A body cools down from 52.5^{circ} C to 47.5^ | vedclass

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

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(C) According to Newton's Law of Cooling,the rate of cooling is proportional to the difference between the average temperature of the body and the surrounding temperature: $\frac{dT}{dt} = -K(T_{av} - T_0)$.For the first interval: $\Delta T_1 = 52.5^{\circ} C - 47.5^{\circ} C = 5^{\circ} C$,$t_1 = 5 \ min$,$T_{av1} = \frac{52.5 + 47.5}{2} = 50^{\circ} C$. Thus,$\frac{5}{5} = K(50 - T_0) \Rightarrow 1 = K(50 - T_0) \dots (i)$.For the second interval: $\Delta T_2 = 47.5^{\circ} C - 42.5^{\circ} C = 5^{\circ} C$,$t_2 = 7.5 \ min$,$T_{av2} = \frac{47.5 + 42.5}{2} = 45^{\circ} C$. Thus,$\frac{5}{7.5} = K(45 - T_0) \Rightarrow \frac{2}{3} = K(45 - T_0) \dots (ii)$.Dividing $(i)$ by $(ii)$: $\frac{1}{2/3} = \frac{K(50 - T_0)}{K(45 - T_0)} \Rightarrow \frac{3}{2} = \frac{50 - T_0}{45 - T_0}$.Cross-multiplying: $3(45 - T_0) = 2(50 - T_0) \Rightarrow 135 - 3T_0 = 100 - 2T_0$.Solving for $T_0$: $T_0 = 135 - 100 = 35^{\circ} C$.

$A$ body cools down from $52.5^{\circ} C$ to $47.5^{\circ} C$ in $5 \ min$ and from $47.5^{\circ} C$ to $42.5^{\circ} C$ in $7.5 \ min$. Then,the temperature of the surroundings is (in $^{\circ} C$)

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