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

Options

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

Answer

C

Explanation

From Stefan's law, the total radiant energy, emitted per second per unit surface area of a black body proportionall 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=400 K`
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`