Question

A black rectangular surface of area A emits energy E per second at \(127^{\circ}\)C. If length and breadth is reduced to half of initial value and temperature is raised to \(527^{\circ}\)C then energy emitted becomes

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

Hint:

Power radiated = sigma A T^4 with T in kelvin.

Answer 1

The Correct Option is C

Step 1: Understanding the Concept
By the Stefan-Boltzmann law, \(E = \sigma A T^4\). Temperature must be in kelvin.

Step 2: Convert and compare
\(T_1 = 127 + 273 = 400\) K and \(T_2 = 527 + 273 = 800\) K, so \(T_2/T_1 = 2\).
Length and breadth are halved, so the area becomes \(\frac14\).
\[ \frac{E'}{E} = \frac{A'}{A}\left(\frac{T_2}{T_1}\right)^4 = \frac14 \times 16 = 4 \]
So the new energy is 4E. Using Celsius directly would give a wrong ratio.

Final Answer:
The energy emitted becomes 4E, option (C). \[ \boxed{4E} \]