JEE Main 2025 Jan 24 Shift 2, Physics Q16: Newton’s Law of Cooling

JEE Main2025Jan 24, Shift 2Physics

The temperature of a body in air falls from $4 0^{\circ} C$ to $2 4^{\circ} C$ in 4 minutes. The temperature of the air is $1 6^{\circ} C$ . The temperature of the body in the next 4 minutes will be :

A
$\frac{14}{3}^{\circ} C$
B
$\frac{28}{3}^{\circ} C$
C
$\frac{56}{3}^{\circ} C$
D
$\frac{42}{3}^{\circ} C$

Solution

Newton's law of cooling in finite-difference form:

$\frac{T _{2} - T _{1}}{t} = - K (\frac{T _{1} + T _{2}}{2} - T_{s})$

Step 1 — find $K$ . First interval: $T_{1} = 4 0^{\circ}, T_{2} = 2 4^{\circ}, t = 4 min, T_{s} = 1 6^{\circ}$ .

$\frac{24 - 40}{4} = - K (\frac{40 + 24}{2} - 16) \Rightarrow - 4 = - K (16) \Rightarrow K = \frac{1}{4} min^{- 1}$

Step 2 — find $T$ at end of next 4 minutes. Now $T_{1} = 2 4^{\circ}, T_{2} = T$ :

$\frac{T - 24}{4} = - \frac{1}{4} (\frac{24 + T}{2} - 16)$

$T - 24 = - (\frac{T + 24}{2} - 16) = - \frac{T + 24}{2} + 16$

Multiply through by 2: $2 T - 48 = - (T + 24) + 32 \Rightarrow 3 T = 56 \Rightarrow T = \frac{56}{3}^{\circ} C$

Correct option: (3)

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JEE Main 2025 Jan 24 Shift 2, Physics Q16: Newton’s Law of Cooling

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