Thermal Dynamics question for you techno science folk

[Jonny69]

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).

 

[jak731]

you're mixing up some principles there.

t^4 is stefan-boltzmann for black body radiation. It is not affected by local temperature.

and "4th power of temp difference" is (t1 - t2)^4, not what you wrote.

Whoops, I should probably burn my bachelors degree now. Not done thermo for a few years in my defence.

 

[Amleto]

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).

I don't get why people think this logically makes sense.

logically, all you can say is:

"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. BUT it starts from a higher temperature than when we add the cooler water at the beginning. Therefore we need to carefully consider both options"

stuff it, i'll write up the proof....

http://img140.imagevenue.com/img.php?image=17632_gif_122_581lo.jpg

http://img168.imagevenue.com/img.php?image=17633_gif2_122_1062lo.jpg

 

[Tefal]

for the love of god someone just go an test this and film it.

 

[Strife212]

The cold water will be heating up to room temperature in those 10 minutes~

Edit: Oh, it already is at room temperature ;_;

 

[vcl_X1]

I don't get why people think this logically makes sense.
.

It's not logic, it's a law of physics

As I've suggested real world the difference is likely to be minimal

 

[aardvark]

It's not logic, it's a law of physics

yep

 

[Amleto]

not according to my maths

 

[Jonny69]

I don't get why people think this logically makes sense.

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

 

[Glaucus]

Havent seen anyone talk about the increased surface area of the mixed option.

 

[Amleto]

Havent seen anyone talk about the increased surface area of the mixed option.

yeah, I delibrately ignored that The equation I used is only valid for constant surface area.

 

[Dave]

yeah, I delibrately ignored that The equation I used is only valid for constant surface area.

You've also assumed that t=t* on your algebra, as you have combined one term in t* with another in t. You're essentially assuming that your heat loss "constant" (/fudge factor) is the same in each case. The last couple of lines of algebra are a little dodgy too - you've taken out a factor of the masses, but it's not been divided through the equation properly by the looks of things?

 

[Amleto]

You've also assumed that t=t* on your algebra, as you have combined one term in t* with another in t. You're essentially assuming that your heat loss "constant" (/fudge factor) is the same in each case.

sorry, all t should be t*, where t* is 10 minutes. the constant is the same, except for the already mentioned case of changing surface area.

and considering that mixing the water can only increase the surface area, which in turn will increase the rate of temperature loss, it can only mean that mixing earlier will result in a lower temperature than mixing later.

e: thats not quite right, it depends on the ratio of volume change to surface area change.

 

[Dave]

sorry, all t should be t*, where t* is 10 minutes

But still, it means that you've assumed that your two tau constants are identical as well. One way to solve it - to CFD! Although first, to bed

 

[Amleto]

because they are otherwise identical

 

[vcl_X1]

and considering that mixing the water can only increase the surface area,.

Do we know anything of the properties of the larger beaker? Are the sides parallel or is the top of it wider than the bottom. If the sides are parallel then the surface area does not change.

Basically rtfq

 

[Glaucus]

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

 

[vcl_X1]

not according to my maths

Your maths is open to (mis)interpretation, laws of phyisics are non negotiable

 

[Amleto]

Do we know anything of the properties of the larger beaker? Are the sides parallel or is the top of it wider than the bottom. If the sides are parallel then the surface area does not change.

Basically rtfq

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.

 

[Amleto]

Your maths is open to (mis)interpretation, laws of phyisics are non negotiable

sorry, maths is not open to interpretation. it's either right or wrong for a given set of assumptions.