Question 642: Thermodynamics & KTG

Question

A black sphere has radius $R$ whose rate of radiation is E at temperature T . If radius is made R / 2 and temperature 3T, the rate of radiation will be

Accepted Answer

Step 1: Formula for power radiated
The rate of radiation (power) from a body is given by
StefanÃ¢ÂÂBoltzmann law
:
$ P = \sigma A T^4 $
where
$\sigma$ = StefanÃ¢ÂÂBoltzmann constant,
$A$ = surface area of the sphere,
$T$ = absolute temperature.
For a sphere of radius $R$:
$ A = 4\pi R^2. $
So,
$ E = \sigma (4\pi R^2) T^4. $
Step 2: New situation
Radius is halved: $R' = 	frac{R}{2}$.
Temperature is tripled: $T' = 3T$.
So new power is
$ E' = \sigma \cdot (4\pi (R/2)^2) \cdot (3T)^4. $
Step 3: Simplify
$ E' = \sigma \cdot 4\pi \cdot \frac{R^2}{4} \cdot 81 T^4 = \sigma \cdot \pi R^2 \cdot 81 T^4. $
Original was
$ E = \sigma (4\pi R^2) T^4. $
So
$ \frac{E'}{E} = \frac{ \sigma \pi R^2 \cdot 81 T^4}{ \sigma (4\pi R^2) T^4 } = \frac{81}{4}. $
Hence,
$ E' = \frac{81}{4} E. $
Ã¢ÂÂ Final Answer:
$ \boxed{\frac{81E}{4}} $
That corresponds to
Option C
.

Answer

$\frac{3 \mathrm{E}}{2}$

Answer

$\frac{27 \mathrm{E}}{8}$

Answer

$\frac{81 \mathrm{E}}{4}$

Answer

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

Thermodynamics & KTG

A black sphere has radius R whose rate of radiation is E at temperature T . If radius is made R / 2 and temperature 3T, the rate of radiation will be

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