Second Law of Thermodynamics Entropy

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

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.

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

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

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

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

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

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| + 0 + |SCR| > 0

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

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

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

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| + 0 + |SHR| > 0