Question

Question: You have 100g of water at 30C to which you add 50g of ice at 0C. Calculate the entropy change for the process.

For this problem I solved for the final temperature, Tf, and got that it is equal to 17.333 C = 290.483 K

I have to solve for the change in entropy, though. How would I go about doing this?

I was going to either use the formula:

Delta S = n*Cp(molar)*ln(Tf/Ti) with n = (150/18.016 mols water) total and Tf = 290.483 K and Ti = 303.15 K

or Delta S = q(reversible)/T = n(DeltaH)/T

Do I say n = 100g + 50g = 150 g then divide by 18.016 to get the mols of what is now all water? My guesses are the answer is either -8191.71 J/K, -26.77 J/K, or 9.56 J/K; are any of these correct? Thank you in advance! :)

Answer 1

The heat absorbed by the ice which melts at Tm and is heated up to Tf is given by

Qice = mice x (dHf + C x (Tf - Tm))

C = 4.18 J/gK = 4.18 J/gC

dHf=333.55 J/g

The heat transferred from the water cooled down from Tb to Tf is

Qwater = mwater x C x (Tf - Tb)

Total energy of system remains constant

Qice + Qwater = 0

Tf = (mwater x Tb + mice x (Tm - dHf/C)) / (mwater + mice)

= ( 100 x 30 + 50( 0 - 333.5/4.18)) / (100 + 50)

Tf = -6.594 C = 266.406 K

Entropy change for melting process is:

dS = dHf/Tm

for Heating/cooling process from T1 to T2 at constant heat capacity:

dS=m x C x ln(T2/T1)

For portion of ice : dSice = mice x (dHf/Tm+C x ln(T2/T1))

For the water : dSwater = mwater x C x ln(Tf/Tb)

The sum of these two will give total change in entropy:

dS = dSice + dSwater

= 100g x (333.55J/g / 273K + 4.18J/gK x ln(266.406/273)) + 50 x 4.18J/gK x ln(266.406/303)

= 85.073 J/K