A body takes 4, text{min} to cool from 61^{ci | vedclass

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(C) According to Newton's Law of Cooling,the rate of cooling is given by $\frac{\Delta T}{\Delta t} = K(T_{avg} - T_s)$,where $T_{avg}$ is the average

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$6$

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(C) According to Newton's Law of Cooling,the rate of cooling is given by $\frac{\Delta T}{\Delta t} = K(T_{avg} - T_s)$,where $T_{avg}$ is the average temperature of the body and $T_s$ is the temperature of the surroundings.For the first case:$\frac{61 - 59}{4} = K \left( \frac{61 + 59}{2} - 30 ight)$$\frac{2}{4} = K(60 - 30)$$0.5 = K(30) \implies K = \frac{0.5}{30} = \frac{1}{60}$.For the second case,let the time taken be $t$:$\frac{51 - 49}{t} = K \left( \frac{51 + 49}{2} - 30 ight)$$\frac{2}{t} = K(50 - 30)$$\frac{2}{t} = K(20)$.Substituting the value of $K$:$\frac{2}{t} = \frac{1}{60} 	imes 20$$\frac{2}{t} = \frac{1}{3}$$t = 6\, 	ext{min}$.

$A$ body takes $4\, \text{min}$ to cool from $61^{\circ} \text{C}$ to $59^{\circ} \text{C}$. If the temperature of the surroundings is $30^{\circ} \text{C}$,the time taken by the body to cool from $51^{\circ} \text{C}$ to $49^{\circ} \text{C}$ is $....\, \text{min}$.

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