# Refrigerator Energy Question **URL:** \ **Category:** Factual Questions\ **Created:** [April 3, 2009, 8:35pm UTC](https://boards.straightdope.com/t/refrigerator-energy-question/491909 "2009-04-03T20:35:21Z")\ **Posts on this page:** 12\ **Page:** 1
**Author:** ![Jackknifed\_Juggernaut](https://avatars.discourse-cdn.com/v4/letter/j/9de053/32.png) [@Jackknifed\_Juggernaut](https://boards.straightdope.com/u/Jackknifed_Juggernaut)\ **Post date:** [April 3, 2009, 8:35pm UTC](https://boards.straightdope.com/t/refrigerator-energy-question/491909/1 "2009-04-03T20:35:21Z")
I purchased 2 Pepsi 24-packs yesterday. Since my fridge is empty, I put both of them in the fridge. Would I save any energy use by waiting for the first 24-pack to finish before putting in the second? ---
**Author:** ![CC](https://avatars.discourse-cdn.com/v4/letter/c/f1d935/32.png) [@CC](https://boards.straightdope.com/u/CC)\ **Post date:** [April 3, 2009, 9:36pm UTC](https://boards.straightdope.com/t/refrigerator-energy-question/491909/2 "2009-04-03T21:36:34Z")
Seems as if the essential calculation has to do with how much energy your refrigerator uses when it’s running to cool something. Taking the heat out of two packs of pop is basically the same whether you do it all at once or one at a time. Of course, there’s all the little factors like opening the door, and so on, but I think you’re not concerned about that. I say, “no appreciable difference.” ---
**Author:** ![Anthony\_N](https://avatars.discourse-cdn.com/v4/letter/a/858c86/32.png) [@Anthony\_N](https://boards.straightdope.com/u/Anthony_N)\ **Post date:** [April 3, 2009, 9:57pm UTC](https://boards.straightdope.com/t/refrigerator-energy-question/491909/3 "2009-04-03T21:57:52Z")
You’re better off filling the fridge with as much food/drinks as possible. Every time you open the door, warm air displaces the cold air that was there and it requires energy to cool this warmer air back down to the set point of the thermostat. Food will not warm up very much each time you open the door, so if you have less air to cool, you will use less energy. ---
**Author:** ![mnemosyne](https://avatars.discourse-cdn.com/v4/letter/m/c4cdca/32.png) [@mnemosyne](https://boards.straightdope.com/u/mnemosyne)\ **Post date:** [April 3, 2009, 10:03pm UTC](https://boards.straightdope.com/t/refrigerator-energy-question/491909/4 "2009-04-03T22:03:50Z")
In theory, I think doing it one at a time would be less efficient. You’re adding heat (Qin(Pepsi1 +Pepsi2) to the inside of your system (fridge) via your cases of Pepsi, and doing work (electricity) to cool it (Win). The fridge coils/compressor then warms up your room (Qout) to dissipate the heat. If nothing else changes, then Qin(Pepsi1+Pepsi2) - Qout = Win If you open the door, though, you have Qin(Pepsi1) +Qin(Pepsi2)+Qin(door opening) - Qout = Win For Win and Qout to be the same in both cases, then you’d have to add no heat to the system by opening the door, which could only happen if your room is in thermal equilibrium with the inside of the fridge, i.e. at the same temperature, and therefore using a fridge to cool the Pepsi isn’t very useful since you could just leave them on the counter and get the same result! For opening the door to be _more_ efficient, your room would have to be colder than the fridge. So doing them together and not opening the door is more efficient. I think. I hope I’m mostly right, since I am actually taking a Thermodynamics class this semester. But it’s Friday, and I’m exhausted, so don’t count on me being right! hehehe ---
**Author:** ![ZenBeam](https://avatars.discourse-cdn.com/v4/letter/z/3ab097/32.png) [@ZenBeam](https://boards.straightdope.com/u/ZenBeam)\ **Post date:** [April 4, 2009, 1:21am UTC](https://boards.straightdope.com/t/refrigerator-energy-question/491909/5 "2009-04-04T01:21:44Z")
If you put the second 12-pack in the basement or garage so that it cools off some before putting it in the refrigerator, it will save a little energy. ---
**Author:** ![Snnipe\_70E](https://avatars.discourse-cdn.com/v4/letter/s/b5a626/32.png) [@Snnipe\_70E](https://boards.straightdope.com/u/Snnipe_70E)\ **Post date:** [April 4, 2009, 1:41am UTC](https://boards.straightdope.com/t/refrigerator-energy-question/491909/6 "2009-04-04T01:41:00Z")
You have the same amount of heat to be removed no matter when you put them in. Plus every time you open the door you have to cool down any worm air that gets in. If you open the door twice the total heat load will be greater. ---
**Author:** ![Jackknifed\_Juggernaut](https://avatars.discourse-cdn.com/v4/letter/j/9de053/32.png) [@Jackknifed\_Juggernaut](https://boards.straightdope.com/u/Jackknifed_Juggernaut)\ **Post date:** [April 4, 2009, 6:58am UTC](https://boards.straightdope.com/t/refrigerator-energy-question/491909/7 "2009-04-04T06:58:29Z")
> [@mnemosyne](#): > > In theory, I think doing it one at a time would be less efficient. > > You’re adding heat (Qin(Pepsi1 +Pepsi2) to the inside of your system (fridge) via your cases of Pepsi, and doing work (electricity) to cool it (Win). The fridge coils/compressor then warms up your room (Qout) to dissipate the heat. If nothing else changes, then > > Qin(Pepsi1+Pepsi2) - Qout = Win > > If you open the door, though, you have > > Qin(Pepsi1) +Qin(Pepsi2)+Qin(door opening) - Qout = Win > > For Win and Qout to be the same in both cases, then you’d have to add no heat to the system by opening the door, which could only happen if your room is in thermal equilibrium with the inside of the fridge, i.e. at the same temperature, and therefore using a fridge to cool the Pepsi isn’t very useful since you could just leave them on the counter and get the same result! For opening the door to be _more_ efficient, your room would have to be colder than the fridge. > > So doing them together and not opening the door is more efficient. > > I think. > > I hope I’m mostly right, since I am actually taking a Thermodynamics class this semester. But it’s Friday, and I’m exhausted, so don’t count on me being right! hehehe I’m going to go with this answer, unless someone objects. ---
