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Twan Belgers December 8, 1998 09:43

As a lay-person I wondered if someone can solve this question : "You poured in a cup of coffee. Just when you wanted to add some milk/creamer someone rings at the door. How do you keep your coffee so warm as possible ? A. By adding quickly the milk before going to the door. B. Add the milk when you get back from the door. C. It doesn't make any difference. I would be thankfull if you could give me the right answer

K December 8, 1998 12:38

Re: Lay-question
This is a heat transfer question. Heat flow occurs due to the difference in temperature. The greater the difference the faster your coffee will lose its heat. You are going to have to add the milk anyway so why not add it early on. This decreases the temperature difference and therefore the heat transfer. The other option is to let the coffee cool and then add the milk. I suppose we are comparing heat transfer from two separate bodies (the coffee and the milk) which both have large temperature differences compared with the ambient to a mixture of the two with a lower temp. difference. I'd go with A. Add the milk before going to the door. Of course the right answer depends on your assumptions. To keep it as warm as possible, don't answer the door.

John C. Chien December 8, 1998 12:50

Re: Lay-question
Well, simply ignore the door bell and enjoy your coffee. It's likely that XYZ company is trying to sell you a new version of the code with numerous new features. And they were unable to get you to the presentation session. They think that you have to attend their training session in order to use the code. The fact is that people will try to break the lock if the code is really good. So, forget about the door bell, enjoy your coffee and write a good code.

jay December 8, 1998 15:03

Re: Lay-question
Q (heat flux) decays exponentially as a function of delta_T (temperature difference.) With this fact in mind: your coffee will be warmer when you return if you add the milk before you answer the door than if you add it after returning from answering it.

Jeff W December 8, 1998 20:15

I'm actually posting a response here!
I am definately not at the level of most of the people on this forum. I've never written code, and have just started using some commercial CFD programs. I'm so happy to finally be able to add something here!

This reminds me of an old wives tale that I dispelled for my father at the age of 10. My dad told me that if you put a bowl of boiling water and a bowl of luke warm water outside in the snow, the bowling water would freeze first. I said I didn't believe him, so we tried it. Of course, the luke warm water froze first!

The idea behind this tale, as mentioned in a previous post, is that temperature difference drives the rate of heat transfer. That's right, the RATE of heat transfer. The boiling water is cooling at a faster rate than the luke warm water and any given instant in time. However, the time it takes to freeze will be longer overall.

Let's say that the luke warm water starts at delta_T_luke and takes 2 hours to freeze. The boiling water starts at delta_T_boil and cools at a quick rate to delta_T_luke. From there it will take 2 more hours to freeze.

-Jeff Waters

Jonas Larsson December 9, 1998 03:32

Re: Lay-question, cooling coffee, freezing water
About the question of whether or not you should add the milk before opening the door - yes, you should at the milk as soon as possible, the heat flux from the warm coffee is proportional to the temperature difference between the surrounding air and the coffee. Thus lowering the temperature of the coffee by adding milk as soon as possible will reduce the total heat-flux from the cup and thus increase the final temeperature.

The question about freezing a certain amout of water is more complex since that problem also involves another effect - evaporation. A boiling cup of water will evaporate more water and thus might freeze quicker than a luke-warm cup of water because the final ice-cup has less mass. I don't know if this will actually happen though, but the problem certainly becomes more complex if you include this effect, which under certain conditions might not be negligeble.

Jeff W December 9, 1998 08:11

Re: Lay-question, cooling coffee, freezing water

I meant to illustrate that temperature difference RATE of heat transfer. You should DEFINITELY add the milk after you answer the door. I don't know what else to say other than try modeling it and see. Better yet, do the calculation by hand... I'm sure it's not that hard. As stated in my last message, the hot coffee will cool to the milk and coffee temperature (assuming the volume and mixture thermal properties haven't changed much by the addition of milk) in a certain amount of time. Both fluids cool from that temperature to, say, ambient air temperature in the same amount of time after that.

Jeff W

Damon December 9, 1998 08:34

Re: Lay-question, cooling coffee, freezing water
Newtons law of cooling says that under forced convection, the rate of heat loss is directly proportional to the temperature of the liquid. Therefore, the hotter the liquid is, the faster it loses heat. As the milk by itself is not very likely to be changing temperature very quickly, then in order to minimise the loss of heat from the whole system, the milk should be put in ASAP so that the average temp. and therefore the loss of heat from the system is brought down. However, as Newtons law is for conditions of forced convection, unless your cup is standing on an open window ledge on a windy day, I doubt if it makes too much difference when the milk goes in.

