# Is the cooling rate of a (very) cold object, sitting next to an AC higher or lower?

In more detail: If i have two soda cans, both are cooled to exactly 4 degrees celsius, And i put one in a 25 degrees room, and the other next to an AC vent set to 16 degrees. After three minutes, which one should be colder than the other and why?

Edit: To clarify - if I have a cold soda can, should I place it near the AC vent or not (if I like my drink cold)? Which location will cause faster heating?

-
What is the velocity of the air from the AC? What are the dimensions of the cans? What temperature/material are the basis which the cans stand on? – Georg Oct 23 '11 at 18:22
I guess the numbers are less important... One can stands in the air, while the other stands in cool wind. – seldary Oct 23 '11 at 20:01
""I guess the numbers are less important."" Why do You ask, if gusssing is sufficient? – Georg Oct 23 '11 at 20:10
@ seldary - This question is confusing. Both cans will heat up to reach eventual thermal equilibrium with their environment, rather than cool. – Richard Terrett Oct 24 '11 at 9:37
@georg - I didn't ask for a guess, rather for an educated explanation. I just don't need an exact number. – seldary Oct 24 '11 at 20:38

At the level you are asking the question it should be obvious that the closer to the cold output of the AC the can is sitting the cooler it will remain, since next to the vent the temperature will be the coldest and not yet mixed with the air in the room, a "refrigerator" set up.

The soda can when coming out of the refrigerator will be 4C but the air coming right out of the AC is much warmer, 16C, the close can will heat up to that in equilibrium. If the other can is at 25C it will reach an equilibrium at that temperature and will be warmer.The argument still holds for 3 minutes, the one at 25C will be incrementally warmer than the one at 16C.

If you are interested in the physics basis of this and real numbers, here is a chapter on heat transfer.: conduction, convection and radiation. Boundary conditions are necessary to get real numbers from differential equations.

-
The problem with this argument is that the convection of the air caused by the AC fan could lead to faster heating in the initial stage, well before equilibrium. – Ron Maimon Oct 25 '11 at 5:01
@Ron Maimon It depends, if the air conditioning is just started or it is already in a steady state in the room. I am conjecturing by the way the question is framed that it is in a steady state, and the cans are brought in. I could equally say that the air from the convection will evaporate the water that must have condensed on the cans faster, and therefore cool further by evaporation and not heat as you suggest. And he talks of three minutes – anna v Oct 25 '11 at 10:49

The answer for the question as posed is easy:

It could be either

This applies for the initial rate of heating. Of course, over time, the story is quite different. The temperature of the one in front of the AC could initially become higher than the one in the static room air. However, given sufficient time, of course the can in front of the AC will be the cooler can and will remain that way. Simple Newton's law of cooling:

$$\frac{dT_{can}}{dt} = C_{air} \left( T_{air} - T_{can} \right)$$

$$\frac{dT_{can}}{dt} = C_{AC} \left( T_{AC} - T_{can} \right)$$

The solution to both of these is simple. Written for both cases, they are:

$$T_{can}(t) = T_{air} - ( T_{air}-T_{can}(0)) exp(-C_{air} t)$$

$$T_{can}(t) = T_{AC} - ( T_{AC}-T_{can}(0)) exp(-C_{AC} t)$$

I can not make any statements about the exact values, and I don't think that's likely to be valuable for this exercise, but allow me to make simple relative statements.

$$T_{air} > T_{AC}$$

$$C_{air} < C_{AC}$$

It's taken as a given that the initial can temperature is the same for both cases. Should the above inequalities have the same direction, then the problem would have an absolute answer. One can would be hotter at all times, $t>0$. In the problem presented, the signs are different leading to two possibilities, which is that the AC case initially leads to a higher temperature, or that the AC case is always of lower temperature. This can be formalized by the following inequality.

$$C_{AC} \left( T_{AC} - T_{can}(0) \right) \stackrel{?}{\le} C_{air} \left( T_{air} - T_{can}(0) \right)$$

If the above inequality is true, then the can in front of the AC is always a lower temperature. If it is false, then the can in front of the AC initially becomes hotter, then the other can becomes hotter.

-
Possibly the most baffling downvote yet. I don't get you downvoters. – Alan Rominger Oct 25 '11 at 20:57
maybye he thinks your making it to complicated, but i agree – Bored915 Oct 25 '11 at 22:34

Tricky. The 25C room is a static environment, whereas in the 16C AC-blowing room, the can environment is forced to be at 16C.

After 3 minutes, which is a short amount of time, I think we need to take into account many factors, like - among others - if there is moisture in the air. The can that is blown on by the AC will have more thermodynamics of condensing-evaporating humidity than the one that is not blown on.

On the other hand the can that is not blown on, will cool the surroundings and depending on how it is placed in the room (suspended? on a surface? which material is the surface made of?) will cool the air around it, dampening temperature gradient between the liquid at 4C and the bulk of the room at 25C. But that cold air will have a different density, so convection comes into play, etc, etc.

-

The air around the soda can effects how fast it transfers heat. If the temperature difference is low it will change gradually because the actual energy it takes to transfer energy is constant. waters specific heat is 4181.3 J/(kg·K), water has the second highest specific heat capacity. This means it takes a certain amount of energy to transfer. The lower the temperature the slower the energy transfered is.

The can in the cold air should remain colder from the exprience from labs i've done with it in the past.

-
What is this? An answer to Your ill posed question? – Georg Oct 26 '11 at 9:39
poor attmept at an answer – Bored915 Oct 26 '11 at 18:04
@Georg - i dont get what that comment about questions from Good Counsel is about are you an Alumni or teacher? – Bored915 Oct 26 '11 at 18:07
Singular is alumnus! – Georg Oct 26 '11 at 18:20
So you are an Alumnus – Bored915 Oct 26 '11 at 18:27

AlanSE's answer is pretty good. I'll try to add to it by keeping it math free and strictly using everyday experiences.

Imagine you're in a pool of cool, still, water. If you keep still you're not too cold but as soon as you move a bit you get very cold. A blanket of warm water forms around you that protects you. As soon as you disturb this blanket by moving, you encourage heat to flow out of your body. The Cs in AlanSEs answer account for differences in fluid flow.

The same thing is happening to your soda cans. If the AC is blowing fast enough, the cooler air will warm the soda can faster - despite the smaller temperature gradient (difference in T). If the forced air current is slow enough, the can at a higher temperature will always warm quicker than the one at a lower temperature.

After a long time, the soda cans will reach equilibrium. The soda placed on the unmoving air has a higher equilibrium temperature, so it will eventually be warmer.

Note that the change in T (dT/dt) is also proportional to the difference in T. You can see this in AlanSE’s answer. If you haven’t taken a differential equations class yet, take it seriously no matter how hard you find the subject because they are beautiful and so powerful.

-