Question

A black rectangular surface of area A emits energy E per unit time at \(27^{\circ}\) C. If length and breadth is reduced to \((\frac{1}{4})^{th}\) of initial value and temperature is raised to \(327^{\circ}\) C, then energy emitted per unit time becomes

• A. \(E\)
• B. \(2E\)
• C. \(\frac{E}{2}\)
• D. \(\frac{E}{4}\)

Hint:

Energy per second goes as area times the fourth power of absolute temperature. Work in kelvin.

Answer 1

The Correct Option is A

Step 1: Understanding the Concept:
A black body emits energy per unit time \(E = \sigma A T^4\), with \(T\) in kelvin.

Step 2: Convert temperatures.
\(T_1 = 27 + 273 = 300\) K and \(T_2 = 327 + 273 = 600\) K, so \(T_2 = 2T_1\).

Step 3: Find the change in area.
Length and breadth are each reduced to \(\dfrac{1}{4}\), so the area becomes \(\dfrac{1}{16}\) of \(A\).

Step 4: Calculate.
\[ \frac{E_2}{E_1} = \frac{A_2}{A_1}\left(\frac{T_2}{T_1}\right)^4 = \frac{1}{16}\times 2^4 = 1 \]

Step 5: Check the options.
The two changes cancel exactly, so the emission is unchanged.

Final Answer:
The new energy per second is \(E\), option (A). \[ \boxed{E} \]