A cup of tea cools from 80^{circ}C to 60^{cir | vedclass

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(D) According to Newton's Law of Cooling: $\frac{	heta_1 - 	heta_2}{t} = K \left[ \frac{	heta_1 + 	heta_2}{2} - 	heta_0 ight]$For the first cas

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$48$

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(D) According to Newton's Law of Cooling: $\frac{	heta_1 - 	heta_2}{t} = K \left[ \frac{	heta_1 + 	heta_2}{2} - 	heta_0 ight]$For the first case: $\frac{80 - 60}{60} = K \left[ \frac{80 + 60}{2} - 30 ight]$$\frac{20}{60} = K [70 - 30] = 40K$$K = \frac{1}{3} 	imes \frac{1}{40} = \frac{1}{120} 	ext{ s}^{-1}$For the second case: $\frac{60 - 50}{t} = K \left[ \frac{60 + 50}{2} - 30 ight]$$\frac{10}{t} = \frac{1}{120} [55 - 30]$$\frac{10}{t} = \frac{25}{120}$$t = \frac{1200}{25} = 48 	ext{ seconds}$

$A$ cup of tea cools from $80^{\circ}C$ to $60^{\circ}C$ in $1$ minute. The ambient temperature is $30^{\circ}C$. How many seconds will it take to cool from $60^{\circ}C$ to $50^{\circ}C$?

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