A pan filled with hot food cools from 94^{cir | vedclass

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(D) According to Newton's Law of Cooling,the rate of cooling is proportional to the difference in temperature between the body and its surroundings: $

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$42$

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(D) According to Newton's Law of Cooling,the rate of cooling is proportional to the difference in temperature between the body and its surroundings: $\frac{dT}{dt} = K(T_{avg} - T_{surr})$.For the first interval: $\frac{94-86}{2} = K\left(\frac{94+86}{2} - 20ight) \implies 4 = K(90 - 20) \implies 4 = 70K \implies K = \frac{4}{70} = \frac{2}{35}$.For the second interval: $\frac{71-69}{t} = K\left(\frac{71+69}{2} - 20ight) \implies \frac{2}{t} = K(70 - 20) \implies \frac{2}{t} = 50K$.Substituting $K = \frac{2}{35}$ into the second equation: $\frac{2}{t} = 50 	imes \frac{2}{35} \implies \frac{1}{t} = \frac{50}{35} = \frac{10}{7} \implies t = \frac{7}{10} 	ext{ minutes}$.Converting to seconds: $t = \frac{7}{10} 	imes 60 = 42 	ext{ seconds}$.

$A$ pan filled with hot food cools from $94^{\circ} C$ to $86^{\circ} C$ in $2$ minutes when the room temperature is $20^{\circ} C$. How long will it take to cool from $71^{\circ} C$ to $69^{\circ} C$?

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