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JEE Mains · Physics · STD 11 - Thermodynamics
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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 :

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(A) \(\mathrm{m} \ln \left(\frac{\mathrm{T}_2}{\mathrm{~T}_1}\right)\)

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\(\begin{aligned} & d \mathrm{Q}=\mathrm{msdT} \\ & \mathrm{dS}=\frac{\mathrm{dQ}}{\mathrm{T}}=\frac{\mathrm{msdT}}{\mathrm{T}} \\ & \Delta \mathrm{S}=\int \frac{\mathrm{msdT}}{\mathrm{T}}=\mathrm{ms} \ln \frac{\mathrm{T}_{\mathrm{f}}}{\mathrm{T}_{\mathrm{i}}} \\ & \Delta \mathrm{S}=\mathrm{m} \ln \frac{\mathrm{T}_2}{\mathrm{~T}_1}\end{aligned}\)
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