For someone like me who constantly thinks about energy, even getting cold cereal ready to eat in the morning can be a mind-engaging task. As I pull the milk out of the refrigerator to pour on my cold cereal, I know I will probably want the milk again after I finish my cereal in order to pour myself a bowl of fresh milk to finish off my breakfast. The question always pops into my head, “should I leave the milk out, or put it back in the fridge until I need it again?” Of course, the correct answer to this question is, it doesn’t really matter. But for an energy enthusiast like me, I want the real answer. The answer that involves the least amount of energy wasted.
To reiterate, the two options, with respect to what happens with the fridge and the milk, are:
- Option A: Open fridge (assume the fridge is open for about 5 seconds each time it is opened) and take milk out, leave the milk on the counter for 15 minutes, and then open fridge to return the milk.
- Option B: Open fridge, close fridge for about 15 seconds while pouring milk, open fridge again to return the milk; 15 minutes later, open the fridge again to get the milk out, close fridge for 15 seconds while pouring milk, open fridge to return the milk.
So, the way I see it, it comes down to which option wastes more energy. Both options involve some “waste” of energy from the fridge when the fridge is opened to remove and return the milk. The “extra” energy wasted in Option A is the loss of coldness of the milk sitting on the counter for 15 minutes, while the extra energy wasted in Option B is the extra loss of cold air from the fridge when it is opened two additional times. So we just need to calculate the wasted energy.
We will make some assumptions as we go through these calculations, and the answers will vary based on the assumptions we make and the uncertainty in the numbers we use. Keep in mind that for this particular problem, we aren’t really looking for exact answers, we are looking for good enough answers that will allow us to compare the two options. Do both options result in about the same energy waste, or is one of them substantially greater than the other?
Another thing to keep in mind is that we aren’t just looking for the answer to this specific question, we are thinking about how this answer will apply to similar situations. What if what we are removing from the fridge isn’t milk, but something else? What if we are removing something from the freezer instead of the fridge? What if that something is out for an hour instead of just 15 minutes?
While we perform our calculations and compare values, we are looking at the broader picture, for rules to govern our use of the refrigerator in general. Can we come up with some good general ground rules and energy estimates that will make our future use of the refrigerator as efficient as possible? That is the goal.
Now, before we go performing a bunch of calculations, we will check to see what the Internet says about this energy efficiency question. The topic that usually comes up under any searches about whether I should leave the milk out is food safety. The standard answer for how long milk can be left out of a refrigerator without spoiling seems to be about two hours. However, some Europeans normally leave their milk out of the fridge due to a different pasteurization process, though the percentage of milk of this type of milk that is consumed varies from country to country.
But that isn’t what we are looking for about energy efficiency. It turns out there are some ideas out there about how long the refrigerator door should be left open, such as this insightful question and answer post. There are also plenty of general energy saving tips for refrigerators and other kitchen appliances, but they don’t generally get into the details of how long or how often doors should be left open.
Calculation for Option “A” Energy Wasted
So it looks like we will just have to calculate the answers and see how things turn out. In the calculations below, I am assuming a full or nearly full gallon of milk. Let’s start with the extra energy wasted in Option A: the energy wasted with the milk sitting out on the counter. When the milk container sits out for 15 minutes, there are four ways that heat is being transferred into the milk:
- Convection heat transfer from the surrounding air,
- Conduction heat transfer through the surface the container is sitting on,
- Condensation of water vapor on the outside surface of the container, and
- Radiation heat transfer from the surrounding environment.
It is instructive to first consider the total amount of heat transfer required to occur to bring the entire gallon of milk up to room temperature. To calculate this, we simply use the heat equation:
The specific heat capacity of milk is approximately 3,790 joules per kilogram degrees Celsius, and one gallon of milk has a mass of 3.8 kilograms. Thus, the heat required is:
We can expect the heat transferred in 15 minutes to come well short of this value, as we know by experience that it takes at least a few hours for milk to get even close to room temperature. In starting my heat transfer calculations, I can immediately discount the effect of radiation heat transfer, as I know it is relatively very small in cases with relatively minimal difference in temperature (which would include this case with only 20 degrees Celsius difference).
