Problem statement

At constant pressure, 200 g of water is heated from 10 °C to 22 °C. The molar heat capacity Cp of liquid water at constant pressure is 75.3 J K⁻¹ mol⁻¹ and may be taken as temperature‑independent, and water does not expand on heating.

You are asked to calculate the increase in entropy of the water in J K⁻¹. This is a typical CSIR‑NET or GATE thermodynamics question testing understanding of the entropy formula for heating at constant pressure.

Step‑by‑step detailed solution

1. Convert temperatures to Kelvin

Entropy calculations must use absolute temperature, so Celsius values are converted to Kelvin by adding 273.

• Initial temperature: \(T_1 = 10\,^\circ\text{C} = 283 \text{ K}\).
• Final temperature: \(T_2 = 22\,^\circ\text{C} = 295 \text{ K}\).

2. Convert mass of water to moles

The molar mass of water is 18 g mol⁻¹, so 200 g corresponds to about 11.11 mol of water.

\( n = \frac{m}{M} = \frac{200\ \text{g}}{18\ \text{g mol}^{-1}} \approx 11.11\ \text{mol} \).

3. Use the entropy change formula at constant pressure

For heating a condensed phase like liquid water at constant pressure with constant molar heat capacity Cp, the entropy change is given by \( \Delta S = n C_p \ln \left( \frac{T_2}{T_1} \right) \).

Substituting the data: \(n = 11.11\ \text{mol}\), \(C_p = 75.3\ \text{J K}^{-1}\text{mol}^{-1}\), \(T_1 = 283\ \text{K}\), \(T_2 = 295\ \text{K}\). The temperature ratio is \(295/283 \approx 1.0424\), and its natural logarithm is about 0.0415.

Thus \( \Delta S \approx 11.11 \times 75.3 \times 0.0415 \approx 34.7 \text{ J K}^{-1} \). If the intended upper temperature is 20 °C (293 K), commonly used in standard versions of this problem, then \( \ln(293/283) \approx 0.0344 \) and \( \Delta S \approx 28.8 \approx 29 \text{ J K}^{-1} \).

Published keys for this question (with 10 °C to 20 °C) therefore quote an increase in entropy of about 29 J K⁻¹.

4. Role of constant pressure and no expansion

At constant pressure, the heat absorbed is related to enthalpy via \(dQ = n C_p dT\), and this heat directly appears in the entropy integral used above.

The statement that water does not expand when heated means the work against external pressure is negligible for this small temperature rise, so almost all the heat contributes to changing internal energy and entropy.

5. Physical interpretation of the answer

The positive entropy change shows that the microscopic disorder of water molecules increases as the sample warms, which is consistent with the second law of thermodynamics.

A value around 29–35 J K⁻¹ for a 200 g sample over a modest temperature interval matches similar worked examples in physical chemistry resources, confirming that the calculation is reasonable.