Thermodynamics Question - Entropy etc

Click For Summary
SUMMARY

The discussion focuses on solving a thermodynamics problem related to entropy and the coefficient of performance (COP) of a refrigeration cycle. Key equations include the entropy change formula dS = dQ/T and the COP formula COP = Qh / (Qh - Qc). The user attempts to calculate entropy changes for a temperature range from 293K to 273K and evaluates the COP for an ice-maker, concluding that the entropy of the universe increases during the process. The user seeks clarification on the calculations and concepts involved.

PREREQUISITES
  • Understanding of thermodynamic principles, specifically entropy and its calculation.
  • Familiarity with the laws of thermodynamics, particularly the second law.
  • Knowledge of the coefficient of performance in refrigeration cycles.
  • Basic calculus for integrating thermodynamic equations.
NEXT STEPS
  • Study the derivation and application of the entropy change formula dS = dQ/T.
  • Learn about the second law of thermodynamics and its implications for entropy.
  • Research the concept of the coefficient of performance (COP) in refrigeration and heat pump systems.
  • Explore practical examples of entropy calculations in real-world thermodynamic systems.
USEFUL FOR

Students studying thermodynamics, engineers working on refrigeration systems, and anyone looking to deepen their understanding of entropy and thermodynamic cycles.

doubleB
Messages
19
Reaction score
0

Homework Statement


I've attached a J-peg with the question in, laid out much more clearly than I could expect to type it here. I hope the attachment is ok.


Homework Equations





The Attempt at a Solution



So far I have:

(1) dS = dQ / T
Entropy is a measure of a system's disorder.

(2) I've drawn a flow diagram but can't get a picture of that on here. Suffice to say I have tried to do this part quite hard but can't get it to work.

(3)

(a) dS = dQ /T ... dQ = 3/2 N K dT ... so dS = 3/2 N K dT/T. Then I can work out the integral and do this first part for temperature going from 293 to 273? Does mass not matter here?

(b) dS = dQ/T ... We know Q from the question as 333kJ/kg so S = Q/T goes to S = 333 * 1 kilogram of water / 273 Kelvin. Is this right?

(c) I have no idea how to do this.

The entropy change of the universe will increase during this process, as it will do for all processes above 0 Kelvin.

(4) Coefficient of Performance = Qh / Work = Qh / Qh - Qc ... this can then be written as Th / Th - Tc = 323K / 323K - 273K = 6.46.

To improve the coefficient of performance of the ice-maker you could decrease the temperature of the hot reservoir? This doesn't seem to make sense...

If anyone has the time to look at this and help me I would very much appreciate it as I have lost all my second year notes on thermodynamics and can't see how to do this question properly.

DoubleB
 

Attachments

  • Thermo Question.JPG
    Thermo Question.JPG
    44.6 KB · Views: 504
Physics news on Phys.org
Can anyone help me?
 

Similar threads

Change in entropy of the surroundings in an irreversible process
  • · Replies 14 ·
Replies
14
Views
1K
What is the Relationship Between Entropy and Temperature?
  • · Replies 2 ·
Replies
2
Views
2K
Change in Entropy When Mixing Water at Different Temperatures
  • · Replies 5 ·
Replies
5
Views
2K
Entropy change for water in contact with a reservoir
  • · Replies 4 ·
Replies
4
Views
5K
What is the Entropy Change When Water Freezes to Ice?
  • · Replies 4 ·
Replies
4
Views
2K
How to Calculate Entropy and Enthalpy in a Thermodynamics Problem
  • · Replies 4 ·
Replies
4
Views
1K
How to calculate the entropy for two sources
  • · Replies 5 ·
Replies
5
Views
2K
Change in entropy when mixing cold and hot water
  • · Replies 12 ·
Replies
12
Views
4K
Entropy and the Helmholtz Free Energy of a Mass-Piston System
  • · Replies 8 ·
Replies
8
Views
2K
Entropy change and reversible/irreversible processes
  • · Replies 1 ·
Replies
1
Views
2K