Question 942: Thermodynamics & KTG

Question

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

Accepted Answer

The energy emitted by a black body per unit area per unit time (or power per unit area) is given by Stefan-Boltzmann Law :
$ \sigma T^4 $
Where :
$ \sigma $ is the Stefan-Boltzmann constant.
$ T $ is the absolute temperature of the body.
Given :
Initial energy emitted $ E $ for area $ A $ at temperature $ 27^\circ C = 273 + 27 = 300 	ext{K} $
$ E = A \sigma (300^4) $
Now, the area is reduced to $(1/3)^2$ of its original value :
$ A' = \left(\frac{1}{3}ight)^2 A = \frac{A}{9} $
The temperature is raised to $ 327^\circ C = 273 + 327 = 600 	ext{K} $
The new energy emitted $ E' $ will be :
$ E' = A' \sigma (600^4) $
Dividing the two equations :
$ \frac{E'}{E} = \frac{A' \sigma (600^4)}{A \sigma (300^4)} $
Given $ A' = \frac{A}{9} $ and using $ (600^4)/(300^4) = 2^4 = 16 $ :
$ \frac{E'}{E} = \frac{16}{9} $
So, $ E' = \frac{16E}{9} $
The correct answer is :
Option C
$\frac{16E}{9}$

Answer

$$\frac{20 \mathrm{E}}{9}$$

Answer

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

Answer

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

Answer

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

Thermodynamics & KTG

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

Choose the correct answer: