Will a refrigerator cool down a room?

[l0cke]

I hope I am not unleashing another "plane on a treadmill" here, but anyways.

During a recent 10th grade science test, there was a relatively simple looking question:

If you open a refrigerator in a room, will the room cool down?

My Opinion: Of course not. The fridge is putting out heat too, so the room will not cool down. Also the room is a source of heat, so it would be the same as if you were to put a computer in a refrigerator.

My Friends Opinion: He thinks that the question meant if you open a fridge in a normal kitchen, will the kitchen cool down. Here is where another problem comes in. He lives in a castle (lol), where apparently his refrigerator vents it's hot air into somewhere else besides in the kitchen. In my house, we have a separate refrigerator that vents right into the kitchen. The question does not specify anything about the matter.

Also proof would be helpful so I can show him the right answer later.

I have looked at things such as
http://home.howstuffworks.com/refrigerator.htm
but they do not really provide the solid answer I am looking for.

 

[CycloWizard]

If you can vent the heat somewhere else, then of course it will cool down a kitchen. It would essentially be an inefficient air conditioner at that point.

 

[Modelworks]

As efficient as the newer model refrigerators are , they still produce more heat than they remove from the air.
Its actually going to get warmer in the room than cooler.

 

[Rubycon]

Basic law of refrigeration is gain=loss. Of course the heat of compression lift plus inefficiencies will always add heat. Small systems with hermetic compressors depend on the refrigerant vapor on the suction side to cool the windings. The greatest efficiency will result in locating the compressor, condenser and condenser fan motor outside of the cooled space. (Classic "split system")

The old trick of opening the fridge door and putting a fan in to "cool the kitchen" so to speak may seem to work if you're standing in the airflow. Of course the compressor will run forever as the sensor bulb inside will never get to satisfaction temperature. The head pressure will run higher making the condenser hotter. The compressor is in the same space too. So actually it will make it hotter in the room in the end!

A dehumidifier works the same way - air is drawn in over a cool evaporator. Moisture forms and drips off the coil into a collector or goes into a drain hose to a sump, etc. The cool air is then fed over the condenser heating it back up before it it is returned to the room. A small dehumidifier (they're usually measure d in pints) will heat the room slightly however this heat is not felt as much since the air in the space is drier.

 

[QuixoticOne]

It is kind of a meaningless trick question whose answer depends on the circumstances.

Let's say the refrigerator is cold inside, you could certainly *unplug* it, open the door,
and the room it is in will most certainly diffuse its heat into the cold sink making the room colder and the refrigerated areas warmer. This is entirely consistent with the original question which did not ask if the refrigerator would as a result of its continued operation cool the room on an indefinitely ongoing basis.

If the refrigerator has a heat sink outside / somewhat isolated from the room in question then of course it could continue to cool the room acting an an active air conditioning type of unit.

It also depends on what you mean by 'refrigerator' .. one could certainly contrive a device that, with the application of sufficient energy / intelligence could cool its warm surroundings at the expense of some kind of input 'energy' / 'entropy or whatever. It is a tacit assumption that the 'energy' powering the cooling flows in from 'outside' the system and returns there as well. If you're allowing such an open system then it's merely a question of engineering as to how cleverly you want to use the thermodynamics / conservation laws to cause cooling of the space. One could sink waste 'heat' into the electromagnetic waves flowing out on the power cord, for instance, though it isn't obviously a practical / typical design.

 

[PolymerTim]

As others have pointed out, there are a lot of possibilities you can come up with, but I think it is safe to say that for a 10th grade science test, the correct answer is probably "no". My guess is that the purpose of the question is just to show that, not only is the heat removed from the refrigerator being put back into the room (in most standard refrigerators), but that the inefficiencies involved in the transfer causes a net heat gain in the room (the excess usually coming from the motor).

I usually find on tests that, if you reason the question too deeply such that there are multiple correct answers depending on unstated specifics, you have probably reasoned too deeply and should simplify your thinking. It can be a fun thought experiment to go as deep as you can, but "it depends" kinds of questions usually aren't asked on tests unless it is more of an essay-based test as opposed to short answer and the purpose is really to determine your reasoning skills.

 

[beansbaxter]

Define refrigerator?

 

[Modelworks]

You could really throw a wrench into the whole thing.
Yes, it will make the room colder and not add any heat at all.
What ?

