(no title)
dmlerner | 2 years ago
The steps would be:
1) Drink > 140, PCM solid
2) energy flows drink -> PCM, drink cools toward 140, PCM liquifies gradually
Then two possibilities:
3a) drink cools to 140-epsilon before PCM liquifies fully
4a) PCM gives energy to drink in dynamic equilibrium, while also losing energy to environment, solidifying
5a) PCM is entirely solid at 140
6a) PCM drops below 140. drink gives energy to PCM and PCM to environment. drink -> PCM thermal conductivity is presumably much higher than PCM -> env, so drink and PCM remain at same temp
OR
3b) PCM liquifies fully before drink hits 140-epsilon
4b) drink and PCM stay at thermal equilibrium (see 6a) while cooling toward 140. energy flows drink -> PCM and PCM -> environment. The former is faster, so the PCM continues liquifying
5b) PCM is entirely liquid at 140. Due to thermal equilibrium, drink is also at 140.
6b) drink stays at 140 while PCM solidifies
7b) see 5a
8b) see 6a
newaccount74|2 years ago
So there will be energy going from the drink to the environment through the lid, which in turn allows energy to flow back from PCM -> drink.