Question 952: Thermodynamics & KTG

Question

Two spherical black bodies of radius $$r_1$$ and $$r_2$$ with surface temperature $$T_1$$ and $$T_2$$ respectively, radiate same power, then $$r_1: r_2$$ is

Accepted Answer

The power radiated by a black body can be described by the formula:
$ E = \varepsilon \sigma A T^4 $
Where:
$ \varepsilon $ is the emissivity of the body,
$ \sigma $ is the Stefan-Boltzmann constant,
$ A $ is the surface area,
$ T $ is the surface temperature.
For a spherical black body, the area $ A $ is given by $ 4 \pi r^2 $.
Given that two spherical bodies radiate the same power, we have:
$ E_1 = E_2 $
So:
$ \varepsilon \sigma (4 \pi r_1^2) T_1^4 = \varepsilon \sigma (4 \pi r_2^2) T_2^4 $
Since emissivity, the Stefan-Boltzmann constant, and $ 4\pi $ are common, they cancel out:
$ T_1^4 r_1^2 = T_2^4 r_2^2 $
Rearranging gives:
$ \frac{r_1}{r_2} = \left(\frac{T_2}{T_1}\right)^2 $

Answer

$$\left(\frac{T_2}{T_1}\right)^4$$

Answer

$$\left(\frac{T_2}{T_1}\right)^2$$

Answer

$$\left(\frac{T_1}{T_2}\right)^2$$

Answer

$$\left(\frac{T_1}{T_2}\right)^4$$

Thermodynamics & KTG

Two spherical black bodies of radius $r_1 and r_2 with surface temperature T_1 and T_2 respectively, radiate same power, then r_1: r_2$ is

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