Question 762: Thermodynamics & KTG

Question

A black sphere has radius $R$ whose rate of radiation is E at temperature T . If radius is made half and temperature 4 T , the rate of radiation will be

Accepted Answer

To determine the change in the rate of radiation when the radius and temperature of a black sphere are altered, consider these steps:
Initial Formula for Radiation
:
The rate of radiation $ E $ is given by the formula:
$    E = eA \sigma T^4    $
where $ e $ is the emissivity, $ A $ is the surface area, $ \sigma $ is the Stefan-Boltzmann constant, and $ T $ is the temperature.
Surface Area of a Sphere
:
For a sphere, the surface area $ A $ is:
$    A = 4\pi R^2    $
Thus, the initial radiation rate is:
$    E = e(4\pi R^2) \sigma T^4    $
Revised Conditions
:
The sphere's radius is reduced to $ \frac{R}{2} $, and the temperature is increased to $ 4T $.
New Radiation Formula
:
With the new radius and temperature, let the new rate of radiation be $ E_2 $. Substitute the new values into the radiation formula:
$    E_2 = e \left(4\pi \left(\frac{R}{2}\right)^2\right) \sigma (4T)^4    $
Comparing Both Rates
:
To find the ratio of the initial and new radiation rates:
$    \frac{E}{E_2} = \frac{R^2 T^4}{\left(\frac{R}{2}\right)^2 (4T)^4}    $
Calculation
:
Simplify the ratio:
$    \frac{E}{E_2} = \frac{R^2 T^4}{\frac{R^2}{4} \cdot 256 T^4} = \frac{1}{\frac{1}{4} \cdot 256} = \frac{1}{64}    $
Therefore, $ E_2 = 64E $.
From these calculations, when the radius is halved and the temperature is increased by four times, the rate of radiation increases to $ 64 $ times the original rate.

Answer

64 E

Answer

32 E

Answer

16 E

Answer

8 E

Thermodynamics & KTG

A black sphere has radius R whose rate of radiation is E at temperature T . If radius is made half and temperature 4 T , the rate of radiation will be

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