A metallic sphere cools from

Question:

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

Solution:

(2) From Newton's Law of cooling,

$\frac{T_{1}-T_{2}}{t}=K\left[\frac{T_{1}+T_{2}}{2}-T_{0}\right]$

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)$....(ii) Dividing eqn. (ii) by (i), we get$\frac{40-T}{10}=\frac{40+T-40}{50+40-40}=\frac{T}{50}\Rightarrow 40-T=\frac{T}{5} \Rightarrow 200-5 T=T\therefore T=\frac{200}{6}=33.3^{\circ} \mathrm{C}\$