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: Understanding the Question:
A black body of area \(a\) emits power \(E\) at \(27^\circ \text{C}\). Length and breadth are each reduced to \(1/3\) of the original, so area changes. Temperature is raised to \(327^\circ \text{C}\). We need the new emitted power.

Step 2: Key Formula or Approach:
Stefan‑Boltzmann law: \(P = \sigma A T^4\), where \(\sigma\) is Stefan’s constant, \(A\) is surface area, \(T\) is absolute temperature.

Step 3: Detailed Explanation:
Initial: \(T_1 = 27 + 273 = 300\ \text{K}\), \(A_1 = a\), \(P_1 = E = \sigma a (300)^4\).
New: length and breadth become \(1/3\) each, so new area \(A_2 = \left(\frac{1}{3} \times \frac{1}{3}\right) a = \frac{a}{9}\).
New temperature: \(T_2 = 327 + 273 = 600\ \text{K}\).
New power: \[ P_2 = \sigma A_2 T_2^4 = \sigma \left(\frac{a}{9}\right) (600)^4. \] Write \(600 = 2 \times 300\), so \((600)^4 = 16 \times (300)^4\). Thus: \[ P_2 = \sigma \frac{a}{9} \times 16 \times (300)^4 = \frac{16}{9} \left[ \sigma a (300)^4 \right] = \frac{16}{9} E. \]

Step 4: Final Answer:
The new energy emitted per second is \(\frac{16E}{9}\), option (A).