Q 11-11-051JEE MainJEE Main 2025 (23 Jan, Shift 2)Easy
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 $1\ \text{J kg}^{-1}\text{K}^{-1}$, is
Answer: (A) $m\ln\left(\dfrac{T_2}{T_1}\right)$
For slow (reversible) heating, $dS = \dfrac{dQ}{T} = \dfrac{mc\,dT}{T}$:
$$\Delta S = mc\int_{T_1}^{T_2}\frac{dT}{T} = mc\ln\frac{T_2}{T_1}$$
With $c = 1$: $\Delta S = m\ln\left(\dfrac{T_2}{T_1}\right)$
Solution by Sreeraj P, M.Sc Physics