Question: 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)
Options:
2260 J/K
2260 kJ/K
0 J/K
1130 J/K
Correct Answer: 2260 J/K
Exam Year: 2021
Solution:
ΔS = Q/T = 2260 kJ / 373 K = 6.06 kJ/K.
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.
Entropy Change – The change in entropy (ΔS) is calculated using the formula ΔS = Q/T, where Q is the heat added or removed and T is the absolute temperature in Kelvin.
Latent Heat of Vaporization – The latent heat of vaporization is the amount of heat required to convert a unit mass of a substance from liquid to gas at constant temperature.
Thermodynamic Temperature – The temperature must be in Kelvin for thermodynamic calculations, as it is an absolute scale.
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