In the comments to this answer, there is some discussion as to whether putting significant mass of hot food into a refrigerator will unacceptably warm the other food already inside.

Elsewhere on Seasoned Advice, Athanasius argues passionately that with modern refrigerators, this is no longer an issue, including a personal story of measuring the temperature in their own fridge over time after putting in hot stock. This is indicative, but only represents one particular refrigerator, one method, and one trial.

So the question is, is there substantive scientific or engineering data (as from manufacturers) indicating whether or not modern home-type refrigerators--and I do mean the more common makes (in the US, this would be brands like Kenmore, GE, or Whirlpool, not the premium "restaurant quality" brands like Sub Zero)--to address this issue? (I failed completely to find such data using terms like "refrigerator recovery time" when googling myself.)

Can an average refrigerator from the last 5 to 10 years handle, for example:

  • One gallon (4 liters) of hot stock
  • or a full sized hot casserole dish, like a lasagna

put in hot (say 170-180 F), without allowing the temperatures of nearby foods to rise past the 40 F level (or at least not far past it, and not for long)?

Do the have sufficient flow of air to convect the heat away, and enough capacity in the heat engine to provide the necessary cooling?

I will admit, I have believed that putting volumes of hot food in the a home fridge is a bad idea, even for modern equipment since home refrigerators are not blast chillers.

Note: this question is not about how fast the introduced hot food cools, and whether that is safe or wise. This question is about the effect on other foods in the fridge.

  • 1
    Not an advisable practice. Best to use an ice bath first.
    – zanlok
    Jan 28, 2013 at 19:02
  • 1
    @zanlok We know it is not advisable, that is not the point of this question :-) See the note at the bottom of the question.
    – SAJ14SAJ
    Jan 28, 2013 at 19:09
  • Understood, but thought it should be stated. FDA has industry regulations for cold holding that may be instructive as to "how", though it doesn't answer your question of "why".
    – zanlok
    Jan 28, 2013 at 19:14
  • 1
    As I referenced in my comment that is linked here, official websites of food safety organizations agree-- FDA: "Despite what some people believe, putting hot food in the refrigerator doesn't harm the appliance." USDA: "Hot food can be placed directly in the refrigerator or it can be rapidly chilled in an ice or cold water bath before refrigerating." I don't know what technical research such recommendations are based on, but they agree about the practice.
    – Athanasius
    Feb 4, 2013 at 2:53
  • 1
    And again, please note that I am not advocating putting large quantities of hot food in the fridge. Use proper cooling methods (ice baths, etc.). However, if for some reason you only have two choices: (1) put directly in fridge, or (2) leave out on counter to cool, I'd generally choose the first in most modern fridges -- unless the hot container would of necessity be in direct contact with other food... in which case you have a food safety problem no matter which you choose.
    – Athanasius
    Feb 4, 2013 at 2:57

7 Answers 7


In the answer linked in the question, I already provided the results of a simple experiment I carried out a few years ago with an infrared thermometer. However, tonight I decided to try something slightly better with something closer to a worst-case scenario. I don't think it definitively answers the question, but it gives another few data points.

I heated 4 quarts of water in a 6-quart stainless pot (with a glass lid) to a rolling boil. I chose water since I didn't want to risk spoiling a large quantity of food. Also, in some ways, water is a worst-case scenario. It doesn't hold as much heat as, say, the equivalent volume of chili, but the heat circulates better in a thin liquid. That means that the entire pot will stay at roughly the same hot temperature as it cools, rather than developing a cooler outer layer (as in a pot of chili), which will start to transfer heat more slowly after the initial burst.

Meanwhile, I inserted a digital probe thermometer with a cable to the display (usually for measuring meat temps in an oven) into a quart container of yogurt. The probe was stuck through the seal in the top of the container, so very little air should have been able to get in or out. The probe measures temperatures down to 32F accurately. I taped the probe in position so the tip was immersed in the yogurt about 1/8 inch in from the edge of the container.

At the start of the experiment, the temperature of the yogurt was 38F. Using an infrared thermometer, I could measure surface temps of many other items in the fridge, which varied from about 33F to 40F. (There were a couple outliers, due to inaccuracies about the way infrared thermometers deal with reflective surfaces.)

