Can we cool a room by opening the door of a refrigerator? [duplicate]

The room cools for a moment but then it becomes warmer. Why?

The refrigerator is a heat pump which uses mechanical work to move heat out of the interior of the refrigerator and dump it into the room it inhabits. That heat dump is accomplished by heat transfer coils at the back of the device which use room temperature air to draw heat out of the refrigerant substance inside the coils.

Because this process is inherently inefficient, the extra work that needs to be applied to the system to drive the process shows up as extra heat which is dumped into the room along with the heat extracted from the inside of the refrigerator.

When you first open the door, the cold air inside the refrigerator flows out into the room, causing its temperature to drop a bit. Sensing the warm air which now is entering the refrigerator, the cooling system turns itself on and tries to chill the air inside it.

The amount of heat the refrigerator is now dumping into the room is greater than the amount of heat being pulled out of the refrigerator because of the inefficiency cited above, and in addition to cancelling the cooling effect of opening the door, the excess heat causes the temperature of the room to rise.

• "cold air inside the refrigerator flows out into the room"..Not correct. There is no flow of air much less from cold to hot. Heat flows hot to cold. Feb 16, 2019 at 12:46
• the cold air, being denser than warm, sinks down out the open door and flows out across the floor. warm air from the room then flows in and takes it place. try this with bare feet standing in front of your refrigerator sometime. Feb 16, 2019 at 13:54
• Don't apply here because the pressure of the air inside the fridge is the same as the pressure outside. What changes is its specific humidity through condensation. You don't feel a rush of air from your fridge when the door is suddenly opened. The rush or air felt in the lower grill is from the condensation fan and is not from convection. Feb 17, 2019 at 15:21
• not on my refrigerator. or in my experience. Feb 17, 2019 at 17:32

This is because refrigerators use energy to lower the entropy in one location, but even more entropy is produced during this process due to dumping of heat outside of the refrigerator. So if you have a closed refrigerator that has air at a lower temperature than the room, when you open the refrigerator the air inside the refrigerator and in the room come to equilibrium (assuming this happens fast enough), which means the room's air will drop in temperature. However, now the refrigerator will be dumping more entropy into the room than it is taking out, and therefore the room's temperature will rise.

• +1 Nice answer. But as per: "...refrigerator will be dumping more entropy into the room than it is taking out" - I don't think the refrigerator takes entropy from the room to somewhere out. Or in other words, it only brings entropy inside the room + refrigerator system at the cost of electric bill. If I'm going wrong, can you please point out through which means the system looses entropy? Apr 12, 2020 at 10:53
• @GuruVishnu The room gains entropy, it doesn't lose it. Apr 12, 2020 at 12:30
• @GuruVishnu It has processes that lower entropy, yes. But with the door open it makes a net increase in entropy. Apr 12, 2020 at 12:45

Your refrigerator is a heat pump.

Heat $$Q_{\rm e}$$ is taken in from inside the refrigerator (ice box) and released outside the refrigerator $$Q_{\rm c}$$ (black fins at the back).

To do this there is a compressor which uses electric energy $$W$$.

Assuming a perfect system and using the conservation of energy $$Q_{\rm c} = Q_{\rm e}+W$$.

So more heat is given out of the back of the refrigerator than taken in inside the refrigerator.

The room cools for a moment . . . .

When the door is first opened the colder interior of the refrigerator absorbs heat from the room.

. . . . but then it becomes warmer

Overall the heat pump converts electrical energy into heat so once the temperature of the inside of the refrigerator has reached the ambient temperature the tempeature of the room will rise.

Entropy is not necessary here. I think the key is just the first law of thermodynamics: energy conservation.

A refrigerator is a "heat pump" and operates in cycles. During each cycle the fridge uses some amount of work $$W$$ to take in some amount of heat $$Q_C$$ from the cold reservoir (i.e. the inside of the fridge) and dump some heat $$Q_H$$ into the hot reservoir (i.e. the room).

The important thing is that at the end of the cycle the internal state of the pump is back where it started. So all the energy that went in has to go out. In other words, $$Q_C+W=Q_H.$$

With the fridge door open the hot and cold reservoirs are the same. The work $$W$$ is supplied by the voltage at the outlet. So in each cycle the fridge just adds a net amount of heat $$W$$ to the room and this warms it up (i.e. it takes in heat $$Q_C$$ and releases heat $$Q_C+W$$).

It looks like the average fridge draws about 600 watts of power from the outlet (that's the work per second). So an open fridge is adding 600 joules of heat to the room per second -- that will warm things up.

• Entropy is all that is at play here. First law exercises an observation of a constitutive equation or net change of different form of energies (chemical, potential,etc..) . Feb 16, 2019 at 12:55
• As you say, it's all about changing energy from one form to another: in this situation electric potential energy is being converted into heat. That's it...
– Alex
Feb 16, 2019 at 22:09
• Electric potential energy into heat?? No.. totally incorrect. You are transfering the internal energy of heat in the air to heat in the cooling fluid inside the compressor coils. You are not producing electricity nor producing heat from electricity in this process. Feb 17, 2019 at 15:04
• In fact, in every refrigerator in the universe heat is being produced from work. In our case the work is provided by an electrical voltage. In a closed fridge $Q_C$ is absorbed from inside the fridge at low temp, $Q_H$ is dumped to the outside at high temp. But this takes work $W$, supplied by the electrical voltage at the outlet. The first law assures us that $Q_H=Q_C+W$. It may sound counterintuitive to you, but more heat is released to the outside than is absorbed from the inside. The ultimate source of this "extra heat" is exactly the work that had to be put in to make the fridge run.
– Alex
Feb 18, 2019 at 0:31