Question

Saturated liquid water (m = 1 kg) initially at 0.101 MPa and 100 °C is heated at constant pressure until the temperature increases to 500 °C. Assume a constant \(C_p\) of steam = 1.9 kJ/kg-K, and enthalpy of vaporization, \(h_{fg}\) = 2257 kJ/kg at 0.101 MPa. The change in entropy of the water is .................... kJ/K (round off to two decimal places)

Hint:

For thermodynamic processes involving phase change, always split the calculation into separate parts: heating/cooling of single phases and the phase change itself. Remember to always use absolute temperatures (Kelvin or Rankine) in entropy calculations.

Answer 1

Step 1: Understanding the Concept:
The process involves heating 1 kg of water at constant pressure from a saturated liquid state to a superheated vapor state. The total change in entropy is the sum of the entropy change during the phase change (vaporization) and the entropy change during the sensible heating of the vapor.
Step 2: Key Formula or Approach:
The total entropy change, \(\Delta S\), is the sum of two parts: 1. Entropy change during vaporization at constant temperature: \(\Delta S_{fg} = \frac{h_{fg}}{T_{sat}}\) 2. Entropy change during superheating of steam at constant pressure: \(\Delta S_{superheat} = C_p \ln\left(\frac{T_{final}}{T_{sat}}\right)\) The total change is \(\Delta S_{total} = \Delta S_{fg} + \Delta S_{superheat}\). All temperatures must be in Kelvin.
Step 3: Detailed Explanation or Calculation:
Given values:
Mass, m = 1 kg (so we calculate specific entropy change)
Initial state: Saturated liquid at \(T_{sat} = 100\) °C. Final state: Superheated steam at \(T_{final} = 500\) °C. \(h_{fg} = 2257\) kJ/kg.
\(C_p\) of steam = 1.9 kJ/kg-K.
1. Convert temperatures to Kelvin:
- Saturation temperature: \(T_{sat} = 100 + 273.15 = 373.15\) K.
- Final temperature: \(T_{final} = 500 + 273.15 = 773.15\) K.
2. Calculate entropy change during vaporization (\(\Delta s_{fg}\)): \[ \Delta s_{fg} = \frac{h_{fg}}{T_{sat}} = \frac{2257 \text{ kJ/kg}}{373.15 \text{ K}} \approx 6.0485 \text{ kJ/kg-K} \] 3. Calculate entropy change during superheating (\(\Delta s_{superheat}\)): \[ \Delta s_{superheat} = C_p \ln\left(\frac{T_{final}}{T_{sat}}\right) = 1.9 \text{ kJ/kg-K} \times \ln\left(\frac{773.15}{373.15}\right) \] \[ \frac{773.15}{373.15} \approx 2.0719 \] \[ \ln(2.0719) \approx 0.7284 \] \[ \Delta s_{superheat} = 1.9 \times 0.7284 \approx 1.3840 \text{ kJ/kg-K} \] 4. Calculate total entropy change (\(\Delta s_{total}\)): \[ \Delta s_{total} = \Delta s_{fg} + \Delta s_{superheat} = 6.0485 + 1.3840 = 7.4325 \text{ kJ/kg-K} \] For 1 kg of water, the total change in entropy is \(\Delta S_{total} = 7.4325\) kJ/K.
Step 4: Final Answer:
Rounding to two decimal places, the change in entropy is 7.43 kJ/K.
Step 5: Why This is Correct:
The problem is solved by correctly identifying the two distinct thermodynamic processes and applying the appropriate formula for entropy change to each. The final sum gives the total change, and the result of 7.43 kJ/K is consistent with the provided answer range.