Question

A black rectangular surface of area 'A' emits energy 'E' per second at \(27^{\circ}\)C. If length and breadth is reduced to half of its initial value and temperature is raised to \(327^{\circ}\)C then energy emitted per second becomes

• A. \(2E\)
• B. \(4E\)
• C. \(8E\)
• D. \(16E\)

Hint:

Energy per second is proportional to area times T^4 with T in kelvin.

Answer 1

The Correct Option is B

Step 1: Understanding the Concept:
Stefan's law for a black body: \(E = \sigma A T^4\), with \(T\) in kelvin.

Step 2: Convert temperatures and area:
\(T_1 = 27 + 273 = 300\) K and \(T_2 = 327 + 273 = 600\) K, so \(T_2/T_1 = 2\).
Length and breadth are both halved, so the area becomes \(\dfrac A4\).

Step 3: Compute the new energy:
\[ \frac{E_2}{E_1} = \frac{A/4}{A}\left(\frac{600}{300}\right)^4 = \frac14\times16 = 4 \]

Step 4: Check:
Option (B) \(4E\). Option (D) \(16E\) forgets the area reduction. Option (A) and (C) do not follow from the factors 1/4 and 16.

Final Answer:
Area drops by 4 and T^4 rises by 16, so energy is 4E. \[ \boxed{\text{(B) }4E} \]