Thermal Dynamics question for you techno science folk

Luke284 said:
Case (a) will be marginally cooler.

In case (a) you immediately drop the temperature to T=(Th*Mh+Tc*Mc)/(Mh+Mc)

Where Th is temperature of the Hot cup and Mh is the mass of water in the hot cup
Tc and Mc are the same for the cold cup. (temperatures need to be in K)

From this point onwards your rate of heat transfer will be increased very slightly due to the increased surface area that the heat can escape from the cup.

In case (b) you don't get that initial drop in temperature and the benefit of slightly increased surface area. Yes, you will have a higher initial rate of heat loss as the temperature will be warmer but that rate will decrease as you drop the temperature. Q=k*Area*(Th-Tair)

I'm also assuming that the density of the water does not change significantly with temperature.
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from your post, are you just guessing a)? Because both of your versions of events are purely qualitative, so how can you make a quantitive distinction?
 
Then your "experimental evidence" is a load of rubbish. Also, if sticking your fingers in was your idea of a good measurement then I would hate to hear what key variables in the experiment you failed to keep constant. At the temperatures we are talking about the differences will be so marginal.
 
What about the OP stating that the cup contains "cold" water, if we can assume that "cold" < T_ambient then I think adding water straight away would cause a lower final temperature due to the larger temperature difference when mixing the cold and hot water
 
Amleto said:
from your post, are you just guessing a)? Because both of your versions of events are purely qualitative, so how can you make a quantitive distinction?
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Because i'm using sound thermodynamic reasoning. We were only asking for a qualitative answer. I could go away and actually calculate the answer but I would have to make some assumptions based on what the thermal constants would be. Even if i was miles out on the thermal constants the equations will still tell me that case (a) will end up cooler.
 
repeat yourself and let me know then, other than a lack of thermometer, my controls were in place. What I stated was that the temperature difference was easily apparent by touch.

P.S. One of us has a degree in Physics and I don't believe it to be you ;)
 
Short answer: B. I'm pretty sure Amleto is correct (Edit: and apparently we have empirical confirmation from SpeedFreak :p).

Longer answer:

In the simplest case where we ignore the change in surface area / volume when doing the mixing, the two options give the same temperature. I've spent a while going through the maths from (almost) first principles on my own and have got expressions for the final temperature for cases A and B (which are probably the same as Amleto's results but the expressions are messy enough that I cba to check). I've tried numerically plugging in lots of combinations of time, amounts of hot/cold water, cooling rate, room temp etc, and the two always come out exactly the same.

Now what if we don't ignore the change in cooling rate due to the area / volume change after the mixing? If we have a flat-bottomed, parallel sided beaker, on adding more water we decrease (I think) the ratio of area:volume thereby decreasing the cooling rate tau, so option B (mixing at the end) will get us cooler water (the only difference between the parameters involved between A and B will be a quicker cooling rate in B).

I think this is good example of a seemingly simple problem which is surprisingly non-trivial to work out, and actually requires going through the maths to be sure of the result.

</physicist>
 
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I have a degree in Chemical Engineering and I work with this sort of stuff all the time! I would repeat your experiment but it would be a little difficult from an airport lounge.
 
If i get board on my 12.5hr flight I might plug some numbers into a spreadsheet. I think the answer is going to be so close though that you would not be able to detect it with even a good household thermometer.
 
Jonny69 said:
The important thing is the temperature difference, as heat transfer is proportional to it. Higher the temp difference, faster the heat transfer. Hotter water will transfer heat to the room faster, so by not adding the cooler water to the beaker of hot water it will lose more heat to the room by itself. Add the cool water in at the last minute. Option b).
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This is correct!
 
Luke284 said:
Because i'm using sound thermodynamic reasoning. We were only asking for a qualitative answer. I could go away and actually calculate the answer but I would have to make some assumptions based on what the thermal constants would be. Even if i was miles out on the thermal constants the equations will still tell me that case (a) will end up cooler.
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no, the answer asked for is quantitative - which one is better. But you only made qualitative descriptions. so how can you come to any conclusion?
 
Okay fine... I still think it shouldn't be necessary to resort to formal mathematics for an intuitive physical problem like this, but here we go. I have written out a solution to the problem, taking into account all relevant factors (including change of surface area etc).

For those who don't want to read the maths, the result is "Answer B" - the unmixed pot cools down more quickly, so long as the usual rules of surface-area to volume ratio are observed* (i.e. the surface-area to volume ratio decreases as the volume increases).

* As a caveat, It MAY be possible to design a strange-shaped pot, where it has a normal shape up to the level filled by the original "hot" tank, and then has a huge surface area in the region just above (lots of fins etc). In this case, adding the cooler water immediately would dramatically increase the overall surface area, leading to the inequality (*) being reversed, and the mixed pot cooling faster. But this is a circumstance that would have to be specifically designed for... under the assumption that the big pot is a simple geometric shape, then the unmixed pot will always cool faster.



... The working is longer than it really needs to be, but I broke everything down into simple steps in order that anyone with an A-level standard knowledge of maths can follow it.







Actually I quite enjoyed tackling a problem like this analytically. Usually when I have heat-transfer problems to solve they are on complex geometry, so we just do CFD models :)
 
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duff, not sure on one of your initial conditions.

if you have two masses at different temperatures (same specific heat), the average is:

(m1T1 + m2T2) / (m1 + m2)
 
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The "T2" in the formula you give there is the ambient temperature of the room (temperature of the smaller pot). Since I'm dealing with T* (the difference from ambient), your "T2" has T* = zero, and our initial condition formulae are equivalent. Note that my "T_h" is also a temperature differential from ambient...

In this case, we're only interested in temperature change from ambient. Of course, if the ambient temperature were to change with time, things would become a little more complex, and the thermodynamics of the smaller pot would come into play (though this would only work to further reinforce that the "mix at the end" configuration decays faster, as the increased surface area of the small pot would enforce this)
 
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