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Picus48
1 year ago

Before I find the wheel again, I prefer to refer to the following Link, where the subject is precisely described. I think that should help.

Hamburger02
1 year ago

(a)

First, the entropy increases with the temperature. This is only linear in good approximation, since the increase in entropy takes place with the logarithm, but it doesn’t matter in the small temperature range.

The water then evaporates at 100° C. whereby the entropy increases accordingly. When the steam is overheated, the entropy then again rises approximately linearly.

(b)

1) Water heating:

ΔS = n * c_p,m * ln(T2/T1)
n= 1 mol
c_p,m = 75,29 J/molK
T1 = 273,15 K
T2 = 373,15 K

ΔS = 1 mol * 75.29 J/molK * ln(373,15/273,15) = 23.49 J/K

2) Evaporating the water:

ΔS = dQ/T = 1 mol * 40.66 kJ/mol / 373,15 K = 108.96 J/K

3) Overheating steam is like 1)

J0T4T4
1 year ago

Entropy for ideal liquid and ideal gas should be created.

The evaporation enthalpy is actually only the evaporation enthalpy divided by the temperature 😉

If necessary, you can derive this from the entropy balance for evaporation.