The temperature of a body in air falls from $40^{\circ} \mathrm{C}$ to $24^{\circ} \mathrm{C}$ in 4 minutes.…

The temperature of a body in air falls from $40^{\circ} \mathrm{C}$ to $24^{\circ} \mathrm{C}$ in 4 minutes. The temperature of the air is $16^{\circ} \mathrm{C}$. The temperature of the body in the next 4 minutes will be:
  1. $\frac{14}{3}{ }^{\circ} \mathrm{C}$
  2. $\frac{42}{3}^{\circ} \mathrm{C}$
  3. $\frac{28}{3}{ }^{\circ} \mathrm{C}$
  4. $\frac{56}{3}{ }^{\circ} \mathrm{C}$

Solution

$\begin{aligned}
& \Delta T=T_2-T_1=16^{\circ} \mathrm{C} \\ & \text { And } T_0=16^{\circ} \mathrm{C} \\ & \frac{\Delta T}{t_1}=-k\left(32-16^{\circ}\right)...(i) \\ & \frac{\left(24-T_3\right)}{4}=-k\left(\frac{24+T_3}{2}-16\right)....(ii) \\ & \frac{16}{4}=-k(16) \\ & \Rightarrow \frac{\left(24-T_3\right)}{4}=-k\left(12+\frac{T_3}{2}-16\right) \\ & \Rightarrow \frac{16}{24-T_3}=\frac{16}{T_3} T_2 \\ & \Rightarrow \frac{T_3}{2}-4=24-T_3 \\ & \Rightarrow \frac{3 T_3}{2}=28 \\ & \Rightarrow T_3=\frac{56}{3}{ }^{\circ} \mathrm{C}
\end{aligned}$

Asked in: JEE Main 2025 (24 Jan Shift 2)

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