Physics · Thermodynamics

JEE Main 2025 — 23 January, Evening Shift — Question 62

Water of mass $m$ gram is slowly heated to increase the temperature from $T_{1}$ to $T_{2}$ . The change in entropy of the water, given specific heat of water is $1 \mathrm{Jkg}^{-1} \mathrm{~K}^{-1}$ , is :

Option A:
zero
Option B:
$\mathrm{m}\left(\mathrm{T}_{2}-\mathrm{T}_{1}\right)$
Option C:
$m \ln \left(\frac{T_{1}}{T_{2}}\right)$
Option D:
$m \ln \left(\frac{T_{2}}{T_{1}}\right)$
Correct

Answer: D

Step-by-step solution

$\mathrm{dQ}=\mathrm{msdT}$

$\mathrm{dS}=\frac{\mathrm{dQ}}{\mathrm{T}}=\frac{\mathrm{msdT}}{\mathrm{T}}$

$\Delta \mathrm{S}=\int \frac{\mathrm{msdT}}{\mathrm{T}}=\mathrm{ms} \ln \frac{\mathrm{T}_{\mathrm{f}}}{\mathrm{T}_{\mathrm{i}}}$

$\Delta \mathrm{S}=\mathrm{m} \ln \frac{\mathrm{T}_{2}}{\mathrm{~T}_{1}}$

Answer key and solution verified before publishing.

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Exam: JEE Main 2025
Paper: 23 January, Evening Shift
Subject: Physics
Chapter: Thermodynamics
Topic: Entropy, Carnot's Engines and Refrigerators