The temperature of a body in air falls from 40 to 24 in 4 minutes. The temperature of the air is 16. The temperature of the body in the next 4 minutes will be: (a) 28/3 (b) 56/3 (c) 42/3 (d) 14/3

Answer: (b). Newton's law: (T₁ - T₂)/(t) = k((T₁ + T₂)/(2) - T₀). First 4 min: 16/4 = k(32 - 16) ⇒ k = 1/4 Next 4 min: (24 - T)/(4) = 1/4((24 + T)/(2) - 16) ⇒ 24 - T = (T - 8)/(2) 48 - 2T = T - 8 ⇒ T = 56/3° C So the answer is (b) 56/3° C.

The temperature of a body in air falls from 40 to 24 in 4 minutes… | JEE Main 2025 (24 Jan, Shift 2)

Q11JEE Main 2025 · 24 JAN · SHIFT 2Medium

Newton's Law of Cooling

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

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Answer: (b)

Newton's law: $\frac{T_1 - T_2}{t} = k\left(\frac{T_1 + T_2}{2} - T_0\right)$ .
First 4 min: $\frac{16}{4} = k(32 - 16) \Rightarrow k = \frac{1}{4}$
Next 4 min: $\frac{24 - T}{4} = \frac{1}{4}\left(\frac{24 + T}{2} - 16\right) \Rightarrow 24 - T = \frac{T - 8}{2}$
$48 - 2T = T - 8 \Rightarrow T = \frac{56}{3}^\circ$ C
So the answer is (b) $\frac{56}{3}{}^\circ$ C.

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