Question 937: Thermodynamics & KTG

Question

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

Accepted Answer

$$\mathrm{E}=\mathrm{e} \sigma \cdot \mathrm{A}\left(\mathrm{T}^4-\mathrm{T}_0^4ight) 	ext { and } \mathrm{A}=\ell \mathrm{b}$$
When $$\ell$$ and $$b$$ change to $$\frac{\ell}{3}$$ and $$\frac{b}{3}$$
$$\begin{aligned}
& A^{\prime}=\frac{A}{9} \
& \frac{E^{\prime}}{E}=\frac{A^{\prime}}{A} \frac{(327+273)^4}{(27+273)^4} ; \quad \frac{E^{\prime}}{E}=\frac{1}{9}\left(\frac{600}{300}ight)^4 \
& 	herefore E^{\prime}=\frac{1}{9} 	imes(2)^4 	imes E \quad \Rightarrow E^{\prime}=\frac{16 E}{9}
\end{aligned}$$

Answer

$$\frac{16 \mathrm{E}}{9}$$

Answer

$$\frac{8 \mathrm{E}}{9}$$

Answer

$$\frac{4 \mathrm{E}}{9}$$

Answer

$$\frac{12 \mathrm{E}}{9}$$

Thermodynamics & KTG

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

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