It takes 10 minutes to cool a liquid from 61^ | vedclass

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(D) According to Newton's Law of Cooling,the rate of cooling is given by $\frac{d	heta}{dt} = k(	heta_{avg} - 	heta_0)$,where $	heta_0$ is the roo

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$15$

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(D) According to Newton's Law of Cooling,the rate of cooling is given by $\frac{d	heta}{dt} = k(	heta_{avg} - 	heta_0)$,where $	heta_0$ is the room temperature.For the first case:$\frac{61 - 59}{10} = k \left( \frac{61 + 59}{2} - 30 ight)$$\frac{2}{10} = k(60 - 30)$$0.2 = 30k \implies k = \frac{0.2}{30} = \frac{1}{150} \ldots (1)$For the second case:$\frac{51 - 49}{t} = k \left( \frac{51 + 49}{2} - 30 ight)$$\frac{2}{t} = k(50 - 30)$$\frac{2}{t} = 20k \ldots (2)$Substituting $k$ from equation $(1)$ into equation $(2)$:$\frac{2}{t} = 20 	imes \frac{1}{150}$$\frac{2}{t} = \frac{2}{15}$$t = 15 \ min$.

It takes $10$ minutes to cool a liquid from $61^{\circ} C$ to $59^{\circ} C$. If the room temperature is $30^{\circ} C$,then the time taken in cooling from $51^{\circ} C$ to $49^{\circ} C$ is .......... $min$.

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