The temperature of a piece of iron is `27^(@)C` and it is radiating energy at the rate of `Q kWm^(-2)` . If its temperature is raised to `151^(@)C` , the rate of radiation of energy will become approximately

Options

A. 2Q `kw//m^(2)`
B. 4Q `kw//m^(2)`
C. 6Q `kw//m^(2)`
D. 8Q `kw//m^(2)`

Answer

B

Explanation

`T_(1)=273+27=300" "T_(2)=273+151=424`
`therefore(R_(2)/R_(1))=(424/300)^(4)=4" "thereforeR_(2)=4QKw//m^(2)`