It takes 1 , min for hot water to cool from 8 | vedclass

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(D) 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 sur

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$50$

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(D) 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: $\frac{d	heta}{dt} = k \left( \frac{	heta_1 + 	heta_2}{2} - 	heta_s ight)$.For the first interval ($80^{\circ}C$ to $60^{\circ}C$ in $60 \, sec$):$\frac{80 - 60}{60} = k \left( \frac{80 + 60}{2} - 30 ight) \implies \frac{20}{60} = k(70 - 30) \implies \frac{1}{3} = k(40) \implies k = \frac{1}{120} \, sec^{-1}$.For the second interval ($60^{\circ}C$ to $50^{\circ}C$ in $t \, sec$):$\frac{60 - 50}{t} = k \left( \frac{60 + 50}{2} - 30 ight) \implies \frac{10}{t} = k(55 - 30) \implies \frac{10}{t} = k(25)$.Substituting $k = \frac{1}{120}$:$\frac{10}{t} = \frac{1}{120} 	imes 25 \implies \frac{10}{t} = \frac{25}{120} \implies \frac{10}{t} = \frac{5}{24}$.$5t = 240 \implies t = 48 \, sec$.Rounding to the nearest option,the answer is $50 \, sec$.

It takes $1 \, min$ for hot water to cool from $80^{\circ}C$ to $60^{\circ}C$. Find the time taken (in $sec$) to cool from $60^{\circ}C$ to $50^{\circ}C$. The temperature of the surroundings is $30^{\circ}C$.

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