Q 11-10-062JEE MainJEE Main 2023 (29 Jan, Shift 1)Medium
A body cools from $60^\circ\text{C}$ to $40^\circ\text{C}$ in $6$ minutes. If temperature of surroundings is $10^\circ\text{C}$, then after the next $6$ minutes, its temperature will be ______ $^\circ\text{C}$.
Numerical value type. Enter your answer.
Answer: 28
Newton's law of cooling (average form): $\dfrac{\Delta T}{t}=k\,(T_{avg}-T_s)$.
First interval: $\dfrac{20}{6}=k(50-10)\Rightarrow k=\dfrac{1}{12}\ \text{min}^{-1}$.
Second interval, final temperature $T$:
$$\frac{40-T}{6}=\frac{1}{12}\left(\frac{40+T}{2}-10\right)\ \Rightarrow\ 80-2T=10+\frac{T}{2}\ \Rightarrow\ T=28^\circ\text{C}$$
Solution by Sreeraj P, M.Sc Physics