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: $\frac{dT}{dt} = -K(T_{avg} - T_s)$.For the first case: $\frac{50.0 - 49.9}{5} = K \left( \frac{50.0 + 49.9}

Vedclass · Accepted Answer

$10$

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(B) According to Newton's Law of Cooling: $\frac{dT}{dt} = -K(T_{avg} - T_s)$.For the first case: $\frac{50.0 - 49.9}{5} = K \left( \frac{50.0 + 49.9}{2} - 30 \right) \Rightarrow \frac{0.1}{5} = K(49.95 - 30) \Rightarrow 0.02 = K(19.95) \dots (i)$.For the second case: $\frac{40.0 - 39.9}{t} = K \left( \frac{40.0 + 39.9}{2} - 30 \right) \Rightarrow \frac{0.1}{t} = K(39.95 - 30) \Rightarrow \frac{0.1}{t} = K(9.95) \dots (ii)$.Dividing $(i)$ by $(ii)$: $\frac{0.02}{0.1/t} = \frac{19.95}{9.95} \Rightarrow 0.2t = 2.005 \Rightarrow t \approx 10 \, s$.

$A$ 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 $39.9^{\circ}C$ (in $, s$)? The temperature of the surroundings is $30^{\circ}C$. Assume Newton's Law of Cooling applies.

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