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

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(B) According to Newton's Law of Cooling,the rate of cooling is given by $\frac{T_1 - T_2}{t} = k \left( \frac{T_1 + T_2}{2} - T_0 \right)$,where $T_0

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$10$

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(B) According to Newton's Law of Cooling,the rate of cooling is given by $\frac{T_1 - T_2}{t} = k \left( \frac{T_1 + T_2}{2} - T_0 \right)$,where $T_0$ is the surrounding temperature.For the first $10\, min$ interval:$\frac{60 - 50}{10} = k \left( \frac{60 + 50}{2} - T_0 \right) \implies 1 = k(55 - T_0)$  --- (Equation $1$)For the next $10\, min$ interval:$\frac{50 - 42}{10} = k \left( \frac{50 + 42}{2} - T_0 \right) \implies 0.8 = k(46 - T_0)$ --- (Equation $2$)Dividing Equation $1$ by Equation $2$:$\frac{1}{0.8} = \frac{55 - T_0}{46 - T_0}$$1.25 = \frac{55 - T_0}{46 - T_0}$$1.25(46 - T_0) = 55 - T_0$$57.5 - 1.25 T_0 = 55 - T_0$$2.5 = 0.25 T_0$$T_0 = 10^{\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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