As for saying that boiling water will freeze faster than water at room temperature, this is a slightly different issue. The fact is that the molecules in the boiling water are in a state that makes it much easier for them to form into the regular lattice structures found in frozen water (ie ice) than those molecules with less energy (ie the water at room temperature).

Jeff Waters December 9, 1998 11:26

Re: Lay-question, cooling coffee, freezing water
I'm not sure if you're saying that boiling water will freeze faster than luke warm water... I believe the answer to be, "luke warm water will freeze in a shorter amount of time".

As for adding the milk ASAP, I must disagree. As for Newtons law only applying to Forced convection, I must disagree. Besides, cooling coffee is under natural convection (so there is bouyant flow).

I. Dotsikas December 9, 1998 22:24

Re: Lay-question, cooling coffee, freezing water
It depends on your Modell: if you just use an exponential cooling equation it should make no difference. If you consider that the properties of your coffee ( like thermal conductivity and heat capacity ) do change while pouring some cream or milk you > might < calculate a
>> non-significant << temperature difference. It could make a difference if you pour slowly your cream or suddenly, or if you consider buyoncy. So start drinking your coffee black. Don't try to drink tee in order not to think about such problems. I tried it some years ago. At this time I had a frient from North Germany who used to place a spoon of cream directly under the surface of the tee and letting the cream going slowly down. At this point you get more problems with surface tension than you can affort. How about drinking water or black coffee and try to write proper code as John suggests?

John C. Chien December 10, 1998 12:21

Re: Lay-question, cooling coffee, freezing water
When I answered this question, I was being invited to attend a CFD presentation about a new version of a commercial code. I didn't even looked at the physics of the problem. Yesterday, I did some simple calculation, and the results showed that the final temperature depends on the rate of the heat loss Q dot which is a function of temperature difference. The heat loss will come from (1) radiation from the surface of liquid ( coffee), (2) evaporation ( mass transfer), (3) heat transfer to the cup wall. This brings up an interesting point. After the coffee is poured, the cup wall temperature will be the same as the coffee itself. That is the cup is as hot as the coffee. Now you add some amount of milk at the room temperature ( or even colder ) to the hot coffee. The temperature of the mixture will drop. Now the cup is hotter than the coffee with milk, and now the coffee will start absorbing heat from the cup until both reach the uniform temperature. ( since the air is a very good insulator, the heat loss from the outside wall of the cup to air is relatively small in this period of time when compared with the heat transfer on the liquid side.) The radiation and the mass transfer from the surface of the coffee is probably important. That's may be the reason why lid was invented for tea cup.

Peng, S-H. December 10, 1998 15:08

Re: Lay-question
I would choose answer B: add the milk after your are back from the door.

Jonas Larsson December 11, 1998 03:53

Re: Lay-question
Why? Heat flux (Qdot) is (at least) proportional to the temperature gradient (Tfluid-Tair). We want to minimize the total heat-energy lost (Q) from the system of milk+coffee, that is minimize the time integral of Qdot. Lowering Tfluid and thereby lowering Qdot as soon as possible is therefore the best approach (assuming that the milk remains in the fridge while you open the door so that the ambient air doesn't warm it more than it cools the coffee if you choose not to add the milk before opening the door).

Peng, S-H. December 11, 1998 04:48

Re: Lay-question
Well, the heat loss is indeed proportional to the temperature difference, as you stated. On the other hand, it is also related to the volume of the fluid (as well as the exposure surface to the ambient). Adding the milk increases the volume and thus the heat loss).

If the time (to answer the door) is not that long, one can readily show that the heat loss due to volume increase (plus the heat loss from a smaller temperature difference by adding milk) is larger than the heat loss resulting only from a larger temperature difference.

Jonas Larsson December 11, 1998 08:00

Re: Lay-question
A yes, I see your point. Let's do a bit of computations on this to see which effect is most important:

Qdot ~ A * DT , where A=Exposed Area, DT=Temperature difference

Assume that the cup is a cylinder:

r = radius of cylinder cup

hc = height of coffee in cup

hm = height of milk in cup

Tc = initial temperature of coffe

Tm = initial temperature of milk

Ta = temperature of ambient air

Lets only look at the side area of the cup - this is the worst case!