Thus, I start by calculating convection heat transfer. Convection heat transfer calculations get fairly involved, so I will spare you the gory details of a slew of dimensionless numbers (including the Prandtl number, the Grashof number, the Nusselt number, the Rayleigh number, the Biot number, and the Fourier number) and strange correlations (including raising a factor to the power of 8/27). If you are interested, though, you can see my spreadsheet with the calculations attached below.
Suffice it to say, a good estimate for heat added to the milk via convection during the 15 minutes sitting on the counter is about 15,000 joules.
I can now calculate the conduction through the surface. I do this by using equations that assume the surface that the milk is sitting on is a semi-infinite solid. This assumption is reasonable in some situations but not others. I could attempt to use finite element analysis and numerical methods to come up with better answers, but instead I am just going to use engineering judgment to modify the answers into something reasonable based on what I know about the scenario. Anyway, it turns out that it matters a great deal what material the surface is made of. The table below shows the calculated energy transfer due to heat conduction in 15 minutes for different surfaces, along with my values modified to be more realistic.
As the calculation indicates, the heat loss when the milk is sitting on steel is much greater (over 150 times greater!) than the heat loss when the milk is sitting on a hot pad, towel, or equivalent insulator. My modifications to make the values more realistic are based on the fact that the materials are not actually semi-infinite (there is a limited volume of material for the heat to be transferred into, which is particularly significant for the steel surface) and to account for the thermal resistance of the milk container and its contact with the surface. At any rate, a reasonable average value for heat loss considering a wide variety of counter top surfaces is approximately 5,000 joules.
Next, I can calculate the heat added to the milk as a result of condensation. The energy transferred as a result of condensation is simply the latent heat of vaporization of the mass of water being condensed from water vapor into liquid water on the surface of the milk container. Of course, the amount of condensation will vary significantly depending on the humidity content of the air. I made some reasonable estimates of the amount of condensation that will occur in my calculations (see full details in spreadsheet attached below) and came up with a figure of approximately 5,000 joules of heat transfer due to condensation.
So, in summary, for Option A, the wasted energy is:
In case you are interested, see attached file for details of heat transfer calculations for this section:
Calculation for Option “B” Energy Wasted
Next, we can calculate how much energy is “wasted” when the fridge is opened. The “waste” that occurs when a refrigerator is opened is a result of the cold air inside the fridge being replaced with room temperature air from outside the fridge. The total interior volume of a typical fridge is approximately 0.3 to 0.5 cubic meters. However, some of that volume is taken up by the contents of the fridge, so maybe a good average number for air volume of a fridge is 0.25 cubic meters. The mass of that air is simply:
V is the volume of the air in the fridge and ρ is the density of the air, with a typical value being 1.2 kilograms per cubic meter. Thus, the mass of air in a fridge is about 0.3 kg. To calculate the energy required to cool this air from room temperature (about 22 degrees Celsius) to fridge temperature (about 2 degrees Celsius), as in the previous section, we simply use the heat equation:
The specific heat capacity of air is approximately 1000 joules per kilogram degrees Celsius. Thus, the heat required is:
Another factor to consider is the water vapor within the air that enters the fridge. While humidity levels in the fridge and in the air outside the fridge can vary significantly, let us examine a typical scenario just to get an idea.
Say the air in your home is at 50% relative humidity and the air in the fridge is at 100% relative humidity (not an uncommon situation if the fridge has been opened recently or if there is anything wet in the fridge). According to this chart, that would mean replacing air in the fridge with 0.005 pounds of water per pound of air with air from the room with 0.008 pounds of water per pound of air. This is an extra 0.003 pounds of water per pound of air. Since we have 0.3 kg of air, that means approximately one gram of extra water vapor in the fridge.
That small amount of water vapor doesn’t seem like a big deal. However, it takes a relatively large amount of energy to condense that one gram of water vapor, about 2,000 joules in fact. If the air in the house is hot and/or humid, the energy required to condense the additional water vapor can exceed the energy required to cool the air itself.
