Water of mass $m$ gram is slowly heated to increase the temperature from $T_1$ to $T_\gamma$. The change in entropy of the water, given specific heat of water is $1 \mathrm{Jkg}^{-1} \mathrm{~K}^{-1}$, is :

<p>To determine the change in entropy when water is heated from temperature $T_1$ to $T_2$ (here, $T_\gamma$ is equivalent to $T_2$), we use the definition of entropy change for a reversible process:</p> <p>$\Delta S = \int_{T_1}^{T_2} \frac{dQ}{T}$</p> <p>Since the water is being heated slowly, the process is reversible, and the heat added is given by:</p> <p>$dQ = mc\,dT$</p> <p>where: </p> <p>• $m$ is the mass of the water, </p> <p>• $c$ is the specific heat capacity (given as $1\,\mathrm{J\,kg^{-1}\,K^{-1}}$).</p> <p>Substitute $dQ$ into the integral:</p> <p>$$ \Delta S = \int_{T_1}^{T_2} \frac{mc\,dT}{T} = mc\, \int_{T_1}^{T_2} \frac{dT}{T}. $$</p> <p>Since $c = 1$, the equation simplifies to:</p> <p>$\Delta S = m \int_{T_1}^{T_2} \frac{dT}{T}.$</p> <p>Evaluating the integral, we have:</p> <p>$$ \int_{T_1}^{T_2} \frac{dT}{T} = \ln{T}\Big|_{T_1}^{T_2} = \ln\left(\frac{T_2}{T_1}\right). $$</p> <p>Thus, the change in entropy is:</p> <p>$\Delta S = m \ln\left(\frac{T_2}{T_1}\right).$</p> <p>Comparing with the options provided, the correct answer is:</p> <p>Option D: $m \ln\left(\frac{T_2}{T_1}\right).$</p>