When the water was boiling, I measured the temperature with a separate probe thermometer: it registered 212F. I quickly put the lid on the pot and whisked it immediately into the fridge and shut the door.

The yogurt was less than 2 inches from the pot. I allowed just enough room for a reasonable amount of air circulation. The yogurt was oriented with the temperature probe toward the hot pot, so it should measure the area of the yogurt that would rise in temperature the most. Also, as noted, the probe was only a fraction of an inch from the edge of the container, so any fluctuations even near the surface of the food should be registered.

Approximate times of temperature change in the yogurt are noted here:

  • 0 minutes: 38F
  • ~13.5 minutes: 39F
  • ~26.5 minutes: 40F
  • ~44.0 minutes: 41F
  • ~64.5 minutes: 42F
  • ~125 minutes: 41F

I was only checking the temperature every 10 minutes or so near the end, so the timing of the move back down to 41F may be slightly off. At 150 minutes (2.5 hours), I stopped the experiment and removed the pot from the fridge, since I didn't want to waste any more time or energy cooling down a large pot of water.

Since little was happening with the temperature of the yogurt, I did open the fridge at 30 minutes to look around. Using an infrared thermometer, I could tell that some container surfaces on the same shelf as the hot pot had reached the upper 40s with a maximum of about 50F. (This included a dark surface container that is gray and black; it was not significantly different in temperature from the surface of the light-colored yogurt container.) However, a probe inserted into these containers showed that no food inside was above 40F after 30 minutes. Note that one large plastic container on that shelf had a large empty space near the top, and the surface temp for the empty portion rose to about 60-65F, but the bottom of container that actually contained juice remained about 40F, just like the yogurt.

Using the infrared thermometer, I measured the surface temperatures of food on shelves above and below the pot -- they barely budged a degree. Nothing on any shelves above or below the pot was above 40F. I checked these again every 30 minutes or so, with the same results.

(Note that 40F is not a hard cut-off point for bacterial growth. Many types of spoilage bacteria grow in the 32-40F range, and they merely grow incrementally faster as the temperature gets warmer above 40F. Spending an hour or two at 41F or 42F or even 45F is unlikely to cause problems -- this is a typical temperature range for most items kept on refrigerator doors -- though to be absolutely safe, avoid putting highly perishable items such as raw meats in areas with temperature fluctuations.)

I could feel warmer air circulating around the pot when the door was open, but it does not seem to have been enough to significantly alter temperatures other than in the items on the same shelf -- and there only by 2-4 degrees.

I also did check the water temperature a few times:

  • 0 minutes: 212F
  • 60 minutes: 156F
  • 120 minutes: 128F
  • 150 minutes: 116F

Since the temperature of the yogurt began to drop a little after 2 hours, it seems that even a gallon of water at about 130F wasn't enough to sustain a temperature increase in the fridge -- even on immediately adjacent items on the same shelf.

So, what do I conclude from this experiment?

Even a very large quantity of very hot food (a gallon of boiling water) was only able to move adjacent food items by a few degrees, and even that only might occur in outer layers of the food. Items on shelves above or below were barely impacted at all.

I would note that I did not place any food directly in contact with the hot pot, because that would obviously cause an unacceptable rise in temperature (the pot continued to feel quite hot to the touch even after a couple of hours). But with only a couple inches of space around the pot, the adjacent foods did not rise significantly in temperature.

I should also emphasize that surface temperatures of containers did rise up to 10-12 degrees on adjacent items in that first hour, even if the interior of the food varied much less. (By about 1-1.5 hours, the surface temps had settled back down to within a degree of the internal food temps.) I think this observation suggests that caution should be applied to keep highly perishable foods (e.g., uncooked meats) away from any very hot containers, though this seems like common sense.

Perhaps the most surprising result from my perspective is that the temperature rise was halted by the time the water temperature got down to maybe 140F or so. I doubt many people are placing foods a lot hotter than 140F directly in the fridge. Also, from a food safety perspective, the food could be cooled outside to 140F (which is when bacteria may begin to grow again), and then placed in the fridge for the rest of the cooling. In my fridge, anyway, it seems doubtful that even a relatively large quantity of food 140F or lower would cause things to heat up around it.

