If a body is heated to 110^{circ} C and place | vedclass

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(C) According to Newton's Law of Cooling,$\frac{dT}{dt} = -K(T - T_s)$,where $T_s = 10^{\circ} C$ is the surrounding temperature.Integrating this,we g

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$\left(\frac{\log 5}{\log 2}-1ight) 	ext{ hrs}$

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(C) According to Newton's Law of Cooling,$\frac{dT}{dt} = -K(T - T_s)$,where $T_s = 10^{\circ} C$ is the surrounding temperature.Integrating this,we get $T(t) = T_s + (T_0 - T_s)e^{-Kt}$.Given $T_0 = 110^{\circ} C$ and $T_s = 10^{\circ} C$,the equation becomes $T(t) = 10 + 100e^{-Kt}$.At $t = 1 	ext{ hour}$,$T = 60^{\circ} C$:$60 = 10 + 100e^{-K(1)} \Rightarrow 50 = 100e^{-K} \Rightarrow e^{-K} = \frac{1}{2} \Rightarrow K = \log 2$.Now,we find the total time $t$ required to reach $T = 30^{\circ} C$:$30 = 10 + 100e^{-(\log 2)t} \Rightarrow 20 = 100e^{-(\log 2)t} \Rightarrow \frac{1}{5} = e^{-(\log 2)t}$.Taking the natural logarithm on both sides:$-\log 5 = -(\log 2)t \Rightarrow t = \frac{\log 5}{\log 2}$.The additional time required is $t - 1 = \frac{\log 5}{\log 2} - 1 	ext{ hours}$.

If a body is heated to $110^{\circ} C$ and placed in air at $10^{\circ} C$,and after $1 \text{ hour}$ its temperature is $60^{\circ} C$,then the additional time required for it to cool to $30^{\circ} C$ is

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