Q 11-10-059JEE MainJEE Main 2025 (29 Jan, Shift 2)Medium
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 the same room temperature is
Answer: (D) $\dfrac{13}{5}t$
Using the average form of Newton's law of cooling, $\dfrac{\Delta\theta}{\Delta t} = k(\bar\theta - \theta_0)$:
First: $\dfrac{10}{t} = k(85 - 20) = 65k$.
Second: $\dfrac{20}{t'} = k(70 - 20) = 50k$.
Dividing: $\dfrac{20}{t'}\cdot\dfrac{t}{10} = \dfrac{50}{65} \Rightarrow t' = \dfrac{2\times65}{50}t = \dfrac{13}{5}t$
Solution by Sreeraj P, M.Sc Physics