# Refrigerator temperature - compartment gas volume exchange, time to equilibrate?

I open the door for 30s of a 140L capacity 80W (240VAC, 0.8A) barfridge with a 20L freezer compartment (partition open to main compartment at one end) at a temperature of -5℃ with an internal main compartment temperature at 2℃, with initially 30L of 50-50 glass and plastic containers container mainly water inside, in a room at an ambient temperature of 41℃ and 85% humidity (Sydney today), to take out a 6-pack of medicinal beer.

Presumably, the cold, denser air from inside the fridge flows out, and is replaced by the hot, lighter air from the room.

• How much of the fridge and freezer compartment air is exchanged in that time period?

• How long does it take the exchanged air in the fridge to equilibrate to its initial temperature of 2℃? What equation describes this? Is there an initial temperature rise?

• How many grams of ice will precipitate from the exchanged air into the freezer compartment?

EDIT UPDATE

After @Chris Johns' answer, I rolled the barfridge out and actually looked at the manufacturer's plate, and then updated my question from my original eyeball estimates.

1. Changed 40L to 140L

2. Added power info: "80W (240VAC, 0.8A)".

From my calculations, $P = V­_{rms} * I_{rms}$ = 240VAC * 0.8A = 192 W,

so efficiency, $\eta = P_{out} / P_{in} \%%$ $= {80W \over 192W} = 41.7\%%$

3. Changed freezer compartment capacity eyeball estimate from 10L to 20L

4. The refrigerant is $CCl_2F_2-R12$, 85g.

5. Clarified the internal contents of the fridge as containing a total of 30L of plastic and glass containers (50-50) containing mostly water.

6. Clarified the temperature of the freezer compartment at -5℃.

I presume you can calculate the coefficient of performance (CoP) from this data?

• Could you please edit your question showing us what you've done so far? As is, this looks like a "homework question", which are out-of-scope for this site without demonstration of effort from your part. – Wasabi Feb 25 '16 at 10:32
• What I have done so far is get a beer out of bar fridge on a hot day, and that got me thinking of this questn, so I'm asking out of curiosity. It is not a "homework question", tho u're welcome to use if you think suitable for ur course material. I'm an electronics engr assoc, not a mech engr, and kno nothing of thermodynamics, heat-transfer, airflow or gases, other than frm BSc 1st year phys, maths & chem courses I did 35 yr ago, & the therml managmnt eqns I use for calcg heatsink sizes 4 an elecc cct, tho I hv taught myself diff eqns & Laplace+Fourier txforms. Will start BSc+BEE 2018. – My Other Head Feb 25 '16 at 10:57
• I get the feeling that this question requires simulation in something like COMSOL Multiphysics, Modelica or Rhino 3D, etc., neither of which I've been trained in, rather than calculation from a closed form set of equations? – My Other Head Feb 25 '16 at 11:21
• @wasabi define "homework question". – Fennekin Feb 25 '16 at 12:43
• @Fennekin, this isn't the place to discuss this, but read here for the community's opinion on homework questions (which has some definitions as well). – Wasabi Feb 25 '16 at 13:30

If you open the door of a small upright fridge for 30 seconds it's reasonable to assume that virtually all of the cold air inside will be replaced with warm air from outside.

The missing information in this question is the power and coefficient of performance of the fridge, this tells you the rate at which it is able to move heat from inside to outside.

As a first approximation you are looking at the heat exchange required to cool 40L of air from 41 deg C to 2 deg C. This is simply the specific heat capacity of air multiplied by the mass of air multiplied by the temperature difference. And works out at around 40 Joules so basically very little.

A more detailed approach would consider the change in enthalpy of condensing out and freezing the water vapour present and also the heat transfer from the fridge body and contents during the time that the door is open, but again we are talking about quite small numbers.

You will put much more demand on the fridge when you put in cans of beer which are at ambient room temperature as the heat capacity of the liquid is much greater than that of the air that it replaces.

Eg the heat capacity of 1 litre of water is around 4200 J/C so a 40 deg C temperature difference would require 168 KJ of heat transfer to achieve.

Say the fridge has a power consumption of 100W and a CoP of 3 then is can transfer 300W of heat and so would take at least 10 minutes to cool 1 litre of water, although this figure will be further limited by the heat transfer rate between the water container and the air.

So in summary :

• Just opening the door for a short and taking something out is fairly minor in terms of load on the fridge.
• The minimum time (and energy requirement) to cool an object from its current temperature to fridge temp and return to equilibrium depends on the performance of the fridge, the initial temperature difference and heat capacity of the object being cooled.
• The actual time taken will depend on the rate of heat transfer between the cold air in the fridge and the object.
• I have updated the question parameters from when you saw them, from observation. You may wish to update your calcs. I think it would be a bit presumptuous of me to do it. – My Other Head Feb 25 '16 at 13:10
• The numbers are very much back of the envelope stuff to give you a sense of how the problem might be approached. A full analysis could go into far more detail than this Q&A format allows but (as the comments the the original question suggest) this is a fairly standard undergraduate thermodynamics problem. Deducing CoP is not straightforward and i is usually determined empirically but values of around 3.5 are typical. – Chris Johns Feb 25 '16 at 19:05