A pan filled with hot food cools from 94^{cir | vedclass

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Question

(B) According to Newton's Law of Cooling, $\frac{dT}{dt} = k(	heta - 	heta_0)$, where $	heta$ is the average temperature of the body and $	heta_0$

Vedclass · Accepted Answer

$2.8$ minutes

Vedclass · Answer

(B) According to Newton's Law of Cooling, $\frac{dT}{dt} = k(	heta - 	heta_0)$, where $	heta$ is the average temperature of the body and $	heta_0$ is the room temperature.For the first case: $\frac{94 - 86}{2} = k \left( \frac{94 + 86}{2} - 20 ight)$.$\frac{8}{2} = k(90 - 20) \Rightarrow 4 = k(70) \Rightarrow k = \frac{4}{70}$.For the second case: $\frac{74 - 66}{t} = k \left( \frac{74 + 66}{2} - 20 ight)$.$\frac{8}{t} = \frac{4}{70} (70 - 20) \Rightarrow \frac{8}{t} = \frac{4}{70} 	imes 50$.$\frac{8}{t} = \frac{200}{70} \Rightarrow \frac{8}{t} = \frac{20}{7}$.$t = \frac{8 	imes 7}{20} = \frac{56}{20} = 2.8$ minutes.

$A$ pan filled with hot food cools from $94^{\circ}C$ to $86^{\circ}C$ in $2$ minutes. When the room temperature is $20^{\circ}C$, how long will it take to cool from $74^{\circ}C$ to $66^{\circ}C$?

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An object is cooled from $75^{\circ} C$ to $65^{\circ} C$ in $2 \text{ min}$. The time it takes to cool from $55^{\circ} C$ to $45^{\circ} C$ is [The temperature of the surrounding is $30^{\circ} C$] (in $\text{ min}$)

$A$ body cools from $70^{\circ} C$ to $40^{\circ} C$ in $5$ minutes. Calculate the time it takes to cool from $60^{\circ} C$ to $30^{\circ} C$. The temperature of the surroundings is $20^{\circ} C$. (in $min.$)

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$A$ body takes $5 \ min$ to cool from $90^{\circ}C$ to $60^{\circ}C$. If the temperature of the surroundings is $20^{\circ}C$,how much time (in $min$) will it take to cool from $60^{\circ}C$ to $30^{\circ}C$?

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