A cup of coffee cools from 90^{circ} C to 80^ | vedclass

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Question

(B) According to Newton's Law of Cooling,the rate of cooling is given by $\frac{dT}{dt} = K(T_{avg} - T_0)$,where $T_{avg}$ is the average temperature

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$\frac{13}{5} t$

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(B) According to Newton's Law of Cooling,the rate of cooling is given by $\frac{dT}{dt} = K(T_{avg} - T_0)$,where $T_{avg}$ is the average temperature of the body and $T_0$ is the surrounding temperature.For the first case,cooling from $90^{\circ} C$ to $80^{\circ} C$ in time $t$:$\frac{90 - 80}{t} = K \left( \frac{90 + 80}{2} - 20 ight)$$\frac{10}{t} = K(85 - 20) = 65K$  --- $(1)$For the second case,cooling from $80^{\circ} C$ to $60^{\circ} C$ in time $t'$:$\frac{80 - 60}{t'} = K \left( \frac{80 + 60}{2} - 20 ight)$$\frac{20}{t'} = K(70 - 20) = 50K$  --- $(2)$Dividing equation $(1)$ by $(2)$:$\frac{10/t}{20/t'} = \frac{65K}{50K}$$\frac{10}{t} 	imes \frac{t'}{20} = \frac{13}{10}$$\frac{t'}{2t} = \frac{13}{10}$$t' = \frac{13}{5} t$

$A$ cup of coffee cools from $90^{\circ} C$ to $80^{\circ} C$ in $t$ minutes,when the room temperature is $20^{\circ} C$. The time taken by a similar cup of coffee to cool from $80^{\circ} C$ to $60^{\circ} C$ at a room temperature same at $20^{\circ} C$ is :

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