What is the change in entropy of 1.00 kg of liquid water at 100°C as it changes to steam at 100°C? Has to be in J/K ?
What is the change in entropy of 1.00 kg of liquid water at 100°C as it changes to steam at 100°C?
What is the change in entropy of 1.00 kg of liquid water at 100°C as it changes to steam at 100°C?
Dr. Zorro
Favorite Answer
ΔS = ∫︎ dQ / T
Note that during a phase transition T is constant, so 1/T can be taken out in front of the integration. The integral of dQ will add up to the heat of vaporization of 1kg of water, L.
So
ΔS = L / T
L = 2.260kJ/kg * 1.00kg = 2260 J
T = 373 K
ΔS = 2260J / 373K = 6.05 J/K
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