Again -- please note that I am NOT advocating this practice, since the hot food itself could take quite a few hours to cool down in the fridge, potentially causing spoilage in the hot food. (For large quantities, use an ice bath, or break down into small containers and allow plenty of air circulation in the fridge.) But, except in extreme circumstances, there should only be a minor impact on the rest of the food in a modern well-functioning fridge.

In any case, putting hot food directly in the fridge is a safer option than leaving it on the counter to cool.


I only know of one home fridge that has a blast chiller in it, and it's from LG (the LFX31935ST). Most manufacturers aren't going to give specs on how well they handle risky behavior for fear of a lawsuit (as they could be seen as promoting risky behaviour).

The only information that I can find about how quickly the LG can transfer heat is from this blurb:

Want a cold drink but nothing’s already in the fridge? Just pop a beverage in LG’s Blast Chiller. It needs less than five minutes to chill, so your ice-cold beverage will be ready in no time.

So, assuming if you put in 12 oz at room temperature, it'll get it down to a typical fridge temp in 5 minutes. I don't know the thermal density of beer, soda or stock, but we'll go with the cross oversimplification of saying they're both mostly water so that we can get a rough estimate.

If our room temperature is near 70F, and fridge temp is 40F, this means that we can chill down 12oz at 6 deg. F per minute. A gallon of stock is 128oz, so it'll take ~10x longer. We're starting from 170-180F, so we've got to move it ~140F, not 40F, so ~3.5x longer.

So, that gallon of stock is going to take:

( 128 / 12 ) * (( 180 - 40 ) / (70 - 40)) * 5 minutes
= ( 32 / 3 ) * ( 14 / 3 ) * 5
= 248 minutes = more than 4 hours

I know what you're thing ... but it says 'under 5 minutes', so it might be 1 minute. That's possible, but if it did then they'd advertise it, so you don't blow up your beer when it hard freezes. They can't know what the starting room temperature is, or how insulative the container is. (a can of beer will chill down faster than a bottle). Or even what the beverage is (sugar solutions If we assume 4 minutes to cold, then we're looking at (4/5) the time, so about 200 minutes (still more than 3 hours).

As a second point of data, we have an early episode of Mythbusters, where they tried to cool a 6 pack. They don't mention their starting temperature, but they said it took 40+ minutes. Using the same assumed 70F start and 40F finish:

( 128 / (12 * 6)) * (( 180 - 40 ) / (70 - 40)) * 40 minutes
= ( 16 / 9 ) * ( 14 / 3 ) * 40
= 331.8 minutes = more than 5.5 hours

They're surprisingly similar, considering that one's for a blast chiller, and one's for a normal fridge. I suspect the '40+' is they stopping at 40 minutes, before it got down to temp. So let's compare it to the Mythbusters' time for being put in the freezer:

( 128 / (12 * 6)) * ((180 - 40) / (70 - 40)) * 25 minutes
= (331.8 * 25 / 40 )
= 207.4 minutes = about 3.5 hours

Maybe the blast chiller times are for a 6 pack. (but then again, with the convection in a blast chiller, maybe that's less relevant and surface to mass ratio matters more)

... but all of these suggest that you're an idiot if you put a gallon of hot stock in the fridge, as even if there's no transfer to other things nearby, the middle of the stock remains in the danger zone too long and probability will catch up to you sooner or later.

We can't actually estimate the effect on the other things in the fridge without knowing a whole lot more:

  • What containers are things in, and their insulative value?
  • Are you opening the fridge repeatedly? (which is going to replace the air in the fridge with room temp. air, which might actually be a benefit in this case)
  • How close are other things to the hot item?
  • What is the specific heat (thermal density) of all of the items? (and are they near a phase change?)
  • What's the mass of the vessel the hot item is in? (180F of cast iron isn't the same as a plastic container).
  • Is there a lid on the stock (rate of evaporative cooling)?
  • What shape are the containers? (surface to mass ratio)
  • Where in the fridge did you put the item? (cold air falls, pushing hot air up)

As such, it's not an answerable question other than to say that yes, there's an effect on the stuff around it, particularly those that may be in contact with it.

ps. Thermo was one of the two classes (along with fluid mechanics) that I almost failed in college ... and that was more than a decade ago, so it's entirely probable that I'm leaving out some other factors that would be signficant to the problem)

