A cup of tea cools from 80,^{circ}C to 60,^{c | vedclass

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(D) According to Newton's Law of Cooling,$\frac{dT}{dt} = -K(T - T_s)$,where $T_s$ is the surrounding temperature.For the first interval: $\frac{80 -

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$48$

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(D) According to Newton's Law of Cooling,$\frac{dT}{dt} = -K(T - T_s)$,where $T_s$ is the surrounding temperature.For the first interval: $\frac{80 - 60}{1} = K \left( \frac{80 + 60}{2} - 30 ight) \implies 20 = K(70 - 30) \implies 20 = 40K \implies K = 0.5\,min^{-1}$.For the second interval: $\frac{60 - 50}{t} = K \left( \frac{60 + 50}{2} - 30 ight) \implies \frac{10}{t} = 0.5(55 - 30) \implies \frac{10}{t} = 0.5 	imes 25 \implies \frac{10}{t} = 12.5$.Solving for $t$: $t = \frac{10}{12.5} = 0.8\,min$.Converting to seconds: $t = 0.8 	imes 60 = 48\,sec$.

$A$ cup of tea cools from $80\,^{\circ}C$ to $60\,^{\circ}C$ in $1\,min$. The ambient temperature is $30\,^{\circ}C$. In cooling from $60\,^{\circ}C$ to $50\,^{\circ}C$,it will take ....... $sec$.

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