A bowl filled with very hot soup cools from 9 | vedclass

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

Vedclass · Accepted Answer

$1.4$

Vedclass · Answer

(B) According to Newton's Law of Cooling,the rate of cooling is proportional to the temperature difference between the body and its surroundings: $\frac{dT}{dt} = -K(T - T_0)$.For small temperature changes,we use the average temperature: $\frac{T_1 - T_2}{\Delta t} = K(T_{avg} - T_0)$.Case $1$: $\frac{98 - 86}{2} = K\left(\frac{98 + 86}{2} - 22ight) \implies 6 = K(92 - 22) \implies 6 = K(70) \implies K = \frac{6}{70}$.Case $2$: $\frac{75 - 69}{\Delta t} = K\left(\frac{75 + 69}{2} - 22ight) \implies \frac{6}{\Delta t} = K(72 - 22) \implies \frac{6}{\Delta t} = K(50)$.Substituting $K = \frac{6}{70}$ into the second equation: $\frac{6}{\Delta t} = \frac{6}{70} 	imes 50$.$\frac{1}{\Delta t} = \frac{50}{70} = \frac{5}{7} \implies \Delta t = \frac{7}{5} = 1.4 	ext{ minutes}$.

$A$ bowl filled with very hot soup cools from $98^{\circ}C$ to $86^{\circ}C$ in $2$ minutes when the room temperature is $22^{\circ}C$. How many minutes will it take to cool from $75^{\circ}C$ to $69^{\circ}C$?

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