A body cools from 50.0^{circ}C to 49.9^{circ} | vedclass

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(B) According to Newton's law of cooling,the rate of cooling is given by $\frac{dT}{dt} = K(T_{avg} - T_s)$,where $T_s$ is the surrounding temperature

Vedclass · Accepted Answer

$10$

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(B) According to Newton's law of cooling,the rate of cooling is given by $\frac{dT}{dt} = K(T_{avg} - T_s)$,where $T_s$ is the surrounding temperature.For the first interval: $\frac{50.0 - 49.9}{5} = K \left( \frac{50.0 + 49.9}{2} - 30.0 ight) \implies \frac{0.1}{5} = K(49.95 - 30.0) = K(19.95)$ ... $(i)$For the second interval: $\frac{40.0 - 39.9}{t} = K \left( \frac{40.0 + 39.9}{2} - 30.0 ight) \implies \frac{0.1}{t} = K(39.95 - 30.0) = K(9.95)$ ... $(ii)$Dividing equation $(i)$ by $(ii)$:$\frac{0.1/5}{0.1/t} = \frac{K(19.95)}{K(9.95)} \implies \frac{t}{5} = \frac{19.95}{9.95} \approx 2$$t = 5 	imes 2 = 10\;s$.

$A$ body cools from $50.0^{\circ}C$ to $49.9^{\circ}C$ in $5\;s$. How long will it take to cool from $40.0^{\circ}C$ to $39.9^{\circ}C$? Assume the temperature of surroundings to be $30.0^{\circ}C$ and Newton's law of cooling to be valid. The time taken is ....... $s$.

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