Question:medium

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

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Power radiated = sigma A T^4 with T in kelvin.
Updated On: Oct 1, 2026
  • \(E\)
  • \(2E\)
  • \(4E\)
  • \(8E\)
Show Solution

The Correct Option is C

Solution and Explanation

Step 1: Kelvin
400 K and 800 K, a ratio of 2.

Step 2: Factors
Area factor $1/4$, temperature factor $2^4 = 16$.

Step 3: Combine
$16/4 = 4$, so the energy is $4E$. Option (C).

Final Answer:
Option (C). \[ \boxed{4E} \]
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