When you said refrigerator, I thought you meant Ice Box.
You know, the ones that only had a huge block of ice

 

[Juno]

will refrigerator cool down a room? lol, not a chance.

 

[silverpig]

Originally posted by: PolymerTim
I usually find on tests that, if you reason the question too deeply such that there are multiple correct answers depending on unstated specifics, you have probably reasoned too deeply and should simplify your thinking. It can be a fun thought experiment to go as deep as you can, but "it depends" kinds of questions usually aren't asked on tests unless it is more of an essay-based test as opposed to short answer and the purpose is really to determine your reasoning skills.

I love this statement, except you should change the beginning to "In AT HT, if you reason..."

Most of the physics problems that get posted here have a very simple answer. The fridge will heat the room up. However, people always over complicate it. What if the heat pumped outside, what if you did it in space, would it get better in an argon atmosphere on the vernal equinox, allowing for the tides, the average wind velocity on that night etc etc...

 

[DrPizza]

The best answer to the original question:

If you open a refrigerator in a room, will the room cool down?

is "Yes."
But, most of us know what the teacher was looking for - long term, what's going to happen?

I find that on a physics test, if the question is that open to interpretation, then it's a poorly formed question. The only time it's a good question is when you're meaning to provoke a discussion. There are some things that are reasonable to assume - i.e. you're on Earth and not on some other planet. "What do you mean how long will it take for a rock to hit the ground when dropped from a height of 45 meters? You didn't say if it was dropped on the moon or on Earth." A student arguing his point is going to say "what if...?" You just need to eliminate the reasonable "what-if's" in the context of your question.

In the original question, it's not reasonable to argue that "what if the refrigerator hadn't been turned on before its door was opened?" Immediately after the door is opened, the room is, indeed, going to cool down. The question in no way seems to imply that the teacher was looking for the longer term effect of leaving the refrigerator door open (which I'm guessing was the original intent of the question.)

The original question would be better if stated this way:
"A refrigerator is operating inside a closed room. What will be the effect of leaving the refrigerator door open for an extended amount of time?"

To the OP: my refrigerator is also vented outside the kitchen, but I don't think it's a reasonable assumption when doing this problem.

 

[l0cke]

Originally posted by: beansbaxter
Define refrigerator?

According to my science book, it is:
A device that takes heat from food on the inside and transfers it to the air outside using a refrigerant.

EDIT: forgot to add that if we can prove the book wrong on any question on the science test then my teacher will count the question as right.

 

[Rubycon]

Originally posted by: DrPizza
I find that on a physics test, if the question is that open to interpretation, then it's a poorly formed question. The only time it's a good question is when you're meaning to provoke a discussion. There are some things that are reasonable to assume - i.e. you're on Earth and not on some other planet. "What do you mean how long will it take for a rock to hit the ground when dropped from a height of 45 meters? You didn't say if it was dropped on the moon or on Earth." A student arguing his point is going to say "what if...?" You just need to eliminate the reasonable "what-if's" in the context of your question.

In the original question, it's not reasonable to argue that "what if the refrigerator hadn't been turned on before its door was opened?" Immediately after the door is opened, the room is, indeed, going to cool down. The question in no way seems to imply that the teacher was looking for the longer term effect of leaving the refrigerator door open (which I'm guessing was the original intent of the question.)

LOL that was great. Reminds me of this. :Q

 

[Taqwus]

Actually the answer is very simply
If your Qout is greater than your Wout than no; if Wout is greater than the room will cool down
Pretty much your comressor would have to be over 50% efficient probably closer to 65% to overcome other losses as well

 

[DrPizza]

Originally posted by: Taqwus
Actually the answer is very simply
If your Qout is greater than your Wout than no; if Wout is greater than the room will cool down
Pretty much your comressor would have to be over 50% efficient probably closer to 65% to overcome other losses as well

Huh?
Here, look at it a different way. Why do modern refrigerators need to be plugged in to an outlet in order to operate? Ans: because they use energy. Follow-up question: where does that energy go? Or, in other words, your efficiency needs to be a tad greater than 100% in order to cool the room.

 

[Cogman]

Depends, What is the starting point for measuring the cool down, how long will the door be open, and does the back of the refrigerator vent anywhere.

If a refrigerator is not running, you start the measurement, open the door for a second, then close it before the thermostat kick the pump on, then yes, the room will be cooler then before the door opened. For 99% of people the own refrigerators, leaving the door open will result in their room becoming warmer (do to the compression pump running).

