Question

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

• A. \(\frac{16E}{9}\)
• B. \(\frac{8E}{9}\)
• C. \(\frac{4E}{9}\)
• D. \(\frac{12E}{9}\)

Hint:

Always convert Celsius to Kelvin in radiation problems. The Stefan‑Boltzmann law uses absolute temperature. Area scaling: if each linear dimension is multiplied by a factor \(f\), area multiplies by \(f^2\). Here \(f = 1/3\), so \(A_2 = A_1/9\).

Answer 1

The Correct Option is A

Step 1: Note the changes.
A black surface of area $a$ gives out power $E$ at $27$ degrees C. Both length and breadth are cut to one third, and the temperature is raised to $327$ degrees C. We want the new power.

Step 2: The radiation law.
Stefan's law says emitted power $P = \sigma A T^4$. So power depends on area and on the fourth power of the absolute temperature.

Step 3: Convert temperatures to kelvin.
$T_1 = 27 + 273 = 300$ K and $T_2 = 327 + 273 = 600$ K. The temperature exactly doubles.

Step 4: Find the new area.
Length and breadth each become one third, so area becomes $\frac{1}{3} \times \frac{1}{3} = \frac{1}{9}$ of $a$, that is $\dfrac{a}{9}$.

Step 5: Handle the temperature factor.
Doubling temperature multiplies $T^4$ by $2^4 = 16$.

Step 6: Combine both effects.
Area gives a factor $\frac{1}{9}$ and temperature gives a factor $16$, so the new power is $\dfrac{16}{9}E$, which is option (A).
\[ \boxed{P_2 = \frac{16E}{9}} \]