A black rectangular surface of area A emits energy E per second at `27^circC`. If length and breadth are reduced to one third of initial value and temperature is raised to `327^circC`, then energy emitted per second becomes

Options

A. `(4E)/(9)`
B. `(7E)/(9)`
C. `(10E)/(9)`
D. `(16E)/(9)`

Answer

D

Explanation

Original area `(A_(1))=LxxB`
and New area `(A_(2))=L/3xxB/3=(LB)/9`
Initial temperature `T_(1)=(27+273)=300K`
and New temperature = 600K
By Stefan's law, `E=sigmaAT^(4)`
`therefore(E')/E=A_(2)/A_(1)(T_(2)/T_(1))^(4)=((LB)/(9xxLB))xx(600/300)^(4)`
`=1/9xx(2)^(4)=16/9`
`thereforeE'=16/9E`