A body cools from 62^{circ}C to 50^{circ}C in | vedclass

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(B) According to Newton's Law of Cooling,the rate of cooling is proportional to the difference between the temperature of the body and the surrounding

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(B) According to Newton's Law of Cooling,the rate of cooling is proportional to the difference between the temperature of the body and the surrounding temperature: $\frac{dT}{dt} = K(T - T_s)$.For the first interval ($62^{\circ}C$ to $50^{\circ}C$ in $10$ min):$\frac{62 - 50}{10} = K \left( \frac{62 + 50}{2} - T_s \right) \implies 1.2 = K(56 - T_s)$  --- $(1)$For the second interval ($50^{\circ}C$ to $42^{\circ}C$ in $10$ min):$\frac{50 - 42}{10} = K \left( \frac{50 + 42}{2} - T_s \right) \implies 0.8 = K(46 - T_s)$  --- $(2)$Dividing equation $(1)$ by $(2)$:$\frac{1.2}{0.8} = \frac{56 - T_s}{46 - T_s} \implies 1.5 = \frac{56 - T_s}{46 - T_s}$$1.5(46 - T_s) = 56 - T_s \implies 69 - 1.5T_s = 56 - T_s$$69 - 56 = 1.5T_s - T_s \implies 13 = 0.5T_s$$T_s = 26^{\circ}C$.

$A$ body cools from $62^{\circ}C$ to $50^{\circ}C$ in $10$ minutes and to $42^{\circ}C$ in the next $10$ minutes. The temperature of the surrounding is ........... $^{\circ}C$.

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