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

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Question

(A) Using the average form of Newton's law of cooling: $\frac{dT}{dt} = k(T_{avg} - T_{room})$.For the first interval ($90^{\circ} C$ to $80^{\circ} C

Vedclass · Accepted Answer

$\frac{13}{5} t$

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(A) Using the average form of Newton's law of cooling: $\frac{dT}{dt} = k(T_{avg} - T_{room})$.For the first interval ($90^{\circ} C$ to $80^{\circ} C$):$\frac{90-80}{t} = k\left(\frac{90+80}{2} - 20ight) \implies \frac{10}{t} = k(85 - 20) = 65k \dots (i)$For the second interval ($80^{\circ} C$ to $60^{\circ} C$):$\frac{80-60}{t'} = k\left(\frac{80+60}{2} - 20ight) \implies \frac{20}{t'} = k(70 - 20) = 50k \dots (ii)$Dividing equation $(i)$ by $(ii)$:$\frac{10/t}{20/t'} = \frac{65k}{50k}$$\frac{10}{t} 	imes \frac{t'}{20} = \frac{65}{50}$$\frac{t'}{2t} = \frac{13}{10}$$t' = \frac{13}{10} 	imes 2t = \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 the similar cup of coffee to cool from $80^{\circ} C$ to $60^{\circ} C$ at the same room temperature is $:$

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