A cup of tea cools from ${80^0}C$ to ${60^o}C$ in one minute. The ambient temperature is ${30^o}C$. In cooling from ${60^o}C$ to ${50^o}C$ it will take ....... $\sec$
Medium
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(d) $\frac{{80 - 60}}{1} = K\left( {\frac{{80 + 60}}{2} - 30} \right)$
==> $K = \frac{1}{2}$
Again $\frac{{60 - 50}}{t} = \frac{1}{2}\left( {\frac{{60 + 50}}{2} - 30} \right)$
==> $t = 0.8 \times 60 = 48$sec.
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