> I imagine seeing all that cold air being sucked out by the opening door and falling on the ground.
Air has negligible thermal mass compared to everything else solid or liquid in the fridge. As long as you don't leave the door open long enough that fridge contents themselves start warming up, the energy losses are minimal.
Quick back-of-the-envelope calculation:
My fridge is around 1.0 x 0.5 x 0.5 m, giving a volume of 0.25 m3.
Outside temp is around 25 C, inside the fridge is around 5 C.
Air specific heat capacity is around 1000 J/m3/K.
If the fridge only contains air (it does not), and all air gets cycled out when I open the door (it does not), the fridge needs to pump out 5 kJ of heat after each door opening.
I don't know what's the typical efficiency, but let's say the fridge pumps out as much heat as you put electrical work into it (COP=2).
That's around 0.0014 kWh wasted per door opening.
It's nothing compared to just steady state consumption of a closed fridge.
Typically your house air will have a dew point above 5C. That means, when it enters your fridge, dew will condense on the inside. The latent heat of condensation is really high. Just 10 grams of water could be 22 kilojoules of energy released.
That water will eventually end up as ice on the evaporator (since the evaporator coil in a fridge-freezer typically has to run sub-freezing because it is shared with the freezer). Thats more energy loss (3 kJ for our 10 grams).
Then the defrost mechanism will kick in to melt it into the drain - which is a resistive heater normally. So 3kJ again. The resistive heater typically heats far more than it needs though - a bunch of heat will be wasted into the metal of the coil and air in the fridge - which in turn will need more refrigeration to correct.
So all in all, the energy loss of opening the door of the fridge is dominated by the water you're letting into the fridge, not the energy loss of the cold air.
This analysis is tricky enough and with enough variables (house humidity, design of fridge, amount of other 'wet' food in the fridge, etc.) that I haven't seen anyone attempt to come up with a cost number, either numerically or experimentally.
I agree with the humidity argument, although even if you 10x the energy loss opening the fridge is still relatively small contribution to total energy use unless you open your fridge tens of times per day. It's nowhere close to what the article claims.
The defrost thing I guess depends on the fridge. Mine has two coils and the one in the 5 C non-freezer section generally stays above the freezing point when the compressor is not running. It does does not have a resistive heater.
10g of water condensation isn't extreme. It's about typical for moderate temperatures, like the 25°C of the OP's example. On the tropics you will have much more than that.
But 10g of water freezing is extreme. There are mechanisms on the fridge to avoid the water freezing.
Think of an ice cold beer sitting out. Quite quickly it gets covered in dew.
But the inside of your fridge and all the goods inside has the surface area of hundreds of ice cold beers.
It isn't just a matter of how much room air you trapped in the fridge, but how much passes the beers in the fridge while you have the door open - perhaps 10x as much.
I also made the same quick calculation, opening the door ~20-30 times is equivalent to placing a bottle of milk in it, in terms of joules. (Milk has ~4x the mass, and ~6x the heat capacity in terms of mass.) Is that negligible? Well, depends on the usage, i guess, but the 10x efficiency from the article certainly can't come from it.
Maybe if you open the door, the moving air heats the things inside much faster?
edit: in a sister thread they state that the humidity of air plays a more significant role.
Air has negligible thermal mass compared to everything else solid or liquid in the fridge. As long as you don't leave the door open long enough that fridge contents themselves start warming up, the energy losses are minimal.
Quick back-of-the-envelope calculation:
My fridge is around 1.0 x 0.5 x 0.5 m, giving a volume of 0.25 m3.
Outside temp is around 25 C, inside the fridge is around 5 C.
Air specific heat capacity is around 1000 J/m3/K.
If the fridge only contains air (it does not), and all air gets cycled out when I open the door (it does not), the fridge needs to pump out 5 kJ of heat after each door opening.
I don't know what's the typical efficiency, but let's say the fridge pumps out as much heat as you put electrical work into it (COP=2).
That's around 0.0014 kWh wasted per door opening.
It's nothing compared to just steady state consumption of a closed fridge.