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:

According to Newton's law of cooling, the rate of cooling is proportional to the temperature difference between the body and the surroundings. It can be approximated as: \( \frac{T_1 - T_2}{t} = K ( \frac{T_1 + T_2}{2} - T_0 ) \). For the first case, cooling from \( 90^\circ\text{C} \) to \( 80^\circ\text{C} \) in time \( t \) with room temperature \( T_0 = 20^\circ\text{C} \): \( \frac{90 - 80}{t} = K ( \frac{90 + 80}{2} - 20 ) \Rightarrow \frac{10}{t} = K(85 - 20) = 65K \). For the second case, cooling from \( 80^\circ\text{C} \) to \( 60^\circ\text{C} \) in time \( t' \) with room temperature \( T_0 = 20^\circ\text{C} \): \( \frac{80 - 60}{t'} = K ( \frac{80 + 60}{2} - 20 ) \Rightarrow \frac{20}{t'} = K(70 - 20) = 50K \). Dividing the two equations: \( \frac{10 / t}{20 / t'} = \frac{65K}{50K} \Rightarrow \frac{t'}{2t} = \frac{13}{10} \Rightarrow t' = \frac{26}{10}t = \frac{13}{5}t \).