> Do you have a reputable source for that?My experience as an HVAC engineer, which was some time ago, hence why I said "off the top of my head". However, it's easy to check by running some numbers.
The cases to be compared are:
A/C -- 38 C x 80% relative humidity (typical for a hot, humid climate) to 23 C x 50% RH (a dewpoint of about 12 C, which is typical as an output for air conditioners).
Heating -- -10 C to 22 C, no change in dewpoint.
We'll need some physical properties of air and water; approximate values will be good enough (i.e., we'll ignore that these values change with temperature and treat them as constant):
Air -- density about 1 kg/m^3, specific heat about 1000 J/kg per degree C
Water -- density about 1000 kg/m^3, specific heat about 4000 J/kg per degree C, latent heat of vaporization about 2.3 x 10^6 J/kg (note how huge that value is--that's key to the comparison we'll be making)
We'll also need to make some assumptions about the volume of air being cooled or heated; I'm going to use a round figure of 300 cubic meters, which I think is fairly average for a home (about 200 square meters of floor area times 1.5 meters of height).
Finally, we'll need the moisture content of air at the two A/C points we gave. We can get that by taking saturation values from the Engineering Toolbox [1] and multiplying by the relative humidity (since that is just the percentage of the saturation value that is actually present in the air). We'll have to interpolate because the table only has values for 20, 30, and 40 C. The values I get are:
38 C -- 0.047 kg/m^3 x .8 = 0.038 kg/m^3
23 C -- 0.021 kg/m^3 x .5 = 0.011 kg/m^3
So the difference in moisture content is 0.027 kg/m^3; that's what has to be condensed out of the air.
Now we can calculate the heat transfer required:
A/C cooling air -- 15 C x 1000 J / kg-C x 1 kg/m^3 x 300 m^3 = 4.5 x 10^6 J
A/C condensing water vapor -- 2.3 x 10^6 J/kg x 0.027 kg/m^3 x 300 m^3 = 18.6 x 10^6 J
So my memory was somewhat off: the condensing heat transfer is only about 4 times the cooling heat transfer, not 6. Still big, though.
Heating air -- 32 C x 1000 J / kg-C x 1 kg/m^3 x 300 m^3 = 9.6 x 10^6 J
So heating for these temperatures is a little more than twice the air cooling heat transfer, but only about half of the condensing heat transfer, so about 2/5 of the total A/C heat transfer.
Note that, as the condensing calculation above makes clear, the relative humidity at the higher temperature is a huge factor. In desert conditions (for example, 38 C x 25% RH, typical for, say, Phoenix, Arizona as opposed to Houston, Texas), the moisture content of the air is only 0.012 kg/m^3, i.e., only a little higher than at the 23 C x 50% RH target point, so the condensation heat transfer goes way down. So if you're used to hot, dry conditions as opposed to hot, wet conditions, then your intuition that the condensation heat transfer isn't that large is correct. But many locations where A/C is used are are hot and wet.
[1] https://www.engineeringtoolbox.com/maximum-moisture-content-...
> contrary to what you say there's absolutely no phase change involved
There most certainly is: water vapor is vapor, and the condensate that comes out of your air conditioner is liquid. Go look at your home air conditioner: there will be a line coming off of it that drains condensate (liquid water) somewhere, usually into your home's sump. Or look under your car after you've been running the A/C for a while: you will see a puddle of water, which is liquid condensed from water vapor in the air.