A certain quantity of water cools from 70,^{c | vedclass

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(A) According to Newton's Law of Cooling,the rate of cooling is given by: $\frac{	heta_{1}-	heta_{2}}{t}=K\left[\frac{	heta_{1}+	heta_{2}}{2}-	he

Vedclass · Accepted Answer

$45$

Vedclass · Answer

(A) According to Newton's Law of Cooling,the rate of cooling is given by: $\frac{	heta_{1}-	heta_{2}}{t}=K\left[\frac{	heta_{1}+	heta_{2}}{2}-	heta_{0}ight]$For the first interval ($70\,^{\circ}C$ to $60\,^{\circ}C$ in $10\, minutes$):$\frac{70-60}{10}=K\left[\frac{70+60}{2}-	heta_{0}ight] \implies 1 = K(65 - 	heta_{0}) \ldots (i)$For the second interval ($60\,^{\circ}C$ to $54\,^{\circ}C$ in $10\, minutes$):$\frac{60-54}{10}=K\left[\frac{60+54}{2}-	heta_{0}ight] \implies 0.6 = K(57 - 	heta_{0}) \ldots (ii)$Dividing equation $(i)$ by $(ii)$:$\frac{1}{0.6} = \frac{65 - 	heta_{0}}{57 - 	heta_{0}}$$\frac{10}{6} = \frac{65 - 	heta_{0}}{57 - 	heta_{0}} \implies \frac{5}{3} = \frac{65 - 	heta_{0}}{57 - 	heta_{0}}$$5(57 - 	heta_{0}) = 3(65 - 	heta_{0})$$285 - 5	heta_{0} = 195 - 3	heta_{0}$$2	heta_{0} = 90 \implies 	heta_{0} = 45\,^{\circ}C$.

$A$ certain quantity of water cools from $70\,^{\circ}C$ to $60\,^{\circ}C$ in the first $10\, minutes$ and to $54\,^{\circ}C$ in the next $10\, minutes$. The temperature of the surrounding is ......... $^{\circ}C$.

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