  • 2
    The question was never about the wisdom of putting the hot item in refrigerator. I am very well aware of the science on that. I would never do so. I like to freeze zip lock bags full of water or 16 oz water bottles to use in rapidly chilling my own stock, because I do know this. The question is about the accuracy of the assertion made elsewhere that it is a myth with modern refrigerators that putting hot items in the refrigerator endangers the other items in the refrigerator. And to that, thus far, we have no information at all.
    – SAJ14SAJ
    Jan 31, 2013 at 10:28
  • @SAJ14SAJ : so either do the experiment yourself, find a high school student looking for ideas for a science fair project, or write into the Mythbusters to see if they'll do it. For home use, you'd likely need a bunch of home weather station type devices (either wired, with the display outside the fridge), or the wireless (once done, recycle them as gifts for friends and family). If you have access to one, there are plenty of industrial data loggers that can take multiple thermocouple probes ... but they're not as cheap. ps. I forgot to mention color of the containers.
    – Joe
    Jan 31, 2013 at 19:12

Danfos, Embraco are some of the world's largest refrigerator compressor suppliers, their web sites have plenty of technical documents. You may notice that, in general, run-time is not listed, as most compressors are designed for continuous use, as they would be in the tropics etc. So they can cool large amounts of hot food, it's just a matter of time

Most domestic refrigerators generally don't have a large internal air flow, so most heat conduction is via every object in the refrigerator. Due to entropy this makes little difference to most objects unless you put a large amount of hot food in contact with a smaller amount of cold food. e.g. placing a pot of hot stock on top of a tray of sausages; the sausage will get quite warm!

A domestic refrigerator will have a cooling capacity of something like 10 to 20°C for a Kg of food in an hour (rough rule of thumb, there are many variables)

Refrigerator compressors multiply their input power by a factor of 2 to 3. So a 500 W refrigerator compressor will remove 1000 W to 1500 W of heat. This assumes the exterior air temperature is within desired operating ranges (this is a simple explanation, not science)

The good reason for not placing hot food in the refrigerator is that it is very power inefficient. So for domestic situations (unless you need maximum refrigeration life for the food in question) just cool it for a hour or two on the bench, or in a water bath before placing it in the refrigerator

  • The core of this answer is the formula 10 C / kg / hr. 10 C delta in what? Per kg of what?
    – SAJ14SAJ
    Jan 30, 2013 at 10:20
  • Did you actually the cooling capacity is (10 C * kg) / hr?
    – SAJ14SAJ
    Jan 30, 2013 at 10:33
  • @SAJ14SAJ don't understand your questions? Kg of food, it a rule of thumb, not an absolute
    – TFD
    Jan 30, 2013 at 19:45
  • Your original formula implies the more food in the refrigerator, the more heat the engine moves out per unit time.
    – SAJ14SAJ
    Jan 30, 2013 at 19:48
  • @SAJ14SAJ ahh, not a formula, but it looks like one, hmmm
    – TFD
    Jan 31, 2013 at 1:49

I'm going to add a separate answer, as I was making some black bean soup today, and decided to perform the experience. (but onyly for about 15 min, then it went into the bath to chill down).

So, the setup : a 6qt lexan container, filled with soup that was between 4L and 4qt (based on the markings on the side), and tightly lidded. It was placed on a half sheet tray, set on top of the lower shelf on my dishwasher. Next to it was 24 oz glass jar, filled with about 10oz of pickle juice. (bread & butter, from my last batch of refrigerator pickles; I save the juice for salad dressings, tuna salad, etc.) The two pickle juice was selected as it had been in the fridge, and was something that I was willing to sacrifice. (this was done in the empty dishwasher, as I wasn't willing to sacrifice the whole fridge).

An indoor/outdoor thermometer had the outdoor probe end taped onto the outside of the jar, on the side facing the hot soup container. It was attached with a small strip of gaffer's tape (note; see problems at 4:46pm)

I did not get a starting temperature on the soup; my instant read thermometer was at my neighbor's (poor planning on my part, as I was getting ready to put the soup away, I remembered the thermometer in my greenhouse, and decided to try this)

Timings are based on my cell phone:

4:36pm : 51.2F
4:37pm : --- (none taken, realized my pen didn't write and had to go get one)
4:38pm : 58.6F
4:39pm : 60.8F
4:40pm : 62.7F
4:41pm : 64.9F
4:42pm : 69.6F
4:43pm : 74.8F
4:44pm : 78.8F
4:45pm : 82.2F **
4:46pm : 77.5F
4:47pm : 75.3F
4:48pm : 75.2F
4:49pm : 76.1F
4:50pm : 76.4F

At 4:45pm I had planned to end the experiment, and retrieve the soup to put into the ice bath. When I opened the dishwasher, I found that the temperature probe had fallen off the jar, and was sitting about 1" from the soup, but more importantly was near the sheet tray, not up the side of the jar. So I re-attached it, and continued to record times 'til it looked like it was heading up again.

