Q 11-10-045NEETJEE MainMedium
A body cools from $80°$C to $60°$C in $10$ min in a room at $20°$C. Using the average form of Newton's law of cooling, the time it takes to cool from $60°$C to $40°$C is about
Answer: (D) $16.7$ min
$\dfrac{20}{10} = K(70 - 20) \Rightarrow K = 0.04\ \text{min}^{-1}$.
$\dfrac{20}{t} = 0.04(50 - 20) = 1.2 \Rightarrow t \approx 16.7$ min. Cooling slows as the body approaches room temperature.
Solution by Sreeraj P, M.Sc Physics