A pot contains 257 g of water at 99°C. If this water is heated and all evaporates to form steam at 100°C, what is the change in the entropy of the H2O?

1556.906J/K=ΔS

Explanation:

Entropy is a thermodynamic property that measures the level of molecular disorder in a substance.

This property is already calculated for all pressure and temperature values.

Therefore, to solve this problem we must use thermodynamic tables for water and calculate the specific entropy in the two states, finally multiply by the mass to find the entropy change.

the entropy change is given by the following equation

ΔS=m (s2-s1)

where

ΔS=change in the entropy

s = especific entropy

m=mass=257g=0.257Kg

for the state 1: entropy for liquid water at 99°C

s1=1296J/kgK

for the state 2: entropy for steam at 100°C

s2=7354J/kgK

solving

ΔS=0.257 (7354-1296) = 1556.906J/K