What is the change in entropy when 1 kg of water at 100°C is converted to steam
Practice Questions
Q1
What is the change in entropy when 1 kg of water at 100°C is converted to steam at 100°C? (Latent heat of vaporization = 2260 kJ/kg) (2021)
2260 J/K
2260 kJ/K
0 J/K
1130 J/K
Questions & Step-by-Step Solutions
What is the change in entropy when 1 kg of water at 100°C is converted to steam at 100°C? (Latent heat of vaporization = 2260 kJ/kg) (2021)
Step 1: Identify the mass of water being converted to steam, which is 1 kg.
Step 2: Note the temperature at which the conversion occurs, which is 100°C. Convert this temperature to Kelvin by adding 273.15, resulting in 373 K.
Step 3: Identify the latent heat of vaporization for water, which is given as 2260 kJ/kg.
Step 4: Calculate the total heat (Q) required for the conversion by multiplying the mass of water (1 kg) by the latent heat of vaporization (2260 kJ/kg). This gives Q = 2260 kJ.
Step 5: Use the formula for change in entropy (ΔS), which is ΔS = Q/T.
Step 6: Substitute the values into the formula: ΔS = 2260 kJ / 373 K.
Step 7: Perform the calculation to find ΔS, which equals approximately 6.06 kJ/K.
