A body cools from $60^{\circ} C$ to $40^{\circ} C$ in $6$ minutes. If, temperature of surroundings is $10^{\circ} C$. Then, after the next 6 minutes, its temperature will be $.........{ }^{\circ} C$.
JEE MAIN 2023, Medium
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By average form of Newton's law of cooling
$\frac{20}{6}= k (50-10)$
$\frac{40- T }{6}= K \left(\frac{40+ T }{2}-10\right)$
From equation $(i)$ and $(ii)$
$\frac{20}{40- T }=\frac{40}{10+ T / 2}$
$10+\frac{ T }{2}=80-2\,T$
$\frac{5 T }{2}=70 \Rightarrow T =28^{\circ}\,C$
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