 

[Taqwus]

You do not need it over 100% it depends on how the energy is used. If 40% is given off to heat and out of the remaining 60% of kj/kg of energy left you should only need enough work output in the form of cooling to over come the 40%
I.E.
If you are putting in 1000 j/kg of work into the system and with an efficiency of 60% you would get 600j/kg of cooling and 400j/kg being rejected out and that should give you a total of 200j/kg of cooling left over
Which is what I would guess would be a correct answer for a 10th grader since the question would (I assume) come from a physics teacher and they would not be worried about cycle analysis

 

[DrPizza]

Originally posted by: Taqwus
You do not need it over 100% it depends on how the energy is used. If 40% is given off to heat and out of the remaining 60% of kj/kg of energy left you should only need enough work output in the form of cooling to over come the 40%
I.E.
If you are putting in 1000 j/kg of work into the system and with an efficiency of 60% you would get 600j/kg of cooling and 400j/kg being rejected out and that should give you a total of 200j/kg of cooling left over
Which is what I would guess would be a correct answer for a 10th grader since the question would (I assume) come from a physics teacher and they would not be worried about cycle analysis

You don't seem to understand how a refrigerator works. It cools what's inside the refrigerator by taking the energy (heat) from that environment and moving to the coils in the back were it releases that heat into the room. With the door left open and operating at exactly 100% efficiency, all it would do is remove 1000 joules of heat from the air in front of it and release those 1000 joules of heat into the air behind it. The room wouldn't change in temperature. If it operated at 99% efficiency then that 1% would heat the room. Operating at 100% efficiency is impossible. And, operating at above 100% is silly.

 

[Nathelion]

LOL at how many posts this thread is getting

 

[CycloWizard]

Originally posted by: Nathelion
LOL at how many posts this thread is getting

No joke. I thought mine would be just about the only post, so I didn't say much.

The simplest way to define efficiency in this case is as heat removed from the fridge divided by electrical energy input from the wall. ALL of the energy input from the wall will eventually become heat in the kitchen via entropic compression, pumping inefficiency, even valve inefficiency. In other words, the input energy all becomes heat. Thus, the efficiency of the system is heat out divided by heat in. Unless this quantity is greater than one, which it can never be, then the total energy in the room will always increase.

There are ways in which one could look at the system and say that the temperature decreases. If one considers only a subsection of the room (say, just outiside the refrigerator door) on short timescales (say, 1 second after the door is open), then yes, the refrigerator will decrease the temperature of that part of the room temporarily.

Like DrPizza said, this isn't a very good question. However, it's asked to a middle school class, to whom the meaning of the question should be fairly obvious due to the limits of their understanding. If, on the other hand, I saw this question on one of my exams, I'd have to answer it a little differently than someone in middle school. The context of the question is, therefore, not trivial in addressing the question itself.

 

[PolymerTim]

Originally posted by: CycloWizard

Originally posted by: Nathelion
LOL at how many posts this thread is getting

No joke. I thought mine would be just about the only post, so I didn't say much.

Hehe, My favorite part is that it is also practically a Who's Who of HT-AT. Pretty much all the regulars managed to get into this one (which I rarely see).

 

[AeroEngy]

No net change. I don't know about you but my kitchen is climate controlled. So any additional heat load is compensated for by the central AC

 

[PlasmaBomb]

Nice one aero

 

[Throckmorton]

A fridge is a heat pump. It pumps heat from the inside of the fridge to the back. That means it's impossible to cool down a room with a fridge that is in the room. End of discussion.

 

[gururu2]

the best way to answer such questions is to K.I.S.S (keep it simple stupid).

the motion in air is not destroyed by a refrigerator. it is transferred through an efficient conduit (refrigerants). a crazy amount of air has to cycle over a refrigerant (which traps heat like noones business) to strip its heat off. all that air is released into the surroundings.

a refrigerator has enough refrigerant surface area to cool a certain volume of air. if you open the door, the refrigerant will begin to attempt to cool much more air than it was designed to. the refrigerant will absorb its maximum heat capacity and will stay maxed out even though the fan/air system is working at maximum capacity. at some time, the refrigerator will simply stop cooling.

The temperature change in the room will be related to the volume and heat capacity of the refrigerant. Obviously the molecules of refrigerant can be expected to hold more energy than air molecules, thus the overall energy of the room will decrease by that much, which may or may not result in a detectable temperature change.