Of course, this was measured from the outside of the container, so it doesn't accurately reflect the temperature of the juice itself; it would've been at best the temperature of the outside wall of the container, and at worst the air temp next to the container.. It wasn't in a fridge, but it was in a similar environment (closed, white reflective walls) albeit with no other items in there, and no compressor to chill the air. Different distances from the container would likely have show different temperature curves; direct contact, as may be the case when trying to cram a large container into an otherwise occupied fridge, would have increased temperature faster (as shown by the first 10 min).

So ... if we just look at the period between 4:40 and 4:50, that's a 13.7F increase.

Oh... and ambient air temp raised from 62.6F to 64.6F during this, based on the 'indoor' reading on the probe. I have no idea if that was heat radiating out from the dishwasher (which would've been retained better by a fridge), or that I hadn't allowed the probe enough time to come up to temperature after taking it from my greenhouse (as there's snow on the ground right now).

And I have no idea how well the probe is calibrated ... I had it in the fridge while I was typing this, and it's reading 41.7F, which is higher than the 39F reported by my fridge thermometer ... I'm assuming that it's precise, but not accurate. (so the change in temp is good, the absolute temp might not be)

  • 1
    Joe, I admire the initiative for an experiment. However, I think the chilling effect of the refrigerated air coupled with the air circulation deliberately produced in the refrigerator would significantly alter these results. Also, as you noted, you were mostly measuring the temperature of the air, which had been warmed by convection currents that would be produced around a large hot mass sitting in an enclosed space (which would be cold circulating air in a fridge). I doubt your pickle juice moved by more than a degree or two in those 15 minutes.
    – Athanasius
    Feb 4, 2013 at 3:11
  • @Athanasius: a degree or two could be enough when your fridge is at 40F. (as mine might be). And this was just one single experiment. There are way too many variables, as I stated in my above answer, and for full scientific rigor, you'd have to attempt to eliminate all of the variables. I could have done many, many things worse. (a dark annodized pot to increase radiant heat; placing the hot pot in the fridge, to increase the hot mass, a dryer item (lower specific heat), placing them closer together, etc.) This is but one data point; take your complaints about my method, and add another.
    – Joe
    Feb 4, 2013 at 15:18
  • 1
    40 F is NOT a magical number. Most spoilage bacteria continue to grow below 40 F (generally down to near freezing), and bacteria do not magically begin growing significantly faster at 40 F. It's just an arbitrary round number temperature chosen by many food safety organizations. Food sitting at 41 F or even 45 F will spoil faster than at 40F, but most things would still need days at that temperature to become dangerous. Once you start getting into the 55 or 60F range, you might have a problem in a matter of hours, but a short time at 41F is NOT significantly more dangerous than 40F.
    – Athanasius
    Feb 10, 2013 at 1:00
  • by the way - I already mentioned an experiment I did in the link that was contained in the question which had a methodology much closer to the proposed question here (i.e., it actually took place in a fridge). I was satisfied with the results, but tonight decided to do something slightly better, which I'll write up in an answer here momentarily. Also, frankly, I take offense at your commentary here: I wasn't offering "complaints" about your method. I was merely providing a few reasons why the results may not be accurate, in, I would note, a rather polite manner.
    – Athanasius
    Feb 10, 2013 at 4:22

First perform a heat balance and ignore the heat transfer rate

From what I could find a modern refrigerator is about 700 btu / hour. That information does not seem to be readily published.

How many btu to to cool a gallon of water from 180 to 40
1 btu / F / lb * 140 F * 1 gal * 8.3 lb / gal = 1162 btu

From a raw BTU about about 1.7 hours

Looking at the danger zone (140 - 40)
About 1.2 hours
It is only supposed to be in the danger zone 2 hours

1.2 hours is 100% heat transfer efficiency so need 1.2 / 2.0 efficiency = 0.6.
With decent circulation should get transfer efficiency of 0.6 or more.