So all together, we can estimate that the energy “wasted” when the air in the fridge is replaced with room temperature air is approximately 10,000 joules. Of course, that represents the situation in which the entire volume of air from the fridge is totally replaced with room-temperature air. Since the fridge door is only open for 5-10 seconds, maybe only one quarter to one third of the air is replaced. So, we can estimate that one fridge door opening is approximately 3,000 joules of wasted energy. So two extra fridge door openings result in approximately 6,000 joules of wasted energy.
(Note that in our scenario, both extra door opening occur just 15 seconds or so after the previous opening, so much of the energy loss due to opening the door would have already occurred. That is, the air in the fridge is already somewhere between normal fridge temperature and room temperature. Thus, the amount of energy waste is actually some value lower than 6,000 joules for the two extra fridge openings, but the 6,000 joules is still a reasonable estimate.)
For the specific scenario postulated in the introduction, leaving the milk out of the fridge for 15 minutes wastes more energy than opening the fridge door two extra times to put the milk back and get it out again (25,000 joules compared to 6,000 joules).
If the period of time for the milk to be out was reduced to four minutes, it would be a close call as to which option would waste more energy (both options wasting about 6,000 joules in that case). So four minutes is probably a good cut-off time. If a full gallon of milk will be out for any longer than four minutes, it is better to just open the fridge and put it back in, then get it out again later.
I should note here that the energy waste we have discussed above is not the same as the extra electrical energy required to run the refrigerator. Refrigerators have a typical coefficient of performance of three to four, meaning that for every joule of electrical energy used by the refrigerator, three to four joules of heat energy can be removed from the interior of the fridge. So for scenario above with 6,000 joules of wasted energy, the fridge may use about 2,000 joules of electrical energy. That is about the energy required to power a 60-watt light bulb for about 30 seconds. (Yes, I realize most “60-watt” light bulbs do not actually take 60 watts of electricity to run anymore due to the efficiency of compact fluorescent and LED bulbs).
In the end, perhaps opening the door of the fridge isn’t as bad as we might think in terms of energy efficiency. Once the fridge has been open for a while, though, the cooling system will kick on and the cool air will mostly be wasted immediately to outside the fridge. Since a typical fridge cooling system uses 100-125 watts to run, it only takes about 30 seconds of the fridge running to remove the amount of heat equivalent to replacing all the room temperature air in the fridge with normal fridge temperature air. However, if the fridge is left open for a while, all of that 100-125 watts of the fridge running is almost immediately wasted as cold air flows right out the door.
I should also note that the amount of energy wasted by a refrigerator varies with the season of the year. In the winter time, when energy is being used to heat a home anyway, energy used by the fridge to cool the interior space is put out into the home as heat. So the energy isn’t really wasted at all. A person could, in fact, heat their home just by leaving the door of the refrigerator open all the time, but this isn’t really recommended (as the food wouldn’t stay cold and the fridge probably wouldn’t last long running all the time). Also, for those that have natural gas furnaces, using the furnace to heat the home is much more efficient than using electricity (such as using a refrigerator).
However, in the summer time, when air conditioning is on, the energy wasted by the fridge causes even more energy waste than just that used by the fridge itself. The air conditioning system must work to remove the heat energy (which has been converted from electrical energy by the fridge) from the home. So if you are going to pick a time of year when you are really strict about saving energy for the refrigerator, pick summer time.
Perhaps the most notable energy efficiency tips for use of a fridge that are not immediately obvious are the following:
- There is value in keeping the air in the fridge dry by not leaving standing water or wet, uncovered foods because it takes a relatively large amount of energy to condense water (compared to the relatively small amount of energy to cool air).
- When items are removed from the fridge, consider placing them on a hot pad or towel instead of directly on a counter or table surface (especially if that surface is metal!).
In conclusion, I hope this post has spurred some thoughts about energy efficiency for refrigerators. I hope I have answered the question “should I leave the milk out?” Of course, while each of us acting individually can’t save a whole lot of energy based on how we use our refrigerators, maybe if we all tried to be just a little more efficient, our combined efforts would add up to a big impact in terms of less waste and a better overall world environment. That is the dream, anyway.
And one final note. If you enjoyed this post, you might also enjoy a somewhat related post that answers the question of what appliance wins between a freezer and a toaster. Enjoy.