A cup of coffee cools from 150^{circ} F to 14 | vedclass

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(B) According to Newton's law of cooling, the rate of cooling is proportional to the difference between the average temperature of the body and the su

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$2.14$

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(B) According to Newton's law of cooling, the rate of cooling is proportional to the difference between the average temperature of the body and the surrounding temperature: $\frac{dT}{dt} = K(T_{avg} - T_0)$.In the first case: $T_1 = 150^{\circ} F, T_2 = 144^{\circ} F, t_1 = 1 \ min, T_0 = 72^{\circ} F$.The average temperature is $T_{avg1} = \frac{150 + 144}{2} = 147^{\circ} F$.So, $\frac{150 - 144}{1} = K(147 - 72) \Rightarrow 6 = K(75) \Rightarrow K = \frac{6}{75} = 0.08 \ min^{-1}$.In the second case: $T'_1 = 110^{\circ} F, T'_2 = 104^{\circ} F, T_0 = 72^{\circ} F$.The average temperature is $T_{avg2} = \frac{110 + 104}{2} = 107^{\circ} F$.Using the same law: $\frac{110 - 104}{t_2} = K(107 - 72)$.$\frac{6}{t_2} = K(35)$.Substituting $K = \frac{6}{75}$: $\frac{6}{t_2} = \frac{6}{75} 	imes 35$.$\frac{1}{t_2} = \frac{35}{75} = \frac{7}{15}$.$t_2 = \frac{15}{7} \approx 2.14 \ min$.

$A$ cup of coffee cools from $150^{\circ} F$ to $144^{\circ} F$ in $1 \ min$ in a room at a temperature of $72^{\circ} F$. How much time will the coffee take to cool from $110^{\circ} F$ to $104^{\circ} F$ in the same room (in $min$)?

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