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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Aicharya · Accepted Answer

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:
$\Delta S=\int_{T_1}^{T_2} \frac{d Q}{T}$
Since the water is being heated slowly, the process is reversible, and the heat added is given by:
$d Q=m c d T$
where:
- $m$ is the mass of the water,
- $c$ is the specific heat capacity (given as $1 \mathrm{~J} \mathrm{~kg}^{-1} \mathrm{~K}^{-1}$).
Substitute $d Q$ into the integral:
$\Delta S=\int_{T_1}^{T_2} \frac{m c d T}{T}=m c \int_{T_1}^{T_2} \frac{d T}{T}$
Since $c=1$, the equation simplifies to:
$\Delta S=m \int_{T_1}^{T_2} \frac{d T}{T}$
Evaluating the integral, we have:
$\int_{T_1}^{T_2} \frac{d T}{T}=\left.\ln T\right|_{T_1} ^{T_2}=\ln \left(\frac{T_2}{T_1}\right)$
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 1: $m \ln \left(\frac{T_2}{T_1}\right)$.

Water of mass $m$ gram is slowly heated to increase the temperature from $T_1$ to $T_2$ The change in…

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