1. Second Law of Thermodynamics Entropy

2. Entropy (S) quantitative measure of disorder
adding heat to a body increases entropy
entropy increases in free expansion
*heat is more disordering to a cold object than to a hot object

3. 2nd law of thermodynamics
“The total entropy of an isolated system that undergoes a change cannot decrease.”
*An isolated system cannot exchange energy and matter with its surroundings.

4. decrease in entropy Mix H20 at 0C and H20 at 100C Airconditioner
H20 at 0C increases in entropy
H20 at 100C decreases in entropy
Airconditioner
lowers the entropy of your home
raises the entropy of the outside air

5. total change in entropy
Mix H20 at 0C and H20 at 100C
STOTAL = SC + SH
SC > 0, SH <0
STOTAL = |SC| - |SH|
*|SC| > |SH|
STOTAL > 0

6. Irreversible Process
process that proceed spontaneously in one direction but not the other
flow of heat from warm to cold
free expansion
mechanical energy  friction

7. ENGINE Working substance absorbs heat from the hot reservoir
Performs some mechanical work
Discards the remaining energy in the form of heat into the cold reservoir
*cyclic process
Knight 7

8. engine statement of 2nd law
No device is possible whose sole effect is to transform a given amount of heat completely into work.
Knight

9. ST<0, violates 2nd law ST = -|SHR| + 0 + |SCR| > 0
entropy in engines
If there is no cold reservoir
ST = SHR + SWS
* SHR<0 : HR lost heat to WS
*SWS=0 : after one cycle
ST<0, violates 2nd law
Heat must be introduced to a cold reservoir, SCR>0 , and in such a way that |SCR|>|SHR|
ST = SHR + SWS + SCR
ST = -|SHR| |SCR| > 0

10. since a little heat is disordering to the cold reservoir, much less heat can be delivered to the cold reservoir than that removed from hot reservoir
|QH|>|QC|
the lower the temperature of CR, less heat |QC| is required to satisfy 2nd law

11. refrigerator statement of 2nd law
Heat will not flow spontaneously from a cold object to a hot object
Knight

12. entropy in refrigerators
If there is no work input
ST = SCR + SF + SHR
* SCR<0 : CR lost heat to fluid
* SF=0 : after one cycle
* SHR>0 : HR absorbed heat from fluid
Since |SCR|>|SHR|:
ST < 0, violates 2nd law

13. energy from work is delivered into HR, further increasing its entropy
If there is work input
energy from work is delivered into HR, further increasing its entropy
work is required in order to achieve |SCR|<|SHR|
ST = SCR + SF + SHR
= -|SCR| |SHR|
> 0