**Author:** ![crazyjoe](https://avatars.discourse-cdn.com/v4/letter/c/f14d63/32.png) [@crazyjoe](https://boards.straightdope.com/u/crazyjoe)\ **Post date:** [April 4, 2009, 1:53pm UTC](https://boards.straightdope.com/t/refrigerator-energy-question/491909/8 "2009-04-04T13:53:18Z")
You should take them out of the 12-pack before you put them in, too. It will increase the exposed surface area and help them cool evenly. ---
**Author:** ![mnemosyne](https://avatars.discourse-cdn.com/v4/letter/m/c4cdca/32.png) [@mnemosyne](https://boards.straightdope.com/u/mnemosyne)\ **Post date:** [April 4, 2009, 5:42pm UTC](https://boards.straightdope.com/t/refrigerator-energy-question/491909/9 "2009-04-04T17:42:09Z")
> [@ZenBeam](#): > > If you put the second 12-pack in the basement or garage so that it cools off some before putting it in the refrigerator, it will save a little energy. Not necessarily, since you’d have to know whether the difference between the initial and final heat in the basement Pepsi (Qi-Qf) was greater or less than the heat added to the fridge when you open the door (Qdoor opening) to put the basement Pepsi into the fridge. If Q(door opening) \> Qi-Qf, then you don’t save any electricity at all If Q(door opening \< Qi-Qf, then you would save some electricity The trick is figuring out how much heat transfer occurs when you open the door; in this whole thing, that’s the hardest thing to measure. ---
**Author:** ![Santo\_Rugger](https://avatars.discourse-cdn.com/v4/letter/s/e95f7d/32.png) [@Santo\_Rugger](https://boards.straightdope.com/u/Santo_Rugger)\ **Post date:** [April 4, 2009, 5:54pm UTC](https://boards.straightdope.com/t/refrigerator-energy-question/491909/10 "2009-04-04T17:54:07Z")
Even if we assume all the air in the fridge is replaced with air at ambient temperature, the specific heat of air is so low compared to water (soda) that even cooling the cans a few degrees would more than make up for the small amount of mass of air that needed to be heated up. In a normal fridge, let’s say we have 1m[sup]3[/sup] of air. That’s about 1kg. Air has a specific heat of about 1 kJ/(kg K). So, let’s say the fridge is at 5C and the air is at about 25C. So it’s going to take about 20kJ to cool down the air that got into the fridge when we opened the door. Now, water has one of the highest specific heats of any substance, at 4.18 kJ/(kg K). One 24 pack of 12 oz cans is 288 ounces, or 8.2kg. 20 kJ would change the temperature of this by about a degree and a half (8.2 \* 4.18 / 20). So, even though we can’t easily measure the heat gained by opening the door, even a conservative calculation shows that leaving one or both in the basement first is an energy saving idea. ---
**Author:** ![ZenBeam](https://avatars.discourse-cdn.com/v4/letter/z/3ab097/32.png) [@ZenBeam](https://boards.straightdope.com/u/ZenBeam)\ **Post date:** [April 4, 2009, 6:13pm UTC](https://boards.straightdope.com/t/refrigerator-energy-question/491909/11 "2009-04-04T18:13:46Z")
Also, I can add the second 12-pack when I’m already opening the door for some other reason. ---
**Author:** ![mnemosyne](https://avatars.discourse-cdn.com/v4/letter/m/c4cdca/32.png) [@mnemosyne](https://boards.straightdope.com/u/mnemosyne)\ **Post date:** [April 4, 2009, 7:14pm UTC](https://boards.straightdope.com/t/refrigerator-energy-question/491909/12 "2009-04-04T19:14:17Z")
> [@Santo\_Rugger](#): > > Even if we assume all the air in the fridge is replaced with air at ambient temperature, the specific heat of air is so low compared to water (soda) that even cooling the cans a few degrees would more than make up for the small amount of mass of air that needed to be heated up. In a normal fridge, let’s say we have 1m[sup]3[/sup] of air. That’s about 1kg. Air has a specific heat of about 1 kJ/(kg K). So, let’s say the fridge is at 5C and the air is at about 25C. So it’s going to take about 20kJ to cool down the air that got into the fridge when we opened the door. > > Now, water has one of the highest specific heats of any substance, at 4.18 kJ/(kg K). One 24 pack of 12 oz cans is 288 ounces, or 8.2kg. 20 kJ would change the temperature of this by about a degree and a half (8.2 \* 4.18 / 20). So, even though we can’t easily measure the heat gained by opening the door, even a conservative calculation shows that leaving one or both in the basement first is an energy saving idea. Good point. I wasn’t in the mood to even think about what type of scale we were talking about in comparing those two values. Pure theory and general equations are good enough for me… who cares about actual numbers? hehehe Of course, in my case, I don’t have a basement, so all Pepsi-cooling must be done in a fridge (not that I drink Pepsi, despite being from Québec… Coke Zero for me, thanks!)