A small, perfectly black sphere of radius r is suspended at the center of a large spherical enclosure of radius R ( R ≫ r ). The enclosure walls are maintained at a constant temperature T 0 . A smal

Question

A small, perfectly black sphere of radius r is suspended at the center of a large spherical enclosure of radius R ( R ≫ r ). The enclosure walls are maintained at a constant temperature T 0 ​ . A small circular aperture of radius a ( a ≪ r ) is made on the enclosure wall. A parallel beam of radiation of intensity I falls on the sphere through this aperture. The sphere is in thermal equilibrium. What is the equilibrium temperature of the sphere?

TestHub · Accepted Answer

For the sphere to be in thermal equilibrium, the total power absorbed must equal the total power emitted. The power absorbed by the sphere from the parallel beam of intensity $I$ is $P_{ab s, b e am} = I \times (cross-sectional area of sphere) = I (π r^{2})$ .

Since the sphere is perfectly black, its emissivity $ϵ = 1$ and absorptivity $α = 1$ . The sphere also absorbs radiation from the enclosure walls at temperature $T_{0}$ . The power absorbed from the enclosure walls is $P_{ab s, e n c l os u re} = α σ A T_{0}^{4} = 1 \cdot σ (4 π r^{2}) T_{0}^{4}$ . The power emitted by the sphere at its equilibrium temperature $T_{e q}$ is $P_{e mi t, s p h ere} = ϵ σ A T_{e q}^{4} = 1 \cdot σ (4 π r^{2}) T_{e q}^{4}$ . At thermal equilibrium: $P_{ab s, b e am} + P_{ab s, e n c l os u re} = P_{e mi t, s p h ere}$ . Substituting the expressions: $I (π r^{2}) + σ (4 π r^{2}) T_{0}^{4} = σ (4 π r^{2}) T_{e q}^{4}$ . Divide the entire equation by $4 π r^{2}$ : $\frac{I}{4} + σ T_{0}^{4} = σ T_{e q}^{4}$ . Rearranging for $T_{e q}^{4}$ : $T_{e q}^{4} = \frac{I}{4 σ} + T_{0}^{4}$ . Thus, the equilibrium temperature of the sphere is $T_{e q} = (\frac{I}{4 σ} + T_{0}^{4})^{1/4}$ .

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