JEE Main · 2023 · 25 Jan 2023 · Shift 1 · Heat & Thermodynamics

Newton's law of cooling makes the rate of fall of temperature proportional to the excess over the surroundings, $\frac{\Delta T}{\Delta t}=K\left(T_{\text{avg}}-T_0\right)$. In the first stage the drop is $12^{\circ }$ in $2$ min about a mean of $92^{\circ }\text{C}$, so $\frac{12}{2}=K(92-22)=70K$. In the second stage the drop is $6^{\circ }$ in time $t$ about a mean of $72^{\circ }\text{C}$, so $\frac{6}{t}=K(72-22)=50K$. Dividing the two relations, $\frac{6/t}{6}=\frac{50}{70}$, so $t=\frac{70}{50}=1.4\text{minutes}$.