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

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

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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 temperature difference between the body and its surroundings: $\frac{dT}{dt} = K(T - T_s)$.For the first case:Average temperature $T_{avg1} = \frac{94 + 86}{2} = 90\,^{\circ}C$.Temperature difference $\Delta T_1 = 90 - 20 = 70\,^{\circ}C$.Rate of cooling $\frac{dT_1}{dt_1} = \frac{94 - 86}{2} = 4\,^{\circ}C/min$.So,$4 = K(70) \implies K = \frac{4}{70} = \frac{2}{35}$.For the second case:Average temperature $T_{avg2} = \frac{71 + 69}{2} = 70\,^{\circ}C$.Temperature difference $\Delta T_2 = 70 - 20 = 50\,^{\circ}C$.Rate of cooling $\frac{dT_2}{dt_2} = \frac{71 - 69}{t} = \frac{2}{t}$.Using the law: $\frac{2}{t} = K(50)$.Substituting $K = \frac{2}{35}$:$\frac{2}{t} = \frac{2}{35} 	imes 50 = \frac{100}{35} = \frac{20}{7}$.$t = \frac{2 	imes 7}{20} = 0.7\,min$.Converting to seconds: $0.7 	imes 60 = 42\,s$.

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

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