Thermal Properties of Matter

JEE Main level, single correct, medium

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Question 3 of 126

The temperature of a body in air falls from 40∘C40^{\circ} \mathrm{C} to 24∘C24^{\circ} \mathrm{C} in 4 minutes. The temperature of the air is 16∘C16^{\circ} \mathrm{C}. The temperature of the body in the next 4 minutes will be:

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Answer: (D) 563∘C\frac{56}{3}{ }^{\circ} \mathrm{C}

ΔT=T2−T1=16∘C And T0=16∘CΔTt1=−k(32−16∘)...(i)(24−T3)4=−k(24+T32−16)....(ii)164=−k(16)⇒(24−T3)4=−k(12+T32−16)⇒1624−T3=16T3T2⇒T32−4=24−T3⇒3T32=28⇒T3=563∘C\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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