KTG & Thermodynamics | JEE Main 2025 Solution

JEE MAIN 2025

23-01-2025 SHIFT-2

Question

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

Select the correct option:

zero

$\mathrm{m}\left(\mathrm{T}_{2}-\mathrm{T}_{1}\right)$

$m \ln \left(\frac{T_{1}}{T_{2}}\right)$

$m \ln \left(\frac{T_{2}}{T_{1}}\right)$

✓ Correct! Well done.

✗ Incorrect. Try again or view the solution.

Solution

$\quad \mathrm{dQ}=\mathrm{msdT}$
$\mathrm{dS}=\frac{\mathrm{dQ}}{\mathrm{T}}=\frac{\mathrm{msdT}}{\mathrm{T}}$
$\Delta \mathrm{S}=\int \frac{\mathrm{msdT}}{\mathrm{T}}=\mathrm{ms} \ln \frac{\mathrm{T}_{\mathrm{f}}}{\mathrm{T}_{\mathrm{i}}}$
$\Delta \mathrm{S}=\mathrm{m} \ln \frac{\mathrm{T}_{2}}{\mathrm{~T}_{1}}$

Question Tags

JEE Main

Physics

Medium

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