The rate of cooling at 600 K is R,given the | vedclass

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(A) According to Stefan-Boltzmann Law,the rate of cooling $R$ is proportional to $(T^4 - T_0^4)$,where $T$ is the temperature of the body and $T_0$ is

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$\frac{16}{3}R$

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(A) According to Stefan-Boltzmann Law,the rate of cooling $R$ is proportional to $(T^4 - T_0^4)$,where $T$ is the temperature of the body and $T_0$ is the temperature of the surroundings.So,$\frac{R_1}{R_2} = \frac{T_1^4 - T_0^4}{T_2^4 - T_0^4}$.Given $T_1 = 600 \ K$,$T_2 = 900 \ K$,$T_0 = 300 \ K$,and $R_1 = R$.Substituting the values:$\frac{R}{R_2} = \frac{(600)^4 - (300)^4}{(900)^4 - (300)^4} = \frac{(300)^4 (2^4 - 1^4)}{(300)^4 (3^4 - 1^4)} = \frac{16 - 1}{81 - 1} = \frac{15}{80} = \frac{3}{16}$.Therefore,$R_2 = \frac{16}{3}R$.

The rate of cooling at $600 \ K$ is $R$,given the surrounding temperature is $300 \ K$. What is the rate of cooling at $900 \ K$?

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