Water at 100^{circ} C cools in 15 text{ minut | vedclass

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

Vedclass · Accepted Answer

$\left(\frac{175}{3}\right)^{\circ} C$

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(D) Let $	heta^{\circ} C$ be the temperature of water 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 $t = 0, 	heta = 100^{\circ} C$,so $\ln(75) = C$. Thus,$\ln\left(\frac{	heta - 25}{75}ight) = -Kt$. At $t = 15, 	heta = 75^{\circ} C$,so $\ln\left(\frac{75 - 25}{75}ight) = -15K \Rightarrow \ln\left(\frac{50}{75}ight) = -15K \Rightarrow \ln\left(\frac{2}{3}ight) = -15K$. Therefore,$K = -\frac{1}{15} \ln\left(\frac{2}{3}ight) = \frac{1}{15} \ln\left(\frac{3}{2}ight)$. Now,for $t = 30 	ext{ min}$,$\ln\left(\frac{	heta - 25}{75}ight) = -30 	imes \left(\frac{1}{15} \ln\left(\frac{3}{2}ight)ight) = -2 \ln\left(\frac{3}{2}ight) = \ln\left(\left(\frac{3}{2}ight)^{-2}ight) = \ln\left(\frac{4}{9}ight)$. So,$\frac{	heta - 25}{75} = \frac{4}{9} \Rightarrow 	heta - 25 = \frac{4 	imes 75}{9} = \frac{300}{9} = \frac{100}{3}$. $	heta = 25 + \frac{100}{3} = \frac{75 + 100}{3} = \frac{175}{3}^{\circ} C$.

Water at $100^{\circ} C$ cools in $15 \text{ minutes}$ to $75^{\circ} C$ in a room temperature of $25^{\circ} C$. Then the temperature of water after $30 \text{ minutes}$ is

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