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

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(D) According to Newton's law of cooling,the rate of cooling is given by $\frac{d	heta}{dt} = K(	heta_{avg} - 	heta_0)$.For the first interval: $\f

Vedclass · Accepted Answer

$48$

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(D) According to Newton's law of cooling,the rate of cooling is given by $\frac{d	heta}{dt} = K(	heta_{avg} - 	heta_0)$.For the first interval: $\frac{80 - 60}{1} = K \left( \frac{80 + 60}{2} - 30 ight)$.$20 = K(70 - 30) = 40K \implies K = \frac{20}{40} = 0.5 	ext{ min}^{-1}$.For the second interval: $\frac{60 - 50}{t} = K \left( \frac{60 + 50}{2} - 30 ight)$.$\frac{10}{t} = 0.5 (55 - 30) = 0.5 	imes 25 = 12.5$.$t = \frac{10}{12.5} = 0.8 	ext{ minutes}$.Since $1 	ext{ minute} = 60 	ext{ seconds}$,$t = 0.8 	imes 60 = 48 	ext{ seconds}$.

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

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