A cup of tea cools from 80 °C to 60 °C in one minute. The ambient temperature is 30 °C . In cooling from 60 °C to 50 °C it will take
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From Newton's law of cooling
$\frac{\mathrm{d}T}{\mathrm{d}t} = -K(T - T_0)$
$\frac{T_1 - T_2}{t} = K \left(\frac{T_1 + T_2}{2} - T_0 \right)$
$\frac{80 - 60}{60} = K \left(\frac{80 + 60}{2} - 30 \right)$
$\frac{1}{3} = K \times 40 \ldots (1)$
and $\frac{60-50}{t} = K \left( \frac{60+50}{2} - 30 \right)$
$\frac{10}{t} = K \times 25 \ldots (2)$
From eqn (1) and (2)
$\frac{t}{30} = \frac{40}{25} \Rightarrow t = 48 sec$
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