A cup of coffee cools from \(90^\circ\text{C}\) to \(80^\circ\text{C}\) in \(t\) minutes, when the room temperature is \(20^\circ\text{C}\). The time taken by a similar cup of coffee to cool from \(80^\circ\text{C}\) to \(60^\circ\text{C}\) at a room temperature same at \(20^\circ\text{C}\) is
Solution:
According to Newton's law of cooling, \(\frac{T_1 - T_2}{\Delta t} = K\left(\frac{T_1+T_2}{2} - T_0\right)\). For the first interval, \(\frac{10}{t} = 65K\). For the second interval, \(\frac{20}{t'} = 50K\). Dividing these equations yields \(t' = \frac{13}{5}t\).
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