Single Choice

A black rectangular surface of area 'A' emits energy 'E' per second at 27$$^o$$C. If length and breadth are reduced to $$\dfrac{1}{3}rd$$ of initial value and temperature is raised to 327$$^o$$C then energy emitted per second becomes

A$$\dfrac{4E}{9}$$
B$$\dfrac{7E}{9}$$
C$$\dfrac{10E}{9}$$
D$$\dfrac{16E}{9}$$
Correct Answer

Solution

From Stefan's law, energy radiated per second $$E = \sigma e A T^4$$
where $$\sigma $$ is Stefan's constant. $$A$$ is the area, $$e$$ and $$T $$ is the emmissivity and temperature of body, respectively.
Initially, temperature of body $$T = 27^o C = 300$$ K
$$\implies$$ $$E = \sigma eA (300)^4$$
Now length and breadth of body is reduced by a factor of $$3$$.
Thus new area of the body $$A' = \dfrac{L}{3}.\dfrac{B}{3}=\dfrac{A}{9}$$
New temperature of the body $$T' = 327^o C = 600 $$ K
New energy radiated per second $$E' = \sigma e.\dfrac{A}{9}. (600)^4$$
$$\therefore$$ $$\dfrac{E'}{E} = \dfrac{\sigma e . A (600)^4/9}{\sigma e . A (300)^4} = \dfrac{2^4}{9}$$
$$\implies$$ $$E' = \dfrac{16 E}{9}$$

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