Does Converting Water to Ice Increase the Entropy of the Universe?

  • Thread starter Thread starter ChronicQuantumAddict
  • Start date Start date
  • Tags Tags
    Entropy Ice Water
ChronicQuantumAddict
Messages
39
Reaction score
0
Entropy - Help! Water --> Ice

Here is my question:
10 kg of h2o at 20 degrees C is converted to Ice at -10 degrees C by being put in contact with a reservoir at -10 degrees C. The process takes place at constant pressure. The heat capacities at constant pressure of water (Cp_h2o) and Ice (Cp_ice) are
4180 and 2190 (J)/(kg)*(K)​
respectively. The heat of fusion of water, l_h2o, is
3.35x10^5 (J)/(kg).​
Calculate the change in entropy of the universe.

I know delta S_universe = delta S_system + delta S_surroundings, but i am a little stuck, any pointers?
 
Physics news on Phys.org
During the fusion process , the heat would be lost to the surroundings which would lead to increase in entropy. The reservoir is at a fixed temperature (-10) and would gain heat ...therefore entropy change for surroundings = Q/T.

For the system , the change in entropy would be due to heat lost to the surroundings As the transformation in from H2o to ice ... the net entropy change would be:

dS = Ca ln (T2/T1) + Cb ln(T2/T3)

BJ
 
Thread 'Need help understanding this figure on energy levels'
This figure is from "Introduction to Quantum Mechanics" by Griffiths (3rd edition). It is available to download. It is from page 142. I am hoping the usual people on this site will give me a hand understanding what is going on in the figure. After the equation (4.50) it says "It is customary to introduce the principal quantum number, ##n##, which simply orders the allowed energies, starting with 1 for the ground state. (see the figure)" I still don't understand the figure :( Here is...
Thread 'Understanding how to "tack on" the time wiggle factor'
The last problem I posted on QM made it into advanced homework help, that is why I am putting it here. I am sorry for any hassle imposed on the moderators by myself. Part (a) is quite easy. We get $$\sigma_1 = 2\lambda, \mathbf{v}_1 = \begin{pmatrix} 0 \\ 0 \\ 1 \end{pmatrix} \sigma_2 = \lambda, \mathbf{v}_2 = \begin{pmatrix} 1/\sqrt{2} \\ 1/\sqrt{2} \\ 0 \end{pmatrix} \sigma_3 = -\lambda, \mathbf{v}_3 = \begin{pmatrix} 1/\sqrt{2} \\ -1/\sqrt{2} \\ 0 \end{pmatrix} $$ There are two ways...

Similar threads

Back
Top