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
Newton's law of cooling (average form): $\dfrac{\Delta\theta}{t} = K(\bar{\theta} - \theta_0)$.
First $5$ min: $\dfrac{20}{5} = K(60 - 20) \Rightarrow K = 0.1\ \text{min}^{-1}$.
Next $10$ min, from $50°$C to $\theta$:
$$\frac{50 - \theta}{10} = 0.1\left(\frac{50 + \theta}{2} - 20\right) \;\Rightarrow\; 50 - \theta = 5 + 0.5\theta \;\Rightarrow\; \theta = 30°\text{C}$$
Solution by Sreeraj P, M.Sc Physics