Discussion Forum

A black rectangular surface of area A emits energy E  per second at 27° C . If length and breadth  are reduced to $\frac{1}{3}$^rd of initial value and temperature is raised  to 327° C . then energy emitted per second becomes

Answer:

Explanation:

According to question ,

   $E= e \sigma A(T^{4}-T_{0}^{4})$

  [ where symbols have their usual meanings]

 when l and b change to $\frac{l}{3}$ and $\frac{b}{3}$ , respectively

 Area becomes $\frac {l}{3} \times \frac{b}{3}=\frac{l b}{3}=\frac {A}{3}$      ($\because$ A=lb)

 Now for two different cases

$\frac{E'}{E}=\frac{A'(227+373)^{4}}{(27+273)^{4}}=\frac{1}{9}(\frac{600}{300})^{4}$

$\therefore$    E'=$\frac{1}{9} \times (2)^{4} \times  E  \Rightarrow  E'= \frac {16 E}{9}$