Hot water cools from 60,^{circ}C to 50,^{circ | vedclass

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(C) According to Newton's law of cooling: $\frac{	heta_{1}-	heta_{2}}{t}=k\left(\frac{	heta_{1}+	heta_{2}}{2}-	heta_{0}ight)$For the first $10\

Vedclass · Accepted Answer

$15$

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(C) According to Newton's law of cooling: $\frac{	heta_{1}-	heta_{2}}{t}=k\left(\frac{	heta_{1}+	heta_{2}}{2}-	heta_{0}ight)$For the first $10\,min$ interval:$\frac{60-50}{10}=k\left(\frac{60+50}{2}-	heta_{0}ight)$$1=k(55-	heta_{0}) \dots(i)$For the total time interval of $20\,min$ (from $60\,^{\circ}C$ to $42\,^{\circ}C$):$\frac{60-42}{20}=k\left(\frac{60+42}{2}-	heta_{0}ight)$$\frac{18}{20}=k(51-	heta_{0})$$0.9=k(51-	heta_{0}) \dots(ii)$Dividing $(i)$ by $(ii)$:$\frac{1}{0.9} = \frac{55-	heta_{0}}{51-	heta_{0}}$$51-	heta_{0} = 0.9(55-	heta_{0})$$51-	heta_{0} = 49.5 - 0.9	heta_{0}$$51 - 49.5 = 	heta_{0} - 0.9	heta_{0}$$1.5 = 0.1	heta_{0}$$	heta_{0} = 15^{\circ}C$.

Hot water cools from $60\,^{\circ}C$ to $50\,^{\circ}C$ in the first $10\,min$ and to $42\,^{\circ}C$ in the next $10\,min$. The temperature of the surrounding is ......... $^{\circ}C$.

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