QJEEM25Phb7e9: JEE Main 2025 Physics - Class 11 JEEM25Phb7e9 Solution: Water of mass gram is slowly heated to increase the temperature …

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