In a room where the temperature is 30^{circ} | vedclass

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(B) According to Newton's Law of Cooling, the rate of cooling is given by $\frac{	heta_1 - 	heta_2}{t} = K \left( \frac{	heta_1 + 	heta_2}{2} -

Vedclass · Accepted Answer

$6$ minutes

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(B) According to Newton's Law of Cooling, the rate of cooling is given by $\frac{	heta_1 - 	heta_2}{t} = K \left( \frac{	heta_1 + 	heta_2}{2} - 	heta_0 ight)$.For the first case: $	heta_1 = 61^{\circ} C$, $	heta_2 = 59^{\circ} C$, $t = 4$ minutes, and $	heta_0 = 30^{\circ} C$.$\frac{61 - 59}{4} = K \left( \frac{61 + 59}{2} - 30 ight) \Rightarrow \frac{2}{4} = K(60 - 30) \Rightarrow 0.5 = 30K \Rightarrow K = \frac{0.5}{30} = \frac{1}{60}$.For the second case: $	heta_1 = 51^{\circ} C$, $	heta_2 = 49^{\circ} C$, and $	heta_0 = 30^{\circ} C$.$\frac{51 - 49}{t} = K \left( \frac{51 + 49}{2} - 30 ight) \Rightarrow \frac{2}{t} = \frac{1}{60} (50 - 30) \Rightarrow \frac{2}{t} = \frac{20}{60} \Rightarrow \frac{2}{t} = \frac{1}{3}$.Therefore, $t = 6$ minutes.

In a room where the temperature is $30^{\circ} C$, a body cools from $61^{\circ} C$ to $59^{\circ} C$ in $4$ minutes. The time taken by the body to cool from $51^{\circ} C$ to $49^{\circ} C$ will be:

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