# Why air conditioner evaporation cools

Preface- I am anoob scum who doesn't know much

How and why does an air conditioner gas cool when it evaporates. Exactly how does the gas's molecular kinetic energy lessen, and more specifically how does the kinetic engery from the room and pipe that holds the gas, get transferred to the gas. I have done a lot of searching on the topic, and I have not found exactly what I am looking for.

## Answers and Replies

Simon Bridge
Science Advisor
Homework Helper
It's not simple - it works by exploiting basic principles.

In general, heat goes from a hot place to a cold place all by itself.
You have to do work to get it to go the other way - this is how a fridge works, so that's what you want to have a look at. See: http://home.howstuffworks.com/refrigerator4.htm

There's quite a lot about this so perhaps if you told us exactly what you are looking for?
i.e. what's wrong with the kinds of descriptions you've found so far?

It's not simple - it works by exploiting basic principles.

There's quite a lot about this so perhaps if you told us exactly what you are looking for?
i.e. what's wrong with the kinds of descriptions you've found so far?

Ok, the freon molecules are moving around at say 10 km/h. They go into the evaporator and expand out. The freon molecules are still moving at the same speed, but they just hit the walls of the pipe less often. So exactly how does the kinetic energy of the evaporators pipes get transferred to the molecules of gas ( after that I can easily imagine how the air's heat can be transferred to the now cool pipe)

Simon Bridge
Science Advisor
Homework Helper
Ok, the freon molecules are moving around at say 10 km/h. They go into the evaporator and expand out. The freon molecules are still moving at the same speed, but they just hit the walls of the pipe less often. So exactly how does the kinetic energy of the evaporators pipes get transferred to the molecules of gas ( after that I can easily imagine how the air's heat can be transferred to the now cool pipe)

The gas cools when it expands - so it is cooler than the room. i.e. the mean kinetic energy is not the same on expansion.

Recall also that the gas is compressed to a liquid and passed through a heat-exchanger before it expands so you are talking about an expanded gas with less energy in it.
http://energyquest.ca.gov/how_it_works/refrigerator.html

The gas cools when it expands - so it is cooler than the room. i.e. the mean kinetic energy is not the same on expansion.

Okay, why does the gas cool when it expands?

Chestermiller
Mentor
Most of the heat goes into the liquid refrigerant, not the gas. Since the pressure within the evaporator is being maintained at a low value, the heat goes into causing the liquid to boil. The vapor comes off at the same temperature as the liquid, but with all the heat of vaporization contained in the gas, which then exists the evaporator. The heat is later removed from the gas and ejected to the air outside the house. This is done by compressing the gas to a higher temperature, and then condensing it with outside air (which is cooler than the gas).

Most of the heat goes into the liquid refrigerant, not the gas. Since the pressure within the evaporator is being maintained at a low value, the heat goes into causing the liquid to boil. The vapor comes off at the same temperature as the liquid, but with all the heat of vaporization contained in the gas, which then exists the evaporator. The heat is later removed from the gas and ejected to the air outside the house. This is done by compressing the gas to a higher temperature, and then condensing it with outside air (which is cooler than the gas).

Are you saying that the liquid goes into the evaporator, and then the gas that boils off takes away heat from the remaining liquid?, so it's the remaining liquid in the evaporator that is cooled, then this too gets boiled away when it takes the kinetic energy from the pipe in the evaporator to turn into a gas?

Chestermiller
Mentor
Are you saying that the liquid goes into the evaporator, and then the gas that boils off takes away heat from the remaining liquid?,

Yes.

so it's the remaining liquid in the evaporator that is cooled, then this too gets boiled away when it takes the kinetic energy from the pipe in the evaporator to turn into a gas?
Not exactly, but almost. The liquid in the evaporator is not cooled. Its temperature remains constant (because the pressure in the evaporator is controlled), but it does get boiled away when it takes the thermal energy from the room (through the pipe) to turn into gas. The heat is conducted in from the room through the wall of the pipe.

1 person
Yes.

Not exactly, but almost. The liquid in the evaporator is not cooled. Its temperature remains constant (because the pressure in the evaporator is controlled), but it does get boiled away when it takes the thermal energy from the room (through the pipe) to turn into gas. The heat is conducted in from the room through the wall of the pipe.

ok, Thanks alot. Just one more Question. Shouldn't the remaining liquid be a bit cooler because the part of the liquid that turned into a gas took kinetic energy from the liquid?

Chestermiller
Mentor
ok, Thanks alot. Just one more Question. Shouldn't the remaining liquid be a bit cooler because the part of the liquid that turned into a gas took kinetic energy from the liquid?

Well, the system is not quite at thermodynamic equilibrium, so there may be some small temperature variations within the liquid. But the liquid is being agitated by the boiling, and the magnitude of these variations in the bulk of the liquid should be very small. The temperature at the interfaces should be equal to the equilibrium temperature at the applied pressure. But, inboard of the interfaces, there should be a temperature gradient in close proximity to the interface to allow the heat to be conducted to the interface where the evaporation is occurring (to supply the heat of vaporization). The temperature gradient at the interface times the thermal conductivity of the liquid must equal to heat of vaporization times the evaporation rate per unit area of interface. So, slightly inboard of the interfaces, the temperature will be a little higher than at the interface. Convection and conduction of heat within the liquid are responsible for transport of the heat from the pipe wall to the bubble interfaces.