A body cools from a temperature of 60^{circ} | vedclass

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

Vedclass · Accepted Answer

$30$

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(A) According to Newton's Law of Cooling, the rate of cooling is given by $\frac{dT}{dt} = -k(T - T_s)$, where $T_s$ is the temperature of the surroundings.For the first interval: $\frac{60 - 50}{10} = k \left( \frac{60 + 50}{2} - T_s ight) \Rightarrow 1 = k(55 - T_s) \quad (1)$For the second interval: $\frac{50 - 40}{15} = k \left( \frac{50 + 40}{2} - T_s ight) \Rightarrow \frac{2}{3} = k(45 - T_s) \quad (2)$Dividing $(1)$ by $(2)$: $\frac{1}{2/3} = \frac{55 - T_s}{45 - T_s} \Rightarrow 1.5 = \frac{55 - T_s}{45 - T_s} \Rightarrow 67.5 - 1.5T_s = 55 - T_s \Rightarrow 0.5T_s = 12.5 \Rightarrow T_s = 25^{\circ} C$.Substituting $T_s = 25$ into $(1)$: $1 = k(55 - 25) \Rightarrow 1 = 30k \Rightarrow k = \frac{1}{30}$.For the third interval from $40^{\circ} C$ to $30^{\circ} C$: $\frac{40 - 30}{t} = k \left( \frac{40 + 30}{2} - T_s ight) \Rightarrow \frac{10}{t} = \frac{1}{30} (35 - 25) \Rightarrow \frac{10}{t} = \frac{10}{30} \Rightarrow t = 30 	ext{ minutes}$.

$A$ body cools from a temperature of $60^{\circ} C$ to $50^{\circ} C$ in $10 \text{ minutes}$ and $50^{\circ} C$ to $40^{\circ} C$ in $15 \text{ minutes}$. The time taken in minutes for the body to cool from $40^{\circ} C$ to $30^{\circ} C$ is

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