A black body is heated from `7^(@)C` to `287^(@)C`. The ratio of radiation emitted is:

Options

A. `1 : 256`
B. `1 : 27`
C. `1 : 16`
D. `1 : 64`

Answer

C

Explanation

Given : Initial temperature , `T_(1) = 7^(@)C = 280 K`
Final temperature = `T_(2) = 287^(@)C = 560K`
According to Stefen's law the radiated energy is given by
`E prop T^(4)`
Hence, `(E_1)/(E_2) = (T_(1)^(4))/(T_(2)^(4)) = (280/560)^(4)=1/16`
`E_(1):E_(2) = 1:16`.