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

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$28^{0} C$

$35^{\circ} C$

$33^{\circ} C$

$31^{\circ} C$

answer is B.

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Detailed Solution

From Newton’s Law of cooling,
$\frac{T_{1} - T_{2}}{t} = K [\frac{T_{1} + T_{2}}{2} - T_{0}]$
Here, $T_{1} = 50^{\circ} C, T_{2} = 40^{\circ} C$
$\begin{array}{l} an T_{0} = 20^{\circ} C, t = 600 s = 5 minutes \\ \Rightarrow \frac{50 - 40}{5} = K (\frac{50 + 40}{2} - 20) . . . (i) \end{array}$
Let T be the temperature of sphere after next 5 minutes.
Then
$\frac{40 - T}{5} = K (\frac{40 + T}{2} - 20)$ ...(ii)
Dividing equation, (ii) by (i), we get
$\begin{array}{l} \frac{40 - T}{10} = \frac{40 + T - 40}{50 + 40 - 40} = \frac{T}{50} \\ \Rightarrow 40 - T = \frac{T}{5} \Rightarrow 200 - 5 T = T \\ ∴ T = \frac{200}{6} = {33.3}^{\circ} C \end{array}$