Question 216: Thermodynamics & KTG

Question

Two spherical bodies of same materials having radii 0.2 m and 0.8 m are placed in same atmosphere. The temperature of the smaller body is 800 K and temperature of the bigger body is 400 K . If the energy radiated from the smaller body is E, the energy radiated from the bigger body is (assume, effect of the surrounding temperature to be negligible),

Accepted Answer

To determine the energy radiated by two spherical bodies, both made of the same material but with different radii and temperatures, we can use the Stefan-Boltzmann law. This law states that the energy radiated by a black body per unit time is proportional to the product of the surface area $ A $ and the fourth power of its temperature $ T $.
The formula for power $ P $ is given by:
$ P = \sigma e A T^4 $
Where:
$ \sigma $ is the Stefan-Boltzmann constant
$ e $ is the emissivity of the material
$ A $ is the surface area of the sphere
$ T $ is the temperature in Kelvin
Expressing this relationship as a proportionality:
$ P \propto A T^4 $
For a sphere, the surface area $ A $ is:
$ A = 4\pi R^2 $
Hence, the power radiated is proportional to:
$ P \propto R^2 T^4 $
For two bodies with radii $ R_1 $ and $ R_2 $, and temperatures $ T_1 $ and $ T_2 $, the ratio of their powers is:
$ \frac{P_1}{P_2} = \left( \frac{R_1}{R_2} ight)^2 \left( \frac{T_1}{T_2} ight)^4 $
Given:
$ R_1 = 0.2 \, 	ext{m} $, $ T_1 = 800 \, 	ext{K} $
$ R_2 = 0.8 \, 	ext{m} $, $ T_2 = 400 \, 	ext{K} $
Substituting into the formula:
$ \frac{E}{E'} = \left( \frac{0.2}{0.8} ight)^2 \left( \frac{800}{400} ight)^4 $
Calculating each part:
$\left( \frac{0.2}{0.8} ight)^2 = \left( \frac{1}{4} ight)^2 = \frac{1}{16}$
$\left( \frac{800}{400} ight)^4 = 2^4 = 16$
Combining these calculations:
$ \frac{E}{E'} = \frac{1}{16} 	imes 16 = 1 $
This implies that the energy radiated by the bigger body $ E' $ equals the energy radiated by the smaller body $ E $. Thus, $ E' = E $.

Answer

64 E

Answer

16 E

Answer

E

Answer

256 E

Thermodynamics & KTG

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