So now lets look at a poor little yogurt. Assume it has the same capacity of water and is 6 oz and starts at 34 F. 1 btu / F / lb * 6 F * 6 oz * 1 lb / 16 oz = 2.25 btu. So from a btu perspective the little yogurt is over powered 516:1.

But for the yogurt is a temperature thing. If the compressor can deliver cold air that is all that matters. A compressor / evaporator is very good at delivering a temperature. It might not be delivering the volume at that temperature but it delivers the temperature. The compressor has to condense the cooling fluid - if it cannot condense it will lock up.

Heat transfer is radiation, conduction, and convection. Don't have the poor little yogurt touching the hot object or even right next to it.

For sure don't have the hot item open with evaporative heat loss. Evaporitve heat loss is rapid and you can over power the compressor with a volume of hot liquid. With evaporative heat loss the compressor must knock the moisture out of the air and that is a lot of work. Even a lasagna should have an air tight seal. Don't use the pan - put it in sealed plastic container.

The hot items and cold items face the same heat transfer so that is kind of a wash. A small item is at a disadvantage as it has less capacity and a greater surface area to mass ratio.

I know you say you don't care about the cooling the hot item but that is really the more important part. You need the btu / hr to get from 140 - 40 in two hours. If you don't have the raw btu then you lose.

5 gallons of stock in an open vessel is too much for a common residential refrigerator.

A liter is easily safe. A gallon seems quite tolerable. At two gallons might start pushing it. 140 is a lot less work than 180. Even if you are in a hurry just put the sealed container in cold water for a few minutes.

Do I have citations no. This is just engineering envelope level calculations.


Assuming nothing is directly adjacent to the item in question, you have to consider what will happen when you put a heated mass in your fridge. Heat will transfer to the air first, then the solid matter. Before much heat can be transferred to the other food items in your refrigerator, the thermostat will trigger and cool the air. The air will then carry more heat from your hot food, until the stat is triggered again. The cycle will continue until the system reaches equilibrium. So I don't think you run much risk of heating the surrounding food substantively.

You do have to bear in mind that in a perfect world, worst case is you shave a few hours off the fridge life of some of your perishables, but our world is far from ideal. You are assuming the food in there wasn't already at the borderline of safety when you put it into the refrigerator, in which case a small warming influence may be enough to put it over the top. This seems unlikely, but you have to decide if it's worth the risk.

  • I am sorry, but this in no way answers the actual question asked, even though most of the facts and reasoning are correct.
    – SAJ14SAJ
    Jan 30, 2013 at 10:23
  • 1
    @SAJ14SAJ : it might not answer the thermodynamics problem, but he's right. Every time you do it you introduce risk and the odds of food poisioning. If you do it enough times, odds are, you'll either kill yourself or get very, very sick.
    – Joe
    Jan 31, 2013 at 9:06
  • Found the link I was looking for -- an NYT editorial (article?) from two days ago talking about repeatedly doing things with a low risk
    – Joe
    Jan 31, 2013 at 9:43
  • 1
    @Joe I never had any intention of actually doing this, and I am completely aware of the nature of risk, the nature of randomness, and the nature of the law of large numbers. My answers in many food safety questions I think demonstrate this. I only want to know if the common wisdom assertion about the capability of the devices is true or not, as it has come up in discussions of other questions.
    – SAJ14SAJ
    Jan 31, 2013 at 10:19

Question is: How effective are modern home refrigerators at keeping other foods cold when hot food is introduced in the space?

This is actually a socio-economic question rather than a technical question. The reason for this is the refrigerator manufacturer has to be economically competitive in the marketplace so he's only going to provide sufficient refrigeration capacity to maintain a 'standard set' of contents at a fixed cool/freezing temperature during steady state conditions plus a little extra in case non-refrigerated items are added. Recent experience with my own refrigerator (Whirlpool W4TXN ... circa 2011) has shown that even adding a fruits and veges from weekly shopping will cause the 'frig to run constantly for two days straight to bring the temperature down. And these weren't even heated - just at room temperature.

