A black reactangular surface of area ' a ' emits energy ' E ' per second at 27. If length and breadth is reduced to (1/3) rd of initial value and temperature is raised to 327 then energy emitted per second becomes (a) 16E/9 (b) 8E/9 (c) 4E/9 (d) 12E/9

Answer: (a). E ∝ A T⁴. Length and breadth become 1/3: A → A/9. T: 300 K → 600 K, factor 16. E' = 16E/9, so the answer is (a).

A black reactangular surface of area ' a ' emits energy ' E ' per… | MHT CET 2021 (21 Sep, Shift 1)

Q94MHT CET 2021 · 21 SEP · SHIFT 1Medium

Thermal Radiation: Black Body, Stefan & Wien

A black reactangular surface of area ' $a$ ' emits energy ' $E$ ' per second at $27^\circ\text{C}$ . If length and breadth is reduced to $\left(\frac{1}{3}\right)$ rd of initial value and temperature is raised to $327^\circ\text{C}$ then energy emitted per second becomes

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Answer: (a)

$E \propto A T^4$ .
Length and breadth become $\frac{1}{3}$ : $A \to \frac{A}{9}$ .
$T: 300\text{ K} \to 600\text{ K}$ , factor $16$ .
$E' = \frac{16E}{9}$ , so the answer is (a).

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Thermal Radiation: Black Body, Stefan & Wien question from MHT CET 2021 (21 Sep, Shift 1)

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