400)^(4)) = (81)/(256)`
Therefore, `E_(1) : E_(2) = 81:256`.