A metallic sphere cools from 50   ° C to 40   ° C in 300   s . If atmospheric…

Using Newton's Law of cooling

$\frac{50-40}{5 \min }=\mathrm{K}\left(\frac{50+40}{2}-20\right)$     ......(i)

Next $5 \min .$

$\frac{40-\mathrm{\theta }}{5}=\mathrm{K}\left(\frac{40+\mathrm{\theta }}{2}-20\right)$     .....(ii)

Dividing (ii)/(i)

$\frac{40-\mathrm{\theta }}{10}=\frac{40+\mathrm{\theta }-40}{50+40-40}=\frac{\mathrm{\theta }}{50}$

$40-\mathrm{\theta }=\frac{\mathrm{\theta }}{5}$

$200-5\mathrm{\theta }=\mathrm{\theta }$

$\therefore \mathrm{\theta }=\frac{200}{6}=33.3 °\mathrm{C}$