Okay, this is back-of-the envelope stuff - you would need a professional refrigeration engineer to actually calculate this out properly. The refrigeration system is a fixed capacity system that turns on and off, but only delivers the constant amount of cooling capacity any time it is running. The refrigerator volume including both the freezer and the cool box are not a heat exchange device - the heat is removed only from the freezer with warm air from the cool box rising through vents up to the freezer section where the heat is rejected to the surrounding air through the refrigeration section. Considering the heat capacity of air is about a quarter that of water (0.24 Btu/lb vs 1 Btu/lb), you would need a substantial air flow to remove the heat of a liquid.

Here's a simple calculation. One pound of water is about 1 pint - so if you have a 4-quart stock pot you have 8 pints or 8 pounds of water. If your 'hot' contents are, say 200 degF, the amount of heat you want to remove is 8 pints x 1 lb/pint x 1 Btu/lb x (200 - 40 degf) = 8 x 160 = 1280 Btu. The air flow rate needed to remove this amount of heat is given by what's called a sensible heat equation, spoken thus: cfm = load/(1.1xdeltaT). If you wanted to remove that much heat in, say, 1 hour, you would need a blower inside the refrigerator with a capacity to move [(1280 Btu/hr)/[1.1 x (20 degF)] = 58 cfm (or cubic feet per minute). In my Whirlpool, the freezer-to-bottom section section airflow openings for convective heat transfer are 2 each @ 1" x 2" free area or 4 square inches = 4/144 = 0.028 sqft. The velocity of air that would have to flow through those openings to remove the heat in one hour would be cfm/Area = 58 ft3/min / 0.028 ft2 = 2070 ft/min which is way higher than normal air handling duct for heating or cooling applications. When I put something warm in my 'frig, and the freezer is closed with its mini-blower is running, there's no way there's anywhere near this velocity of air moving into or out of the vents to the cool box. So we're looking at maybe 4-6 hours (maybe more) to carry away the heat from your stock pot give the meager freezer blower capacity and the limit flow area between the freezer and the cool box below.

But wait, the situation is actually worse than that. The added heat from the warm stock will cause water to evaporate (latent heat) from exposed fruits and veges or any dishes that are not moisture sealed. The heat in this moisture will also have to be removed through the freezer, except now the water will condense and freeze on the coils of the refrigeration system. This in turn will cause the auto-defrost cycle to run more often (or at least provide more work into removing the ice buildup on the coils) slowing the cooling process down even further.

So the simple answer to your question: Not very effective. The rise in temperature of the cool box over an extended period (hours) will result in increase rate of spoilage - which is not healthy, not to mention having to pay for replacement items. The added moisture from your non-sealed stock pot will result in condensation within the cool box providing a breeding ground to increase the rate of decomposition of your fresh foods.

Getting back to the socio-economic statement above ... Residential refrigerator designs have not substantially changed since conception. Most 'innovation' I see is arrangement of shelves the allow putting in and removing items more easily and efficiently. But the actual refrigeration system has not changed. What is being asked for in the problem is a 'blanching' facility that would permit an isolated section of the 'frig to be used strictly for cooling a hot mixture. This takes up storage space, so it would put an appliance manufacturer at a competitive disadvantage - since the great majority of people wouldn't use it - or use it properly anyway. Furthermore, a refrigerator could be designed in an insulated compartmentalized way so that opening one section inside the cool box doesn't affect other sections - thus preventing constant door opening and closing from removing the 'cold' in the entire cool box at once. Again, this would reduce the storage volume so it would place the appliance manufacturer at a competitive disadvantage. Most people who have refrigerators don't consider energy use or efficiency beside what's on the label when originally purchased. They just look at it as a cold box. This essentially prohibits blanching cooling features from being implemented, or a refrigerator from being technically improved because there would be no mass market and therefore no substantial profit. This situation is not about to change anytime soon so you'll have to blanch/cool hot dishes manually using cold water first then ice cubes before placing the hot item in the 'frig.

  • 2
    There is a lot of speculation here, but I see no facts. The purported underlying socio-economic question is in fact not relevant: I was asking what the cooling capacity of modern home fridges is, not why manufacturer's choose a particular capacity. The existence of blast chillers makes it obvious higher capacities are possible.
    – SAJ14SAJ
    Aug 21, 2013 at 22:42

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