A cup of tea cools from $80\,^oC$ to $60\,^oC$ in one minute. The ambient temperature is $30\,^oC$. In cooling from $60\,^oC$ to $50\,^oC$, it will take ....... $\sec$
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$\frac{20}{60}=\mathrm{K}(70-30)$

$\frac{10}{t}=K(55-30)$

$\frac{20}{60 \times 10} \mathrm{t}=\frac{40}{25}$

$\mathrm{t}=\frac{8 \times 6 \times 10}{10}=48 \mathrm{sec}$

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