A metallic sphere cools from $50^{\circ} C$ to $40^{\circ} C$ in $300 \,s.$ If atmospheric temperature around is $20^{\circ} C ,$ then the sphere's temperature after the next $5$ minutes will be close to$.....C$
JEE MAIN 2020, Medium
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$\frac{50-40}{300}=\beta\left(\frac{50+40}{2}-20\right)$
$\frac{40-T}{300}=\beta\left(\frac{40+T}{2}-20\right)$
$\therefore \quad T=\frac{100}{3}$
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