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:

$$ \Delta S = \int_{T_1}^{T_2} \frac{dQ}{T} $$

Since the water is being heated slowly, the process is reversible, and the heat added is given by:

$$ dQ = mc\,dT $$

where:

• $$m$$ is the mass of the water,

• $$c$$ is the specific heat capacity (given as $$1\,\mathrm{J\,kg^{-1}\,K^{-1}}$$).

Substitute $$dQ$$ into the integral:

$$ \Delta S = \int_{T_1}^{T_2} \frac{mc\,dT}{T} = mc\, \int_{T_1}^{T_2} \frac{dT}{T}. $$

Since $$c = 1$$, the equation simplifies to:

$$ \Delta S = m \int_{T_1}^{T_2} \frac{dT}{T}. $$

Evaluating the integral, we have:

$$ \int_{T_1}^{T_2} \frac{dT}{T} = \ln{T}\Big|_{T_1}^{T_2} = \ln\left(\frac{T_2}{T_1}\right). $$

Thus, the change in entropy is:

$$ \Delta S = m \ln\left(\frac{T_2}{T_1}\right). $$

Comparing with the options provided, the correct answer is:

Option D: $$m \ln\left(\frac{T_2}{T_1}\right).$$