Thermo Problem

My guess is it would reach a stasis. Since the refrigeration system takes the heat out of the interior and transfers it to the room, with the doors open it takes the heat out of the room and transfers it back into the room. As soon as everything equalizes it become static.

If the fridge is warmer then its setpoint, the compressor will run. The net difference between the heat generated by the (compressor and the light bulb) of the fridge and the cooling effect it produces will be dumped into the room. If more heat then cooling is produced then the room will warm, if more cooling then heat is produced, then the room will cool. It is a sure bet, based on the published efficiencies of these untis that more heat then cooling will be produced.

If and when (doubtful it ever will) the fridge makes its set point the compressor stops, at that point only the heat from the light bulb will present a net gain to the room. The room will warm and the fridge will cycle on again.

I forgot to add to this equation any heat given off by the conductors flowing current to the fridge.

How much beer is in the fridge?

None, we had to drink it,,,it was getting warm....

Just from an empirical point of view: with or without light-bulb, air temperature in the room will increase. Should this not happen, we would be facing perpetual motion!

Yes of course. I agree.

But... he said "in a prefectly insulated" room. That should mean no transmission of heat into or out of the room. So, negating any impact from the fridge, should not the room temperature remain unchanging?

Agreed... the premise here was to ignore the fridge. Just the empty room, what would it do?

" So, negating any impact from the fridge, should not the room temperature remain unchanging? "

I don´t believe it, because there is an external source of energy (fridge becomes electricity from the outside)

Initially the room will get warmer as the light bulb warms up the room. Assuming that the fridge is 'on', at some point the temp sensor will command the compressor to kick on to cool the interior of the fridge, but since the doors are open, the heat given off by the heat exchanger coils on the back of the fridge will make the overall temperature of the room warmer despite the cooling effect on the 'inside' of the fridge.

You get a condition of positive feedback, where the warmer room temp causes the temp sensor to command the compressor to make the fridge cool, but the net effect of that is to make the overall system hotter.

Eventually the interior of the room gets so hot the fridge fails -- the compressor fails, or the wiring burns up. At which point the fridge dies and the temperature of the room remains in stasis, except for the small amount of heat lost back through the electrical outlet.

After an extremely long time the interior of the room settles back to the same temperature of the space surrounding the room.

The sun then expands to its Red Giant phase, vaporizing the Earth along with the room and the surrounding facility in the superheated plasma around the Sun.

The universe then expands outward to beyond infinity, accelerated by the expansion caused by Dark Energy.

Then... Oh, maybe I've gotten a bit carried away. Let's just stop there.

GA.

Don't overthink it. The room will heat at a rate determined by the mass of the refrigerator and the wattage of the power supply, until something fails due to heat.

Refrigerators don't make cold, they transfer heat. In this case, the heat returns.

The room warms. There is electrical energy coming into the room, because the refrigerator is trying to cool its insides, but all the heat it removes is dumped into the room (from which it is also taken, since the 'fridge door is open). The light and any other waste heat from compressor motors, friction, etc. is also adding heat to the room. Eventually, the room will get hot enough that the heat lost through its walls will equal all that electrical energy turning into heat ... until wiring or bearings overheat and fail, or the whole place catches fire.

The electrical energy used to power both the light and the refrigeration machine will all eventually be converted to heat, which will cause the temperature inside the room to increase. The use of some of this energy to power a heat pumping circuit (viz. the refrigerator's compressor, condenser, and evaporator) and that the evaporator will at first attain a lower temperature than the rest of the room makes no difference.

Heat absorbed by the evaporator plus heat from the motor and from the adiabatic heat of compression will be carried by the refrigerant vapor to the condenser. There it will undergo a phase change from vapor to liquid, releasing heat to the air in the room in the process. This liquid is released into the evaporator through a metering device and boils (or evaporates) in the evaporator, absorbing heat and lowering the temperature. This is what happens in a vapor-compression refrigeration system, used in most household and commercial units, both for food storage and for comfort cooling.

