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: Use Stefan's law.
A black body radiates power $P = \sigma A T^4$. So power scales with area and with the fourth power of absolute temperature.

Step 2: Track the area.
Length and breadth each shrink to a third, so the new area is $\tfrac13\times\tfrac13 = \tfrac19$ of the old.

Step 3: Track the temperature.
Temperatures in kelvin go from $300\ \text{K}$ to $600\ \text{K}$, a doubling, so $T^4$ grows by $2^4 = 16$.

Step 4: Combine the factors.
$P' = \tfrac{1}{9}\times 16\times E = \tfrac{16E}{9}$. \[ \boxed{\tfrac{16E}{9}} \]