Q 11-10-034NEETJEE MainEAMCET 2008 (Medical)Medium
A body cools from $70°$C to $50°$C in $5$ minutes. The temperature of the surroundings is $20°$C. Its temperature after the next $10$ minutes is
Answer: (B) $30°$C
Concept. Newton's law of cooling in average form: $\dfrac{\Delta\theta}{t} = K(\bar\theta - \theta_0)$.
Step 1. Find $K$ from the first $5$ min.
$$\frac{20}{5} = K(60 - 20) \;\Rightarrow\; K = 0.1\ \text{min}^{-1}$$
Step 2. Next $10$ min ($50°$C to $\theta$).
$$\frac{50 - \theta}{10} = 0.1\left(\frac{50 + \theta}{2} - 20\right)$$
Step 3. Solve. $50 - \theta = 5 + 0.5\theta \Rightarrow 1.5\theta = 45 \Rightarrow \theta = 30°$C.
Answer. $30°$C (Option B).
Solution by Sreeraj P, M.Sc Physics