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JEE Mains · Physics · STD 11 - Transmission of heat
  • JEE Mains 2025
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  • MCQ

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:

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Answer & Solution

Correct Answer

(D) \(\frac{56}{3}{ }^{\circ} \mathrm{C}\)

Step-by-step Solution

Detailed explanation

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