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(B) Let $	heta$ be the temperature of the body at any time $t$.According to Newton's law of cooling,$\frac{d 	heta}{dt} = -k(	heta - 20)$,where $k

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$25$

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(B) Let $	heta$ be the temperature of the body at any time $t$.According to Newton's law of cooling,$\frac{d 	heta}{dt} = -k(	heta - 20)$,where $k > 0$.Integrating both sides,we get $\ln(	heta - 20) = -kt + C$.At $t = 0$,$	heta = 100^{\circ} C$,so $\ln(100 - 20) = C \Rightarrow C = \ln(80)$.Thus,$\ln(	heta - 20) = -kt + \ln(80) \Rightarrow \ln\left(\frac{	heta - 20}{80}ight) = -kt$.At $t = 15$ minutes,$	heta = 60^{\circ} C$,so $\ln\left(\frac{60 - 20}{80}ight) = -15k \Rightarrow \ln\left(\frac{1}{2}ight) = -15k \Rightarrow k = \frac{\ln(2)}{15}$.For $t = 1$ hour $= 60$ minutes,we have $\ln\left(\frac{	heta - 20}{80}ight) = -60 	imes \frac{\ln(2)}{15} = -4 \ln(2) = \ln\left(\frac{1}{16}ight)$.Therefore,$\frac{	heta - 20}{80} = \frac{1}{16} \Rightarrow 	heta - 20 = 5 \Rightarrow 	heta = 25^{\circ} C$.

$A$ body cools according to Newton's law of cooling from $100^{\circ} C$ to $60^{\circ} C$ in $15$ minutes. If the temperature of the surrounding is $20^{\circ} C$,then the temperature of the body after cooling down for one hour is (in $^{\circ} C$)

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