Is frequent opening of the fridge really so significant for the power consumption?
That depends on whether the fridge monitors the temperature or not.
Where I work, the large walk-in fridge has a temperature monitor. It starts to cool only when the temperature rises above 4.7°C and stops when it sinks to 3.5°C.
The fridge is very well insulated, meaning the fridge very rarely has to turn on when the door is closed.
I frequently retrieve items from the fridge. This normally means the door is open for less than 15 seconds, but in that time the fridge frequently rises to 5+°C, and you hear the cooler start up again.
For that fridge, energy consumption is close to 0 when not opened and reaches its maximum every time it is opened.
However, you want to know the difference between opening the fridge and cooling 1l of milk.
The Carnot coefficient of refrigeration $$\gamma = {T_c \over T_h-T_c}$$ is the ratio of the heat extracted to the work required to extract this heat.
$T_c$ is the temperature in the fridge (I'll say 2°C = 275 K), and $T_h$ is room temperature (at 22°C = 295 K), So $\gamma = 13.75$. This means to move one joule of heat energy from the milk to outside it takes 0.073 J from the mains.
The energy we want to remove from 1 litre of milk when cooling from $22°C$ to 2°C is (a, b) $$Q=mc\Delta\theta = 1\text{kg} \times 4181 {\text{J} \over \text{kg} °C} \times 20°C = 83620 \text{J}$$ removed from the milk (assuming milk $\approx$ water - it's close, but not perfect). This will take $83620 \text{J} \times 0.073 = 6104 \text{J}$.
My fridge contains about 224l (10 mol) of air. Opening the door raises the temperature from 4°C to around 10°C (I just checked). The $\gamma$ ratio for that is 46.17, so every Joule removed requires 0.02J.
Cooling 224l of air from 10°C to 4°C means moving $Q = 0.288 \text{kg} \times 1000 {J \over kg °C} \times 6°C = 1728 \text{J}$. This will take $1728 \times 0.02 = 34.56$.
However, when I open my fridge, a 15W bulb is turned on. If I open the fridge for 10 seconds, the bulb has already used 4.3x more electricity than will be used cooling the air.
This means you can open the fridge over 175 times before you've reached the energy consumption of cooling your milk (although when including the light bulb, it’s closer to just 33 times). However, at current electricity costs, it's around \$0.00026 to cool that milk - so I doubt the power consumption will ever really matter to you.
If you drink the average amount of milk for a French citizen (260 litres - Wikipedia has bizarre lists) you’re spending just \€0.067 per year on your milk.
Instead of worrying about the milk here’s a few quick suggestions:
- turning lightbulbs off, and changing for energy saving ones - up to €180/yr
- buy a TV which uses very little electricity in standby mode - up to €38/yr
- don’t boil too much water - up to €58/yr
In order to cool 1 liter of room temperature water (why is your milk at room temperature?) from 22 C to 7 C, you need to remove $Q=(1 kg)(4186 \frac{J}{kg ^\circ C})(15^\circ C)\approx 63 kJ$ of energy.
In order to cool 250 liters of air (a reasonable estimate for the empty space in your refrigerator) by the same amount, you need to remove $Q\approx 5 kJ$. It seems that cooling the water requires somewhere around 13x as much energy.
However, there are several issues here. Most importantly, once the water is cooled, it stays cool; on the other hand, much of the cool air escapes every time you open your refrigerator. How many times per day do you open and close your refrigerator? Ten? Twenty? If you have a family with kids, it may be significantly higher than that.
Put a temperature probe in your refrigerator and watch what happens when you open and close the door. The temperature rises remarkably fast, because the cool air falls to the floor and is replaced by the ambient air in the room. The effect is lessened as you go further back into the body of the refrigerator, which is why you should never store things that spoil (like milk) in the door.
Also, don't forget that the kitchen is typically warmer than the rest of the house - especially near the refrigerator, which pumps hot air out the back.