A black body of surface area`5cm^(2)`, is kept at `127^(@)C`. The rate of energy radiated by it is E. If its temperature is increased to `527^(@)C`, then the increase in the rate of energy radiation will be

Options

A. 2 E
B. 8 E
C. 15 E
D. 20 E

Answer

C

Explanation

`E_(2)/E_(1)=(T_(2)/T_(1))^(4)=16`
`E_(2)=16E_(1)`
`therefore` Increase in the rate of radiation = 16E - E = 15E