What Is the Change in Entropy for Condensing and Freezing 45g of H2O?

[ab94]

Homework Statement

45g of H2O(g) are condensed at 100 degrees C, and H2O(l) is cooled to 0 degrees C and then frozen to H2O solid. Find the Change in Entropy


H2O(l): 4.2 J K-1g-1
vaporization at 100 degrees C: 2258 J g-1;
fusion at 0 degrees C: 334 J g

Homework Equations

dS=dq/T

The Attempt at a Solution

q1=(45g)(2258J/g) = -101610J
q2=(45g)(4.2J/K/g)(100K) = -18900J
q3=(45g)(334J/g)= -15030J

Total q=-135540J / 273.15K
delta S = -496.2J/K

I'm not sure if I did that right, can anybody double check? Thank you

 

[ab94]

I think I solved my mistake, I'm not supposed to divide by T at the end, I need to do it for each step.

So q1= -101610J/373.15K = -272.3J/K
q2= 45gx4.2J/K/g x ln(273.15K/373.15K) = -58.96J/K
q3= -15030J/273.15K = -55.02J/K

Total Change in Entropy= -386.3J/K