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
Concept. Newton's law of cooling in average form: $\dfrac{\Delta\theta}{t} = K(\bar\theta - \theta_0)$.
Step 1. Find $K$.
$$\frac{20}{10} = K(70 - 20) \;\Rightarrow\; K = 0.04\ \text{min}^{-1}$$
Step 2. Second interval ($60 \to 40$, average $50°$C).
$$\frac{20}{t} = 0.04(50 - 20) = 1.2 \;\Rightarrow\; t \approx 16.7\ \text{min}$$
Cooling slows as the body nears room temperature.
Answer. $16.7$ min (Option D).
Solution by Sreeraj P, M.Sc Physics