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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.

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  • $\begingroup$ Turning a vertical door freezer into a fridge reportedly is a lot more efficient than a normal fridge: mtbest.net/chest_fridge.html $\endgroup$ – Denis de Bernardy Jul 29 '17 at 18:17
  • $\begingroup$ @DenisdeBernardy A combination of more insulation and the hot air rising out I assume? $\endgroup$ – Tim Jul 29 '17 at 21:09
  • $\begingroup$ @DenisdeBernardy According to the Tim's calculations below, the air cooling is nothing compared to the heat brought with new room-temperature food. It seems that the fridge described in your article has the low consumption thanks rather to the good insulation (I believe freezers have better insulation than common fridges) than to the horizontal door. What do you think? $\endgroup$ – Honza Zidek Jul 30 '17 at 9:20
  • $\begingroup$ @Tim: if memory serves it's primarily because cold air stays in the vertical door fridge when you open it, whereas it flows out of the fridge when you open a normal fridge. There's a linked article to the bottom right of the page with the details. $\endgroup$ – Denis de Bernardy Jul 30 '17 at 9:21
  • $\begingroup$ @Tim Now it's up to you to prove it :) The article mentioned by DesisdeBerdardy tells something different than you, at least it seems to me. $\endgroup$ – Honza Zidek Jul 30 '17 at 9:25
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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
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  • $\begingroup$ I see the issue - my calculation referred to cooling the air from room temperature down to the temperature of the refrigerator, whereas yours uses a (more realistic) intermediate temperature. Between the change in efficiency and the total amount of energy being pumped, that should account for the factor of 10 difference. I would imagine that if you had a small refrigerator (like a mini-fridge) and you held the door open for longer (maybe ~5-10 seconds each time) the answer would get closer to my calculation. $\endgroup$ – J. Murray Jul 28 '17 at 22:44
  • $\begingroup$ @J.Murray That seems likely. A mini fridge would almost certainly have over 90% of the air exchanged as you opened and closed the door. $\endgroup$ – Tim Jul 28 '17 at 22:45
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    $\begingroup$ The time the door is open for might not even be a big deal - as the door is opened and closed complex vortices will be created which will mix the air $\endgroup$ – Tim Jul 28 '17 at 23:51
  • $\begingroup$ So the conclusion is that to save energy we should much more take care about what (= how warm) we put to the fridge than reducing the amount of door opening, am I right? $\endgroup$ – Honza Zidek Jul 29 '17 at 9:12
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    $\begingroup$ Carnot efficiency is only theoretical and can never be reached in practice. The small heat pumps used in typical home refrigerators are in fact rather inefficient. Additionally, I remember seeing a calculation that suggested that it takes more energy for the fridge to condense the moisture in the indoor air than to actually cool the air down. $\endgroup$ – ntoskrnl Jul 29 '17 at 13:52
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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.

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    $\begingroup$ It doesn't require 5kJ to cool the water, that's the amount of energy it moves to the outsides. It uses around 7% of that from the mains. $\endgroup$ – Tim Jul 28 '17 at 21:20
  • $\begingroup$ @Tim Ah right. Poor choice of words on my part. I meant to refer to the amount of heat pumped out, not the amount of energy used by the pump. $\endgroup$ – J. Murray Jul 28 '17 at 21:31
  • $\begingroup$ Yes. Unfortunately, I think it means your estimate for the energy consumption difference is off by 10x ( got a 176 x difference). $\endgroup$ – Tim Jul 28 '17 at 21:59
  • $\begingroup$ Why wouldn't the actual energy to cool the air and water required scale proportionally with the amount of energy being pumped out? $\endgroup$ – J. Murray Jul 28 '17 at 22:09
  • $\begingroup$ cooling gets harder the more you’ve already cooled. Just like heating - the kettle is quick to get to almost boiling then takes ages to do the final 5%. It’s the same with fridges. (But I made a mistake commenting here - It would scale linearly because you’re cooling it from 22°C as with the water, whereas I assumed the air reaches just 10°C). So the 13x more to cool water is correct, but the amount of energy is about 14 times greater than in reality. $\endgroup$ – Tim Jul 28 '17 at 22:11

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