A sphere is at temperature 600 K . In an external environment of 200 K , its cooling rate is ' $R$ '. When the temperature of the sphere falls to 400 K , then cooling rate ' $R$ ' will become

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Aicharya · Accepted Answer

The rate energy emission from a hot surface is given by Stefan-Boltzmann Law. $\therefore \quad \mathrm{R}=\mathrm{e} \mathrm{\sigma A}\left(\mathrm{~T}^4-\mathrm{T}_0^4\right)$
Hence, $\frac{\mathrm{R}^{\prime}}{\mathrm{R}}=\frac{\left(400^4-200^4\right)}{\left(600^4-200^4\right)}=\frac{(256-16) \times 10^8}{(1296-16) \times 10^8}=\frac{3}{16}$ $\therefore \quad \mathrm{R}^{\prime}=\frac{3}{16} \mathrm{R}$

A sphere is at temperature 600 K . In an external environment of 200 K , its cooling rate is ' $R$ '. When…

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