A body takes 5 text{ min} to cool from 80^{ci | vedclass

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(B) According to Newton's law of cooling,the rate of cooling is given by $\frac{dT}{dt} = K(T_{avg} - T_{room})$.In the first case,the body cools from

Vedclass · Accepted Answer

$9$

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(B) According to Newton's law of cooling,the rate of cooling is given by $\frac{dT}{dt} = K(T_{avg} - T_{room})$.In the first case,the body cools from $80^{\circ}C$ to $50^{\circ}C$ in $5 	ext{ min}$.$\frac{80 - 50}{5} = K \left( \frac{80 + 50}{2} - 20 ight)$$\frac{30}{5} = K(65 - 20) \implies 6 = 45K \implies K = \frac{6}{45} = \frac{2}{15} \dots (i)$In the second case,the body cools from $60^{\circ}C$ to $30^{\circ}C$ in $t 	ext{ min}$.$\frac{60 - 30}{t} = K \left( \frac{60 + 30}{2} - 20 ight)$$\frac{30}{t} = K(45 - 20) \implies \frac{30}{t} = 25K \dots (ii)$Dividing equation $(i)$ by $(ii)$:$\frac{6}{30/t} = \frac{45K}{25K}$$\frac{6t}{30} = \frac{45}{25}$$\frac{t}{5} = \frac{9}{5}$$t = 9 	ext{ min}$.

$A$ body takes $5 \text{ min}$ to cool from $80^{\circ}C$ to $50^{\circ}C$. How many minutes will it take to cool from $60^{\circ}C$ to $30^{\circ}C$,if the room temperature is $20^{\circ}C$?

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