Q 11-10-071JEE MainJEE Main 2023 (12 Apr, Shift 1)Medium
A body cools from $80^\circ\text{C}$ to $60^\circ\text{C}$ in $5$ minutes. The temperature of the surrounding is $20^\circ\text{C}$. The time it takes to cool from $60^\circ\text{C}$ to $40^\circ\text{C}$ is
Answer: (C) 500 s
Newton's law (average form): $\dfrac{20}{5}=k(70-20)\Rightarrow k=0.08\ \text{min}^{-1}$.
$\dfrac{20}{t}=0.08(50-20)=2.4\Rightarrow t=\dfrac{25}{3}\ \text{min}=500\ \text{s}$.
Solution by Sreeraj P, M.Sc Physics