Hot water cools from 60^oC to 50^oC in the fi | vedclass

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(B) 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} - 	h

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$10$

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(B) According to Newton's law of cooling,the rate of cooling is given by $\frac{	heta_1 - 	heta_2}{t} = K \left[ \frac{	heta_1 + 	heta_2}{2} - 	heta_0 ight]$,where $	heta_0$ is the surrounding temperature.In the first case,the water cools from $60^oC$ to $50^oC$ in $10$ minutes:$\frac{60 - 50}{10} = K \left[ \frac{60 + 50}{2} - 	heta_0 ight]$$1 = K(55 - 	heta_0)$  --- $(i)$In the second case,the water cools from $50^oC$ to $42^oC$ in the next $10$ minutes:$\frac{50 - 42}{10} = K \left[ \frac{50 + 42}{2} - 	heta_0 ight]$$0.8 = K(46 - 	heta_0)$  --- $(ii)$Dividing equation $(i)$ by equation $(ii)$:$\frac{1}{0.8} = \frac{55 - 	heta_0}{46 - 	heta_0}$$1.25 = \frac{55 - 	heta_0}{46 - 	heta_0}$$1.25(46 - 	heta_0) = 55 - 	heta_0$$57.5 - 1.25	heta_0 = 55 - 	heta_0$$2.5 = 0.25	heta_0$$	heta_0 = 10^oC$.

Hot water cools from $60^oC$ to $50^oC$ in the first $10$ minutes and to $42^oC$ in the next $10$ minutes. The temperature of the surrounding is ......... $^oC$.

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