The unmixed case then gives an intial Qdot of:

Qdot ~ 2*pi*r*hc * (Tc-Ta)

The mixed case gives an intial Qdot of:

Qdot ~ 2*pi*r*(hc+hm) * ((Tc*hc+Tm*hm)/(hc+hm) - Ta)

That gives:

Qdotmixed-Qdotunmixed ~

(Tc*hc+Tm*hm-Ta*(hc+hm)) - hc*(Tc-Ta) =

Tc*hc+Tm*hm-Ta*hc-Ta*hm-Tc*hc+Ta*hc =

Tm*hm-Ta*hm = (Tm-Ta)*hm < 0

(since Tm < Ta - the milk is cooler than the ambient air)

This gives that:

Qdotmixed < Qdotunmixed

Hence, you should add the milk as soon as possible!

Note that this is the "worst case" since I've only looked at the side area of the cup. If you include the top and bottom the results will be even clearer since those areas do not change when you add milk.

Also note that the volume is of no importance since the final volume of the system is the same, irrespective of if you add the milk beofre or after opening the door.

Did I do something wrong?

Peng, S-H. December 11, 1998 10:22

Re: Lay-question
I am afraid that this is not a reasonable derivation. If, for example, Tm=Ta, it would give no point for the comparison you made: something important has been ignored by simply using the simbol "~".

I suppose that the sign of (Qunmixed-Qmixed) is determined by (beta-1-alpha), where

beta=(DT)unmixed/(DT)mixed (normally, beta > 1). [DT is the temp. difference for the fluid after the time period when you answer the door] alpha=volume ratio between those of the milk and of the coffe (normally, alpha<1).

The answer thus depends on the volume of the milk you add, and could be multiple. I suggest B because I usually add milk far more than a few drops to make the strong Swedish coffe light.

Jonas Larsson December 11, 1998 12:36

Re: Lay-question
I assumed that the milk has a lower temperature than the air - seems like the most reasonable assumption to me. But yes, if the milk has the same temperature as the air then my calculation shows that the effect of the increased exposed side area and the lowered tempereture gradient cancels out - the problem then becomes dependent on the balance between the areas unaffected by adding milk (surface and bottom of cup) and the fact that the milk+coffee mixture has a larger mass and thus will cool slower (slower exponential decay) and thereby as the mixture cools also have a higher relative temp. gradient. This problems would need a different analysis method and the problem also become dependent on the time it takes to open the door - you'd have to solve the diff. eq.'s and put in reasonable guesses in the two exponentially decaying temperature solutions that you'll get.

Something is wrong with your reasoning that if you have to add a lot of milk then you should add it after opening the door. Lets say that the Swedish coffee is so strong that you have to add so much milk that if you add it at once then the mixture temperature becomes the same as the ambient air temperature (again assuming that the milk is cooler than the air). Then the final temperature as you come back will of course also be the same as the air. If you on the other hand leave the cup as it is the coffe will cool while you are away and if you when you come back add the same amount of milk the final temperature will be lower than the air temperature. Hence, even in this extreme case it is better to add the milk at once. It should also be noted that the Swedish coffee according to some exchange students can indeed be this strong ;-)

John C. Chien December 11, 1998 17:52

Re: Lay-question
It's interesting indeed. Perhaps, one should be adding coffee to the milk instead in this case. Is it practical to simulate the whole process using CFD code ? Assuming that you pour the milk at the exact center of the cup, then it could be simulated as axi-symmetrical problem. Anyone interested in this simulation problem ?

Jonas Larsson December 12, 1998 12:17

Re: Lay-question
Not a bad idea ... but lets do a full 3D simulation of the whole thing, including evaporation from the free surface, bouancy in the two-component liquid, bouancy in the ambient air and transient heat conduction in the cup. Seems like a solvable problem ;-) I'll bet my money on that it's better to add the milk as soon as possible.

John C. Chien December 14, 1998 11:36

Re: Lay-question
A full 3-D simulation is fine. This can be a virtual CFD project on CFD-online. I guess we need to know more ( to define ) about the geometry, boundary conditions, physical properties etc.. As a first step, we need a coffee cup, a real 3-D coffee cup. Then , perhapes, try to define mesh inside the liquid ( coffee) , the mesh inside the cup wall, and the external mesh. The coffee cup must have a handle similar to the Utah teapot. The initial temperature of the coffee, the room temperature, the milk temperature, ..need to be defined. The nice thing about this problem is that the Reynolds number is low, and there are more options you can use ( methods ) to handle it. Is there a standard size coffee cup ?

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