A cup of coffee cools from 90 ° C to 80 ° C in t minutes when the room temperature is 20 ° C . The time taken by the similar cup of coffee to cool from 80 ° C to 60 ° C at the same room temperature is :

Physics · Heat Transfer

JEE Main 2025 — 29 January, Evening Shift — Question 35

A cup of coffee cools from $90^{\circ} \mathrm{C}$ to $80^{\circ} \mathrm{C}$ in t minutes when the room temperature is $20^{\circ} \mathrm{C}$ . The time taken by the similar cup of coffee to cool from $80^{\circ} \mathrm{C}$ to $60^{\circ} \mathrm{C}$ at the same room temperature is :

Option A:
$\frac{13}{5} t$
Correct
Option B:
$\frac{10}{13} \mathrm{t}$
Option C:
$\frac{13}{10} \mathrm{t}$
Option D:
$\frac{5}{13} \mathrm{t}$

Answer: A

Step-by-step solution

By using average form of Newton's law of cooling

$\frac{90-80}{\mathrm{t}}=\mathrm{k}\left(\frac{90+80}{2}-20\right)$

$\frac{80-60}{\mathrm{t}^{\prime}}=\mathrm{k}\left(\frac{80+60}{2}-20\right)$

(i)/(ii) $\frac{10 \times t^{\prime}}{t \times 20}=\frac{65}{50}$

$\mathrm{t}^{\prime}=\frac{65}{50} \times 2 \mathrm{t}=\frac{65}{25} \mathrm{t}=\frac{13}{5} \mathrm{t}$

Answer key and solution verified before publishing.

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Exam: JEE Main 2025
Paper: 29 January, Evening Shift
Subject: Physics
Chapter: Heat Transfer
Topic: Newton's Law of Cooling