A body cools from 80^{circ} C to 50^{circ} C | vedclass

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(C) According to Newton's law of cooling, $\frac{T_1 - T_2}{t} = K \left( \frac{T_1 + T_2}{2} - T_s \right)$.For the first interval: $\frac{80 - 50}{5

Vedclass · Accepted Answer

$12.5$

Vedclass · Answer

(C) According to Newton's law of cooling, $\frac{T_1 - T_2}{t} = K \left( \frac{T_1 + T_2}{2} - T_s ight)$.For the first interval: $\frac{80 - 50}{5} = K \left( \frac{80 + 50}{2} - 20 ight) \implies 6 = K(65 - 20) \implies 6 = 45K \implies K = \frac{6}{45} = \frac{2}{15}$.For the second interval: $\frac{50 - 30}{t} = K \left( \frac{50 + 30}{2} - 20 ight) \implies \frac{20}{t} = K(40 - 20) \implies \frac{20}{t} = 20K \implies \frac{1}{t} = K$.Substituting $K = \frac{2}{15}$, we get $\frac{1}{t} = \frac{2}{15} \implies t = 7.5 	ext{ min}$.The total time taken is $5 	ext{ min} + 7.5 	ext{ min} = 12.5 	ext{ min}$.

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