Eventually the temperature of the air in the room will become so high that the compressor can no longer raise the pressure of the refrigerant vapor enough to allow the phase change in the condenser to occur. If the system is allowed to continue running, mechanical or electrical failure will stop the process.

If the perfectly insulated room remains intact and unopened, this elevated temperature will persist indefinitely. Don't try this at home.

Since you are bringing energy into the room and it cannot escape the temperature has to rise to some degree. You have the heat from the motor in the compressor and the light bulb contributing to temperature rise. Any cooling taking place is neutralized by energy consumption and conversion to heat but not 1 to 1 ratio.

↑∞

Many correct answers here.

The refrigerator was just obfuscation, the simple and best answer is that energy is flowing into the doom but none is leaving.

I don't think it is going to take all that long to get quite hot

All the energy coming in blah, blah

Moving the heat out of the fridge is actually very inefficient, more heat you try to move more heat you generate

Probably not as good as a radiant heater, but this room is getting intolerable very very fast.

a refrigerator is a device that transfers heat at one place to another (from evaporator to condenser) at 100% eff. the room will be as is. Unfortunately, there is none 100% process in real world.

Judging from a carnot cyle, since refrigeration is just some sort of ideal reverse carnot cycle. The only attainable effeciency would only about 30% . Therefore, if you invest a 100% effort to transfer heat from the evaporator the effect will be just 30%.

The 70% actually escapes in the surrounding-in this case is the room. Definitely, the scenario is, the room will be getting hotter and hotter since thermostats wont attain set points, motor keeps on running. Waste energy will be released as heat in the surrounding

Depends on the temperature of the room and set point of the fridge.

A lot of interesting answers above.

Fridge 150 - 750 W (including spurious globe ~25 W) is no different to a light of the same wattage.

Of course it has a W defrosting heater, one would have to consider the relevant duty cycles to find the rate of rise.

Answer is; the room will heat _{(until a balance is achieved of Watts in verses losses to the outside ambient. There being no such thing as perfect insulation)}

If everything about the fridge was 100% effecient then the room temp. will stay the same, but because it is not ideal the room will warm up.

And a 100% efficient light globe?

So long as you are putting power into the room it will heat.

To be "the same" you need an over-unity fridge that additionally makes as much as the light wastes.

Get a good one off you tube and when it starts exporting power, the room will cool.

Yes, I agree that the room will heat up.

However, I think that the main point of the question is not so much aimed at the light bulb and electric generator, which give off heat as they operate. The point, as I see it, is that a compression/expansion (Carnot) cycle of the refrigerator (or heat pump for that matter) is designed to move heat from one location to another. If there is only one location, a single room, whatever heat is drawn out of the open refrigerator (i.e., the room) is also put back into the room. The compressor heats the refrigerant, and that heat gets transfered into the room. Then the expansion valve (or expansion motor) allows the refrigerant to expand, which cools the fluid and draws that same heat back out of the room.

Technically, if the movement of heat is perfectly isothermic (i.e. the heat sink is large enough to keep everything at the same temperature - absorbing the temperature differential as it happens) the system will be perfectly efficient. However, if the system is not completely efficient, one could argue that some heat would be used up as work (the Rankine Cycle), which would cause the room to cool. But again, from my vantage point, I don't see the heat actually disappearing in Rankine Cycle engines, but rather simply being spread out over a greater volume of the working fluid (gas or liquid) as it expands. Consequently, the heat then needs to be exhausted into the environment and cooler and more condensed air needs to be put back into the system and reheated in order to get more work out of the system.

In short, a perfect refrigeration cycle (less the heat of the supplimental components) would neither cool nor heat the room. I'm open to feedback...

Yeah, but there is no such thing as a "perfect" refrigeration cycle. The evaporator sucks some heat out of the room, but then the condenser rejects this same heat back into the room, plus the energy of compression (and the light bulb). Thus the room heats up.

(This was all discussed in a similar thread about two years ago, and is well known to all informed persons.)