Question

The rectangular surface of area $8cm\times 4cm$ of a block body at a temperature of ${127}^{o}C$ emits energy at the rate of $E$ per second. If the length and breath of the surface are each reduced to half of the initial value and the temperature is raised to ${327}^{o}C$, the rate of emission of energy will become

• A. $\frac{3}{8}E$
• B. $\frac{81}{16}E$
• C. $\frac{9}{16}E$
• D. $\frac{81}{64}E$

Answer 1

Answer: (D)

$\frac{81}{64}E$

Emitted power $\left(P\right)=E/s=\sigma eA{T}^4$ where symbols

have their usual meanings

$P_i=\sigma eA{T}^4$

$P_f=\sigma e{A}_f{T}_f^4...\left(1\right)$

Now, $T={127}^{\circ }C=400K$

$T_f={327}^{\circ }C=600K$

$\Rightarrow T_f=\frac{3}{2}T$

Now, $A=5\times 4=32c{m}^2$

$A_f=\frac{8}{2}\times \frac{4}{2}=8c{m}^2$

$\Rightarrow A_f=\frac{A}{4}$

Put in (1) ve get,

$P_f=\sigma e\left(\frac{A}{4}\right){\left(\frac{3T}{2}\right)}^4$

$=\sigma e\frac{A}{4}\times \frac{81}{16}$

$=\frac{81}{64}\left(E/s\right)$ (D)