A body takes 5 min to cool from 90^{circ}C t | vedclass

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(C) According to Newton's Law of Cooling,$\frac{dT}{dt} = K(T_{avg} - T_s)$.For the first interval: $\frac{90 - 60}{5} = K\left( \frac{90 + 60}{2} - 2

Vedclass · Accepted Answer

$11$

Vedclass · Answer

(C) According to Newton's Law of Cooling,$\frac{dT}{dt} = K(T_{avg} - T_s)$.For the first interval: $\frac{90 - 60}{5} = K\left( \frac{90 + 60}{2} - 20 ight) \Rightarrow 6 = K(75 - 20) \Rightarrow 6 = 55K \Rightarrow K = \frac{6}{55}$.For the second interval: $\frac{60 - 30}{t} = K\left( \frac{60 + 30}{2} - 20 ight) \Rightarrow \frac{30}{t} = \frac{6}{55}(45 - 20) \Rightarrow \frac{30}{t} = \frac{6}{55} 	imes 25$.$\frac{30}{t} = \frac{6 	imes 5}{11} \Rightarrow \frac{30}{t} = \frac{30}{11} \Rightarrow t = 11 \ min$.

$A$ body takes $5 \ min$ to cool from $90^{\circ}C$ to $60^{\circ}C$. If the temperature of the surroundings is $20^{\circ}C$,how much time (in $min$) will it take to cool from $60^{\circ}C$ to $30^{\circ}C$?

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