Question

A spherical ball A of surface area 20 cm2 is kept at the centre of a hollow spherical shell B of area 80 cm2. The surface of A and the inner surface of B emit as blackbodies. Both A and B are at 300 K. (a) How much is the radiation energy emitted per second by the ball A? (b) How much is the radiation energy emitted per second by the inner surface of B? (c) How much of the energy emitted by the inner surface of B falls back on this surface itself?

Answer 1

Explanation:

Given in the question  

Spherical Ball surface area, $\mathrm{S}_{\mathrm{A}}=20 \mathrm{cm}^{2}=20 \times 10^{-4} \mathrm{m}^{2}$

Spherical shell  Surface area, $\mathrm{S}_{\mathrm{B}}=80 \mathrm{cm}^{2}=8 \mathrm{p} \times 10^{-4} \mathrm{m}^{2}$

The spherical ball Temperature, $\mathrm{T}_{\mathrm{A}}=300 \mathrm{K}$

The spherical shell Temperature , $\mathrm{T}_{\mathrm{B}}=300 \mathrm{K}$

(a)The radiation energy emitted per second is given by the spherical ball A

$E_{A}=\sigma S_{A} T_{A}^{4}$

$E_{A}=6.0 \times 10^{-8} \times 20 \times 10^{-4} \times 300^{4}$

$E_{A}=120 \times 10^{-12} \times 81 \times 10^{8}$

$E_{A}=120 \times 10^{-4} \times 81$

$E_{A}=9720 \times 10^{-4}$

$E_{A}=0.97 J$

(b)The radiation energy emitted per second from the spherical shell B's inner surface is given by

$E_{B}=\sigma S_{B} T_{B}^{4}$

$E_{B}=6.0 \times 10^{-8} \times 80 \times 10^{-4} \times 300^{4}$

$E_{B}=480 \times 10^{-12} \times 81 \times 10^{8}$

$E_{B}=480 \times 10^{-4} \times 81$

$E_{B}=38880 \times 10^{-4}$

$E_{B}=3.8 \mathrm{J}$

(c)Energy emitted from the inner surface of B falling back on its surface is provided by

$E=E_{B}-E_{A}=3.8-0.97$

$E=2.83 \mathrm{J}$