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 : [JEE Main 2025, 23 Jan (Shift 2)]

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Correct answer: a) $\mathrm{mln}\left(\frac{{\mathrm{T}}_{2}}{{\mathrm{T}}_{1}}\right)$. $I=dQ\ dQ=TdS\ dS=\frac{dQ}{T}\ \int dS={\int }_{{T}_{1}}^{T}mC\frac{dT}{T}\ \Delta S=mC[\ln T{]}_{{T}_{1}}^{{T}_{2}}$ $\Delta S=mC\ln \frac{{T}_{2}}{{T}_{1}}$

Water of mass \(m\) gram is slowly heated to increase the temperature from \({T}_{1}\) to \({T}_{2}\). The change in ent…

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