Q 11-10-057JEE MainJEE Main 2025 (24 Jan, Shift 2)Medium
The temperature of a body in air falls from $40^\circ\text{C}$ to $24^\circ\text{C}$ in 4 minutes. The temperature of the air is $16^\circ\text{C}$. The temperature of the body in the next 4 minutes will be
Answer: (D) $\dfrac{56}{3}\ ^\circ\text{C}$
By Newton's law of cooling, the excess temperature over the surroundings falls by the same factor in equal time intervals:
$$\frac{\theta_1 - 16}{\theta_0 - 16} = \frac{24 - 16}{40 - 16} = \frac{8}{24} = \frac{1}{3}$$
In the next 4 minutes the excess drops from $8^\circ$ to $\tfrac{8}{3}^\circ$:
$$\theta_2 = 16 + \frac{8}{3} = \frac{56}{3}\ ^\circ\text{C}$$
(The averaged form $\dfrac{\Delta\theta}{t} = k(\bar\theta - \theta_s)$ gives the same answer here.)
Solution by Sreeraj P, M.Sc Physics