A sphere is at temperature 600 K. In an exte | vedclass

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(A) According to Stefan-Boltzmann Law,the rate of cooling $R$ of a body at temperature $T$ in an environment at temperature $T_0$ is given by: $R = e

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$\frac{3}{16} R$

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(A) According to Stefan-Boltzmann Law,the rate of cooling $R$ of a body at temperature $T$ in an environment at temperature $T_0$ is given by: $R = e \sigma A (T^4 - T_0^4)$ For the first case,$T = 600 \ K$ and $T_0 = 200 \ K$: $R = k (600^4 - 200^4)$ For the second case,$T' = 400 \ K$ and $T_0 = 200 \ K$: $R' = k (400^4 - 200^4)$ Taking the ratio: $\frac{R'}{R} = \frac{400^4 - 200^4}{600^4 - 200^4} = \frac{(4^4 - 2^4) 	imes 10^8}{(6^4 - 2^4) 	imes 10^8}$ $\frac{R'}{R} = \frac{256 - 16}{1296 - 16} = \frac{240}{1280} = \frac{24}{128} = \frac{3}{16}$ Therefore,$R' = \frac{3}{16} R$.

$A$ 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 the cooling rate $R'$ will become:

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