Question:

Please Help with the Second Part

by  |  earlier

0 LIKES UnLike

It is possible, with care, to cool water below its freezing point without causing the water to freeze. This is referred to as supercooling. Suppose that you did this, and had 129.0 grams of supercooled water at -8.50deg C in a perfectly insulated container. Now suppose that the water suddenly and spontaneously changes to a mixture of ice and water (it partially freezes). Assume a constant volume, a constant pressure of 1 atm, and no loss or gain of heat to or from the surroundings. How much of the water freezes? Enter your answer as a mass in grams with two decimal places. The heat of fusion of water is 333 J/g. Assume that the specific heat of H2O(l) is constant at 4.184 JK-1g-1.

Answer is 13.78g

Part 2)

What is the entropy change for the process described in the previous question? Enter your answer in J/K.

PLEASE help me!!! I have no idea what I keep doing wrong. Could you please show me how to do this!

Thank You!!!

 Tags:

   Report

1 ANSWERS


  1. We assume the entropy of crystallization to be the same as the reversible process of melting.

    ΔSmelt =ΔHmelt/Tmelt

    ΔHmelt = 333 J/g; Tmelt = - 8.5°C = 264,5 °K

    Specific entropy is ΔSmelt = 1,26 J/(K*g)

    multiplied with the mass of ice formed ΔSmelt = 13.8 J/K

    No guaranty, long time ago my physical chem.

Question Stats

Latest activity: earlier.
This question has 1 answers.

BECOME A GUIDE

Share your knowledge and help people by answering questions.