The rate of dissipation of heat by a black body at temperature `T` is `Q`. What will be the the rate of dissipation of heat by another body at temperature `2T` and emissivity `0.25`?

Options

A. `16 Q`
B. `4 Q`
C. `8 Q`
D. `4.5 Q`

Answer

B

Explanation

By Stephen's law
`Q = sigma T^(4)`
The rate of dissipation of heat by a body emissivity `e` is given by
`Q' = e sigmaT'4`
There `T' = 2T , e = 0.25`
`:. Q = 0.25xxsigma(2T)^(4)`
`=4 sigma T^(4) = 4Q`.