A possible case of an irreversible process in which total entropy decreases

Click For Summary
SUMMARY

The discussion centers on the thermodynamic principles surrounding entropy changes when a hot stone is dropped into a lake. It is established that this scenario does not represent a case of total entropy decrease. The entropy decrease in the stone is outweighed by the entropy increase in the lake, even if the temperature change in the lake is negligible. The calculations involve the equations dS_stone = -|dQ|/T_stone and dS_lake = |dQ|/T_lake, confirming that the total change in entropy (dS) remains positive.

PREREQUISITES
  • Understanding of the second law of thermodynamics
  • Familiarity with entropy calculations
  • Knowledge of heat transfer concepts
  • Basic thermodynamic equations and their applications
NEXT STEPS
  • Study the second law of thermodynamics in detail
  • Learn about entropy change calculations in thermodynamic processes
  • Explore heat transfer mechanisms and their effects on entropy
  • Investigate real-world examples of irreversible processes in thermodynamics
USEFUL FOR

Students of thermodynamics, physicists, and anyone interested in understanding the principles of entropy and irreversible processes in heat transfer.

ticktock
Messages
1
Reaction score
0
When a hot stone is dropped into a lake the change in temperature of the lake is negligible, but the stone cools down and so its entropy decreases. Is this therefore a case of an irreversible process in which total entropy decreases?

This isn't for homework but for revision towards my resit. I just can't find anything in my notes which is helping me understand.
 
Physics news on Phys.org
Hi ticktock, welcome to PF. No, this is not an example of total entropy decreasing. The entropy increase in the lake is finite (and greater than the entropy decrease in the stone) even if you consider the temperature increase in the lake to be negligible.
 
When an amount of heat |dQ| moves from the stone to the lake, the decrease in entropy of the stone is[tex]dS_\mathrm{stone} = -\frac{|dQ|}{T_\mathrm{stone}}[/tex] and the increase in entropy of the lake is [tex]dS_\mathrm{lake} = \frac{|dQ|}{T_\mathrm{lake}}[/tex]The total change in entropy is dS = dSlake+dSstone. Given the relative temperature of the lake and stone, what can you say about the sign of dS? Does assuming Tlake is a constant make a difference?
 

Similar threads

  • · Replies 3 ·
Replies
3
Views
3K
  • · Replies 29 ·
Replies
29
Views
6K
  • · Replies 1 ·
Replies
1
Views
2K
  • · Replies 22 ·
Replies
22
Views
6K
  • · Replies 6 ·
Replies
6
Views
1K
  • · Replies 4 ·
Replies
4
Views
8K
  • · Replies 3 ·
Replies
3
Views
26K
  • · Replies 4 ·
Replies
4
Views
3K
  • · Replies 1 ·
Replies
1
Views
2K