Q 11-10-123JEE MainJEE Main 2018 (15 Apr, Shift 2)Medium
A body takes 10 minutes to cool from $60^\circ\text{C}$ to $50^\circ\text{C}$. The temperature of surroundings is constant at $25^\circ\text{C}$. Then, the temperature of the body after next 10 minutes will be approximately
Answer: (A) $43^\circ\text{C}$
Using the average form of Newton's law of cooling, $\dfrac{\Delta T}{\Delta t} = k\left(\bar T - T_0\right)$.
First 10 min:
$$\frac{10}{10} = k(55 - 25) \quad\Rightarrow\quad k = \frac{1}{30}\ \text{min}^{-1}$$
Next 10 min, final temperature $T$:
$$\frac{50 - T}{10} = \frac{1}{30}\left(\frac{50 + T}{2} - 25\right) = \frac{T}{60}$$
$$300 - 6T = T \quad\Rightarrow\quad T = \frac{300}{7} \approx 43^\circ\text{C}$$
Solution by Sreeraj P, M.Sc Physics