Thermal Dynamics question for you techno science folk

[vcl_X1]

Area will get larger regardless, as the contact area with the beaker increases.

You're right, but is the beaker a better insualtor than air? If it is then does this have much of an effect? If it's not then how much effect does it have then? If it's equivalent to air then does it have any effect at all? Either way it seems to have a small bearing on the outcome.

My main point is the 4th power of the temp difference is the largest factor at work in the scenario and the remining variables seem relatively negligible in comparision.

 

[Dave]

because they are otherwise identical

They are identical if you assume that the start and final temperatures are identical for the time taken, as of course it is solely dependent on the boundary conditions of the problem. I thought the usual procedure for mathematical proof is that you show that the opposite of your assumed result is false, thereby leaving the assumption as true?

 

[Tefal]

Probably because that's what they taught me in first year thermodynamics?

I'm not going to pretend to understand your maths. I last did that stuff over 12 years ago and yours is undoubtedly correct because this is an internet forum. I'll take your word for it but I won't understand why

Damn it Jonny you are undoubtedly the kind of man who owns a thermometer a stopwatch and if not a beaker an old paint tin.

Test it man!

but an end to this insanity.

 

[vcl_X1]

maths is not open to interpretation. .

Not my words..I said your maths...

 

[Glaucus]

ok, if we take the beaker to be at the environment temperature, and all sides (ok not the bottom ) to be parallel to mavity etc, then surface area is constant, volume increases, and now mixing earlier will keep it hotter than mixing later.

That's amazing, you have managed to masdively comress water under atmospheric pressure. Doesn't matter what shape the beaker is, surface area will increase.

 

[aardvark]

not according to my maths

which is wrong.


take a room at, say, zero degrees, then have 2 liquids, one at 10 degrees and one at 2000 degrees - do they both fall 10 degrees at the same speed? of course not.

 

[Amleto]

which is wrong.


take a room at, say, zero degrees, then have 2 liquids, one at 10 degrees and one at 2000 degrees - do they both fall 10 degrees at the same speed? of course not.

never suggested that they would

 

[aardvark]

never suggested that they would

i misunderstood then - what are you saying?

 

[Amleto]

That's amazing, you have managed to masdively comress water under atmospheric pressure. Doesn't matter what shape the beaker is, surface area will increase.

It was implicit that since I was using Newtons law of cooling (for convection), that heat transfer through the beaker (conduction) wasn't considered. hence the surface area for convection would be constant in e.g. a constant radius beaker.

 

[Glaucus]

So your ignoring a significant way of cooling.

 

[Amleto]

They are identical if you assume that the start and final temperatures are identical for the time taken, as of course it is solely dependent on the boundary conditions of the problem. I thought the usual procedure for mathematical proof is that you show that the opposite of your assumed result is false, thereby leaving the assumption as true?

ok, then read the bit in my first maths link as "my assertion is that ... != ..." (ie change = for !=)

Then read all the subsequent '=' as '!='. What you end up with is that Tenv != Tenv.

this cannot be true, therefore the 'assertion that ... != ...' cannot be true.

 

[Amleto]

So your ignoring a significant way of cooling.

yes, in an attempt to bring some accuracy into this thread using the principles that others were quoting. Namely Newtons law of cooling.

I used that law of cooling an applied it, as best I could, to the problem given.


e: if you re-do the algebra with different \tau which considers surface area as constant, and volume as larger in the mix-at-the-beginning case, then the answer comes out that mixing earlier will keep a higher temperature.

 

[SpeedFreak]

experimental proof confirms that adding it first retains a higher temperature

 

[Tefal]

For the love of the zombie jebus test it then work out why later, that's the way to do it!

 

[SpeedFreak]

For the love of the zombie jebus test it then work out why later, that's the way to do it!

see post above yours

 

[vcl_X1]

experimental proof confirms that adding it first retains a higher temperature

Great news, how many times did you repeat the experiment?

 

[SpeedFreak]

seriously, this isn't uni physics any more lol.

 

[Luke284]

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.

 

[SpeedFreak]

experimental evidence has proven otherwise

 

[Luke284]

lets have the results then! I would imagine that it would be so close that it would be impossible to tell with the accuracy of any measurement devices you would have around the house.