Question

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

• A. $\frac{4E}{9}$
• B. $\frac{7E}{9}$
• C. $\frac{10E}{9}$
• D. $\frac{16E}{9}$

Answer 1

Answer: (D)

$\frac{16E}{9}$

From Stefan's law, energy radiated per second $E=\sigma eA{T}^4$

where $\sigma$ is Stefan's constant. $A$ is the area, $e$ and $T$ is the emmissivity and temperature of body, respectively.

Initially, temperature of body $T={27}^{o}C=300$ K

$⟹$ $E=\sigma eA(300{)}^4$

Now length and breadth of body is reduced by a factor of $3$.

Thus new area of the body $A^{\prime }=\frac{L}{3}.\frac{B}{3}=\frac{A}{9}$

New temperature of the body $T^{\prime }={327}^{o}C=600$ K

New energy radiated per second $E^{\prime }=\sigma e.\frac{A}{9}.(600{)}^4$

$\therefore$ $\frac{E^{\prime }}{E}=\frac{\sigma e.A(600{)}^4/9}{\sigma e.A(300{)}^4}=\frac{2^4}{9}$

$⟹$ $E^{\prime }=\frac{16E}{9}$