If a body cools from 80^{circ} C to 60^{circ} | vedclass

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(C) Let $	heta$ be the temperature of the body at any time $t$. According to Newton's Law of Cooling,the rate of change of temperature is proportiona

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$48$

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(C) Let $	heta$ be the temperature of the body at any time $t$. According to Newton's Law of Cooling,the rate of change of temperature is proportional to the difference between the body temperature and the surrounding temperature. $\frac{d	heta}{dt} = -k(	heta - 30)$ Integrating this,we get $\ln(	heta - 30) = -kt + C$. At $t = 0$,$	heta = 80^{\circ} C$,so $\ln(80 - 30) = C \Rightarrow C = \ln(50)$. Thus,$\ln(	heta - 30) = -kt + \ln(50) \dots (i)$. At $t = 30 	ext{ min}$,$	heta = 60^{\circ} C$: $\ln(60 - 30) = -k(30) + \ln(50) \Rightarrow \ln(30) - \ln(50) = -30k \Rightarrow \ln(3/5) = -30k$. So,$k = -\frac{1}{30} \ln(3/5) = \frac{1}{30} \ln(5/3)$. Now,for $t = 60 	ext{ min}$ (one hour): $\ln(	heta - 30) = -\left(\frac{1}{30} \ln(5/3)ight)(60) + \ln(50)$ $\ln(	heta - 30) = -2 \ln(5/3) + \ln(50) = \ln((3/5)^2 	imes 50)$ $\ln(	heta - 30) = \ln(9/25 	imes 50) = \ln(18)$ $	heta - 30 = 18 \Rightarrow 	heta = 48^{\circ} C$.

If a body cools from $80^{\circ} C$ to $60^{\circ} C$ in a room temperature of $30^{\circ} C$ in $30 \text{ min}$,then the temperature of the body after one hour is (in $^{\circ} C$)

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