A body is heated to 110^{circ} C and placed i | vedclass

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(C) According to Newton's Law of Cooling,$\frac{d	heta}{dt} = -k(	heta - 	heta_0)$,where $	heta_0 = 10^{\circ} C$.Integrating this,we get $	heta(

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$\left(\frac{\log 5}{\log 2}-1\right)$ hours

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(C) According to Newton's Law of Cooling,$\frac{d	heta}{dt} = -k(	heta - 	heta_0)$,where $	heta_0 = 10^{\circ} C$.Integrating this,we get $	heta(t) = 	heta_0 + Ce^{-kt}$.At $t = 0$,$	heta = 110^{\circ} C$,so $110 = 10 + C \Rightarrow C = 100$.Thus,$	heta(t) = 10 + 100e^{-kt}$.At $t = 1$ hour,$	heta = 60^{\circ} C$,so $60 = 10 + 100e^{-k} \Rightarrow 50 = 100e^{-k} \Rightarrow e^{-k} = \frac{1}{2}$.Taking the natural logarithm,$-k = \ln(1/2) = -\ln 2$,so $k = \ln 2$.Now,we find the total time $t$ when $	heta = 30^{\circ} C$:$30 = 10 + 100e^{-kt} \Rightarrow 20 = 100e^{-kt} \Rightarrow e^{-kt} = \frac{1}{5}$.Taking the natural logarithm,$-kt = \ln(1/5) = -\ln 5$,so $kt = \ln 5$.Since $k = \ln 2$,we have $t = \frac{\ln 5}{\ln 2}$ hours.The additional time required is $t - 1 = \frac{\ln 5}{\ln 2} - 1$ hours.

$A$ body is heated to $110^{\circ} C$ and placed in air at $10^{\circ} C$. After $1$ hour,its temperature is $60^{\circ} C$. The additional time required for it to cool to $30^{\circ} C$ is

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