Water of mass m gram is slowly heated to increase the temperature from T₁ to T₂. The change in entropy of the water, given specific heat of water is 1 Jkg^–1K^–1, is :

The change in entropy ($\Delta S$) of water when it is heated slowly from temperature $T_1$ to $T_2$ can be found using the thermodynamic formula:

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

Since the heat added is:

$$dQ = mc \, dT$$

where:

• $m$ = mass of water (in grams),
• $c$ = specific heat capacity of water ($1$ J/g·K),
• $dT$ = small temperature change.

Substituting $dQ$ into the entropy formula:

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

Since $m$ and $c$ are constants, they come out of the integral:

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

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

Final Answer:

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

Since the specific heat of water is given as 1 J/g·K, we substitute $c = 1$:

$$\Delta S = m \ln \left(\frac{T_2}{T_1} \right) \quad \text{J/K}$$