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 :

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

(A) $\frac{14}{3}{ }^{\circ} \mathrm{C}$

(B) $\frac{28}{3}{ }^{\circ} \mathrm{C}$

(D) $\frac{42}{3}{ }^{\circ} \mathrm{C}$

JEE Main · 2025 · 24 Jan 2025 · Shift 2 · Heat & Thermodynamics

The temperature of a body in air falls from

4 0^{\circ} C

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 :

Answer: (C)

Using the average form of Newton's law of cooling,

\frac{T _{1} - T _{2}}{t} = k [\frac{T _{1} + T _{2}}{2} - T_{s}],

where

T_{s}

is the surrounding temperature.

First 4 minutes:

T_{1} = 4 0^{\circ} C

T_{2} = 2 4^{\circ} C

t = 4

min,

T_{s} = 1 6^{\circ} C

\frac{40 - 24}{4} = k [\frac{40 + 24}{2} - 16] ⟹ 4 = k (32 - 16) = 16 k ⟹ k = \frac{1}{4} .

Next 4 minutes: the body cools from

2 4^{\circ} C

to some temperature

T

\frac{24 - T}{4} = \frac{1}{4} [\frac{24 + T}{2} - 16] .

Multiplying both sides by 4:

24 - T = \frac{24 + T}{2} - 16.

Multiplying through by 2:

48 - 2 T = 24 + T - 32 = T - 8.

56 = 3 T ⟹ T = \frac{56}{3}^{\circ} C .

JEE Main · 2025 · 24 Jan 2025 · Shift 2 · Single correct option