Hot water cools from 60,^oC to 50,^oC in the | 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} -

Vedclass · Accepted Answer

$10$

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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.For the first $10$ minutes: $\frac{60 - 50}{10} = K \left[ \frac{60 + 50}{2} - 	heta_0 ight] \implies 1 = K(55 - 	heta_0) \dots (i)$For the next $10$ minutes: $\frac{50 - 42}{10} = K \left[ \frac{50 + 42}{2} - 	heta_0 ight] \implies 0.8 = K(46 - 	heta_0) \dots (ii)$Dividing equation $(i)$ by $(ii)$:$\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 surroundings is ...... $^oC$

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