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. 8Q
B. 4Q
C. 16Q
D. 32Q

Answer

B

Explanation

Rate of dissipation by a black body is given by
`Q=sigmaAT^(4)`
For the black body emissivity = 1
The other body has temperature 2T,
Area A and emissivity = 0.25
`therefore` Its rate of dissipation `Q'=sigmaA(2T)^(4)xxe`
`therefore(Q')/Q=e((2T)/T)^(4)=025xx(2)^(4)=16xx0.25=4`
`thereforeQ'=4Q`