The temperature of a body in air falls from 40 ° C to 24 ° C in 4 minutes. The temperature of the air is 16 ° C . The temperature of the body in the next 4 minutes will be :

Physics · Heat Transfer

JEE Main 2025 — 24 January, Evening Shift — Question 60

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 :

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

Answer: C

Step-by-step solution

We use Newton’s law of cooling:

\frac{T_2 - T_1}{t} = K \left(T_{\text{avg}} - T_s\right).

Substitute values:

\frac{40 - 24}{4} = K \left(\frac{24+40}{2} - 16\right) \;\Rightarrow\; K = \tfrac{1}{4}.

Now for cooling from $24^\circ C$ to $T$ :

\frac{24 - T}{4} = \tfrac{1}{4}\left(\tfrac{T+24}{2} - 16\right).

Solving gives:

24 - T = \tfrac{T-8}{2} \;\Rightarrow\; \tfrac{3T}{2} = 28 \;\Rightarrow\; T = \tfrac{56}{3}{}^\circ C.

Answer key and solution verified before publishing.

Practise Heat Transfer

Start with this question, then two more from the same chapter — with a tutor that explains every step. Free.

Solve a similar one free

Exam: JEE Main 2025
Paper: 24 January, Evening Shift
Subject: Physics
Chapter: Heat Transfer
Topic: Newton's Law of Cooling