A body cools from 60^{circ} C to 40^{circ} C in 6 minutes. If, temperature of surroundings is 10^{ci

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10-2.Transmission of Heat

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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 $.........{ }^{\circ} C$.

A

$28$

B

$22$

C

$20$

D

$21$

(JEE MAIN-2023)

Solution

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 $(ii)$

$\frac{20}{40- T }=\frac{40}{10+ T / 2}$

$10+\frac{ T }{2}=80-2\,T$

$\frac{5 T }{2}=70 \Rightarrow T =28^{\circ}\,C$

Standard 11

Physics

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