JEE Mains · Physics · STD 11 - 10.2 transmission of heat
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:
- A \(\frac{14}{3}{ }^{\circ} \mathrm{C}\)
- B \(\frac{42}{3}^{\circ} \mathrm{C}\)
- C \(\frac{28}{3}{ }^{\circ} \mathrm{C}\)
- D \(\frac{56}{3}{ }^{\circ} \mathrm{C}\)
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) \\ &…
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