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(B) Let $	heta$ be the temperature of the body at time $t$. The temperature of the surrounding is $	heta_s = 20^{\circ} C$. According to Newton's la

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(B) Let $	heta$ be the temperature of the body at time $t$. The temperature of the surrounding is $	heta_s = 20^{\circ} C$. According to Newton's law of cooling,$\frac{d	heta}{dt} = -K(	heta - 	heta_s)$.Integrating this,we get $\ln(	heta - 20) = -Kt + C$.At $t = 0$,$	heta = 100^{\circ} C$,so $\ln(100 - 20) = C \Rightarrow C = \ln(80)$.Thus,$\ln\left(\frac{	heta - 20}{80}ight) = -Kt$.At $t = 20 	ext{ min}$,$	heta = 60^{\circ} C$,so $\ln\left(\frac{60 - 20}{80}ight) = -K(20) \Rightarrow \ln(0.5) = -20K \Rightarrow K = \frac{-\ln(0.5)}{20}$.We need to find $	heta$ at $t = 60 	ext{ min}$ (one hour).$\ln\left(\frac{	heta - 20}{80}ight) = -\left(\frac{-\ln(0.5)}{20}ight)(60) = 3 \ln(0.5) = \ln(0.5^3) = \ln(0.125)$.$\frac{	heta - 20}{80} = 0.125 = \frac{1}{8}$.$	heta - 20 = 10 \Rightarrow 	heta = 30^{\circ} C$.

$A$ body cools according to Newton's law of cooling from $100^{\circ} C$ to $60^{\circ} C$ in $20 \text{ minutes}$. If the temperature of the surrounding is $20^{\circ} C$,then the temperature of the body after one hour is: (in $^{\circ} C$)

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