कॉफी का एक कप \( 90^\circ\text{C} \) से \( 80^\circ\text{C} \) तक \( t \) मिनट में ठंडा होता है, जब कमरे का तापमान \( 20^\circ\text{C} \) है। इसी प्रकार के कॉफी के कप को \( 80^\circ\text{C} \) से \( 60^\circ\text{C} \) तक ठण्डा होने में (कमरे का तापमान \( 20^\circ\text{C} \) रहने पर) कितना समय लगेगा?

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: \( \frac{10 / t}{20 / t'} = \frac{65K}{50K} \Rightarrow \frac{t'}{2t} = \frac{13}{10} \Rightarrow t' = \frac{26}{10}t = \frac{13}{5}t \).