If a body cools from a temperature of 62^{cir | vedclass

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(B) According to Newton's Law of Cooling, the rate of cooling is given by $\frac{dT}{dt} = -k(T - T_s)$, where $T$ is the temperature of the body, $T_

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$26$

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(B) According to Newton's Law of Cooling, the rate of cooling is given by $\frac{dT}{dt} = -k(T - T_s)$, where $T$ is the temperature of the body, $T_s$ is the temperature of the surroundings, and $k$ is a constant.For the first interval: $\frac{62 - 50}{10} = k \left( \frac{62 + 50}{2} - T_s \right) \implies 1.2 = k(56 - T_s) \quad (1)$For the second interval: $\frac{50 - 42}{10} = k \left( \frac{50 + 42}{2} - T_s \right) \implies 0.8 = k(46 - T_s) \quad (2)$Dividing equation $(1)$ by equation $(2)$:$\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$.

If a body cools from a temperature of $62^{\circ} C$ to $50^{\circ} C$ in $10 \, \text{minutes}$ and to $42^{\circ} C$ in the next $10 \, \text{minutes}$, then the temperature of the surroundings is (in $^{\circ} C$)

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