The temperature of hot water falls from 60^oC | vedclass

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(B) According to Newton's Law of Cooling,the rate of cooling is given by: $\frac{d	heta}{dt} = K \left( \frac{	heta_1 + 	heta_2}{2} - 	heta_s ig

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(B) According to Newton's Law of Cooling,the rate of cooling is given by: $\frac{d	heta}{dt} = K \left( \frac{	heta_1 + 	heta_2}{2} - 	heta_s ight)$,where $	heta_s$ is the surrounding temperature.For the first interval ($60^oC$ to $50^oC$ in $10 \, min$):$\frac{60 - 50}{10} = K \left( \frac{60 + 50}{2} - 	heta_s ight)$$1 = K(55 - 	heta_s) \quad \dots(i)$For the second interval ($50^oC$ to $42^oC$ in $10 \, min$):$\frac{50 - 42}{10} = K \left( \frac{50 + 42}{2} - 	heta_s ight)$$0.8 = K(46 - 	heta_s) \quad \dots(ii)$Dividing equation $(i)$ by $(ii)$:$\frac{1}{0.8} = \frac{55 - 	heta_s}{46 - 	heta_s}$$1.25 = \frac{55 - 	heta_s}{46 - 	heta_s}$$1.25(46 - 	heta_s) = 55 - 	heta_s$$57.5 - 1.25	heta_s = 55 - 	heta_s$$2.5 = 0.25	heta_s$$	heta_s = 10^oC$.

The temperature of hot water falls from $60^oC$ to $50^oC$ in $10 \, min$. It takes another $10 \, min$ to fall from $50^oC$ to $42^oC$. Find the temperature of the surroundings in $^oC$.

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