A metallic sphere cools from


A metallic sphere cools from $50^{\circ} \mathrm{C}$ to $40^{\circ} \mathrm{C}$ in $300 \mathrm{~s}$. If atmospheric temperature around is $20^{\circ} \mathrm{C}$, then the sphere's temperature after the next 5 minutes will be close to :

  1. $31^{\circ} \mathrm{C}$

  2. $33^{\circ} \mathrm{C}$

  3. $28^{\circ} \mathrm{C}$

  4. $35^{\circ} \mathrm{C}$

Correct Option: , 2


(2) From Newton's Law of cooling,


Here, $\mathrm{T}_{1}=50^{\circ} \mathrm{C}, \mathrm{T}_{2}=40^{\circ} \mathrm{C}$

and \mathrm{T}_{\mathrm{o}}=20^{\circ} \mathrm{C}, \mathrm{t}=600 \mathrm{~S}=5$ minutes

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

Let $T$ be the temperature of sphere after next 5 minutes. Then

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

Dividing eqn. (ii) by (i), we get


$\Rightarrow 40-T=\frac{T}{5} \Rightarrow 200-5 T=T$

$\therefore T=\frac{200}{6}=33.3^{\circ} \mathrm{C}$

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