I've read several popular articles telling that frequent opening of the fridge highly increases the power consumption.
Is it really so significant? Isn't the heat in the room-temperature food which is brought to the fridge so much more relevant that some air which goes into the fridge upon opening the door is nothing compared to that?
To make it more concrete: How many times do I have to open and close the fridge so the effect is comparable with putting there a 1 litre box of milk at the room temperature? Let's say the room has 22°C, the fridge 7°C.
Best Answer
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: