The temperature of two solid spheres of same material of diameter… - Vidyadip

Question

The temperature of two solid spheres of same material of diameters 10 cm and 8 cm are $327^{\circ}\text C $ and $227^{\circ}\text C.$ Temperature of the environment is $27^{\circ}\text C.$ With the help of Stefan's law compare the rates of cooling of the two spheres.

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Answer

The rate of cooling of any object
$\begin{aligned}& R & =\frac{\sigma e_{r} A}{J}\left[\left(T^4-T_0^4\right)\right] \\\therefore & ~\frac{R_1}{R_2} & =\frac{A_1}{A_2}\left[\frac{T_1^4-T_0^4}{T_2^4-T_0^4}\right]\end{aligned}$
$\begin{array}{l}=\frac{4 \pi r_1^2}{4 \pi r_2^2}\left[\frac{T_1^4-T_0^4}{T_2^4-T_0^4}\right] \\=\left(\frac{r_1}{r_2}\right)^2\left[\frac{T_1^4-T_0^4}{T_2^4-T_0^4}\right]\end{array}$
$\begin{array}{l}\text { Given : } \quad r_1=5 \times 10^{-2} m \\r_2=4 \times 10^{-2} m \\T_1=327^{\circ} C=327+273=600 K \\T_2=227^{\circ} C=227+273=500 K \\T_0=27^{\circ} C=27+273=300 K \\\therefore \frac{R_1}{R_2}=\left(\frac{5 \times 10^{-2}}{4 \times 10^{-2}}\right)^2\left[\frac{(600)^4-(300)^4}{(500)^4-(300)^4}\right] \\\quad=\frac{25}{16} \times\left[\frac{1296-81}{625-81}\right]=\frac{3.49}{1} \\\therefore \quad R_1: R_2=3.49: 1\end{array}$

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