An object is cooled from 75^{circ} C to 65^{c | vedclass

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(C) According to Newton's law of cooling,the rate of cooling is given by: $\frac{dT}{dt} = -k(T_{avg} - T_0)$,where $T_{avg} = \frac{T_1 + T_2}{2}$ an

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$4$

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(C) According to Newton's law of cooling,the rate of cooling is given by: $\frac{dT}{dt} = -k(T_{avg} - T_0)$,where $T_{avg} = \frac{T_1 + T_2}{2}$ and $T_0$ is the surrounding temperature.Case $(1)$: Cooling from $75^{\circ} C$ to $65^{\circ} C$ in $2 	ext{ min}$.$T_{avg} = \frac{75 + 65}{2} = 70^{\circ} C$.Rate of cooling = $\frac{75 - 65}{2} = \frac{10}{2} = 5^{\circ} C/	ext{min}$.So,$5 = k(70 - 30) = k(40) \Rightarrow k = \frac{5}{40} = \frac{1}{8} 	ext{ min}^{-1}$.Case $(2)$: Cooling from $55^{\circ} C$ to $45^{\circ} C$ in $t 	ext{ min}$.$T_{avg} = \frac{55 + 45}{2} = 50^{\circ} C$.Rate of cooling = $\frac{55 - 45}{t} = \frac{10}{t}  ^{\circ} C/	ext{min}$.Using the law: $\frac{10}{t} = k(50 - 30) = k(20)$.Substituting $k = \frac{1}{8}$:$\frac{10}{t} = \frac{1}{8} 	imes 20 = 2.5$.$t = \frac{10}{2.5} = 4 	ext{ min}$.

An object is cooled from $75^{\circ} C$ to $65^{\circ} C$ in $2 \text{ min}$. The time it takes to cool from $55^{\circ} C$ to $45^{\circ} C$ is [The temperature of the surrounding is $30^{\circ} C$] (in $\text{ min}$)

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