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(D) According to Newton's Law of Cooling: $\frac{T_1 - T_2}{t} = K \left( \frac{T_1 + T_2}{2} - T_s \right)$For the first interval ($80^{\circ}C$ to $

Vedclass · Accepted Answer

$15$

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(D) According to Newton's Law of Cooling: $\frac{T_1 - T_2}{t} = K \left( \frac{T_1 + T_2}{2} - T_s ight)$For the first interval ($80^{\circ}C$ to $70^{\circ}C$):$\frac{80 - 70}{12} = K \left( \frac{80 + 70}{2} - 25 ight)$$\frac{10}{12} = K (75 - 25) = 50K \implies K = \frac{10}{12 	imes 50} = \frac{1}{60} \ldots(1)$For the second interval ($70^{\circ}C$ to $60^{\circ}C$):$\frac{70 - 60}{t} = K \left( \frac{70 + 60}{2} - 25 ight)$$\frac{10}{t} = K (65 - 25) = 40K \ldots(2)$Substituting $K = \frac{1}{60}$ into equation $(2)$:$\frac{10}{t} = 40 	imes \frac{1}{60} = \frac{2}{3}$$t = \frac{10 	imes 3}{2} = 15 	ext{ minutes}$.

An object kept in a large room having air temperature of $25^{\circ}C$ takes $12 \text{ minutes}$ to cool from $80^{\circ}C$ to $70^{\circ}C$. The time taken to cool for the same object from $70^{\circ}C$ to $60^{\circ}C$ would be nearly.....$min$

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