Change in entropy of system and universe

Click For Summary
The discussion focuses on calculating the change in entropy for a system where 50.0g of water at 30°C is frozen to ice at -10°C. The relevant equation for entropy change is ΔS=Q/T, and the calculation involves integrating the specific heat capacity over the temperature range. Participants emphasize the need to account for both the water and ice phases in the entropy change. Additionally, they note that the freezer's temperature remains constant, allowing for the calculation of the surroundings' entropy change as -Q/Tfreezer. The conversation highlights the importance of understanding heat flow during phase changes to determine total entropy changes accurately.
sfgoat
Messages
10
Reaction score
0

Homework Statement



50.0g of water (the system) at 30C is frozen to ice at a final temp of -10C in a freezer. Assuming that the volume of water remains the same during the process, calculate the change in entropy of the system and the change of entropy of the thermal universe when the system reaches thermal equilibrium.
constants.jpg


Homework Equations


ΔS=Q/T=∫Ti..Tf(CpdT)/T

The Attempt at a Solution


See attached image for attempt at solution. I already missed the question in class and just want to know how to properly do it at this point. Thanks.
CCI11212014_0001.jpg
 
Physics news on Phys.org
sfgoat said:

Homework Statement



50.0g of water (the system) at 30C is frozen to ice at a final temp of -10C in a freezer. Assuming that the volume of water remains the same during the process, calculate the change in entropy of the system and the change of entropy of the thermal universe when the system reaches thermal equilibrium.

Homework Equations


ΔS=Q/T=∫Ti..Tf(CpdT)/T
This is correct for the entropy change for the water. What about the ice part?
 
sfgoat said:

Homework Statement



50.0g of water (the system) at 30C is frozen to ice at a final temp of -10C in a freezer. Assuming that the volume of water remains the same during the process, calculate the change in entropy of the system and the change of entropy of the thermal universe when the system reaches thermal equilibrium.
View attachment 75716

Homework Equations


ΔS=Q/T=∫Ti..Tf(CpdT)

In order to calculate the change in entropy of the freezer (i.e the surroundings) assume that the freezer is so large that its temperature does not change during the process. In that case, the change in entropy is -Q/Tfreezer where Q is the heat flow out of the water. What is the heat flow out of the water?

AM
 
Thread 'Correct statement about size of wire to produce larger extension'
The answer is (B) but I don't really understand why. Based on formula of Young Modulus: $$x=\frac{FL}{AE}$$ The second wire made of the same material so it means they have same Young Modulus. Larger extension means larger value of ##x## so to get larger value of ##x## we can increase ##F## and ##L## and decrease ##A## I am not sure whether there is change in ##F## for first and second wire so I will just assume ##F## does not change. It leaves (B) and (C) as possible options so why is (C)...

Similar threads

  • · Replies 4 ·
Replies
4
Views
2K
  • · Replies 6 ·
Replies
6
Views
461
  • · Replies 1 ·
Replies
1
Views
2K
  • · Replies 20 ·
Replies
20
Views
3K
  • · Replies 3 ·
Replies
3
Views
2K
  • · Replies 23 ·
Replies
23
Views
3K
Replies
1
Views
2K
  • · Replies 8 ·
Replies
8
Views
2K
  • · Replies 2 ·
Replies
2
Views
2K
  • · Replies 4 ·
Replies
4
Views
4K