If a body cools from 135^{circ} C to 80^{circ | vedclass

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(A) Let $	heta^{\circ} C$ be the temperature of the body at time $t 	ext{ min}$. The room temperature is $T_s = 25^{\circ} C$. According to Newton's

Vedclass · Accepted Answer

$52.5$

Vedclass · Answer

(A) Let $	heta^{\circ} C$ be the temperature of the body at time $t 	ext{ min}$. The room temperature is $T_s = 25^{\circ} C$. According to Newton's law of cooling,$\frac{d	heta}{dt} = -k(	heta - T_s)$.Integrating this,we get $\ln(	heta - 25) = -kt + C$.At $t = 0$,$	heta = 135^{\circ} C$,so $C = \ln(135 - 25) = \ln(110)$.Thus,$\ln\left(\frac{	heta - 25}{110}ight) = -kt$.At $t = 60 	ext{ min}$,$	heta = 80^{\circ} C$,so $\ln\left(\frac{80 - 25}{110}ight) = -60k \Rightarrow \ln(0.5) = -60k \Rightarrow k = -\frac{1}{60}\ln(0.5)$.For $t = 120 	ext{ min}$ $(2 	ext{ hours})$,$\ln\left(\frac{	heta - 25}{110}ight) = -120 	imes \left(-\frac{1}{60}\ln(0.5)ight) = 2\ln(0.5) = \ln(0.5^2) = \ln(0.25)$.Therefore,$\frac{	heta - 25}{110} = 0.25 \Rightarrow 	heta - 25 = 110 	imes 0.25 = 27.5$.$	heta = 27.5 + 25 = 52.5^{\circ} C$.

If a body cools from $135^{\circ} C$ to $80^{\circ} C$ at a room temperature of $25^{\circ} C$ in $60 \text{ minutes}$,then the temperature of the body after $2 \text{ hours}$ is: (in $^{\circ} C$)

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