Any heated liquid will immediantly start transferring its heat to the containment vessel and to the cooler surrounding cooler temperature or air. Heat travels, cold retracts. Both liquids would obtain the same temperature, but only for a very short amount of time. And that would only be after adequate mixing, its like new car depreciation, you don't have it for a long time.

 

Your math is fine unless you take in below freezng. One gram of water equals one calorie per degree. This means that when one gram of water losses one calorie when you lower it one degree. This means thet your math is correct unless you reduce the temperture below 32 degree where it takes 81 calories to go from 32 liquid to 32 frozen.

 

i'm not sure, but i believe you are right on the first one (ie, 146 deg F) but wrong on the second one.

for the second mixing, i believe you have to weight the temp by the mass at that temp. intuitively, one tiny icecube wouldn't cool the ocean, and an iceburg will freeze your pool.

i'm not sure this is correct, but my calc would be:

(2 x 146 + 74) / 3
= (292+74)/3
= 122 deg F

the calcs may be more complicated than this tho. been quite a while since i did physics or thermodynamics.

--sgl

 

Density of H2O @ 74 deg F = .998 g/cc ; 2 gal * .998 g/cc = 1.996 g*gal/cc

Density of H2O @ 212 deg F = .958 g/cc ; 1 gal * .958 g/cc = .958 g*gal/cc

74(1.996 g*gal/cc) = 147.7 F*g*gal/cc
212(.958 g*gal/cc) = 203.1 F*g*gal/cc

Total mass = 1.996 + .958 = 2.954 g*gal/cc

Temperature*mass = 147.7 + 203.1 = 350.8 F*g*gal/cc

350.8 F*g*gal/cc / 2.954 g*gal/cc = 118.8 deg F

Heat loss to the environment not estimated!

Ben and Heidi

 

Caloric energy content of boiling water at 212 is higher than 212 unboiling. So in an isothermic situation, you'll get a higher than estimated temperature in the mixing. But coming contrary to that is the actual thermal losses to the vesels and environment.

 

Density of H2O @ 74 deg F = .998 g/cc ; 2 gal * .998 g/cc = 1.996 g*gal/cc

Density of H2O @ 212 deg F = .958 g/cc ; 1 gal * .958 g/cc = .958 g*gal/cc

74(1.996 g*gal/cc) = 147.7 F*g*gal/cc
212(.958 g*gal/cc) = 203.1 F*g*gal/cc

Total mass = 1.996 + .958 = 2.954 g*gal/cc

Temperature*mass = 147.7 + 203.1 = 350.8 F*g*gal/cc

350.8 F*g*gal/cc / 2.954 g*gal/cc = 118.8 deg F

Heat loss to the environment not estimated!

Ben and Heidi

Okay, I'm impressed!

 

Maybe this will help... lol

Heidis_Goats makes the point that numbers are finite in length and that the set of all possible texts is a countable infinity. I think "It is hard to see how the continuum hypothesis, which concerns nondenumerable infinite sets, could have any application in physics,. "Au contraire, it is easy to see the thought - the rationale is that the meaning of a text cannot be fully established by any text or finite sequence of numbers, ergo (!!) the "meaning" of a text is an infinite text (countably infinite). A given finite length formula has, on this interpretation, a nondenumerable number of potential "meanings".

-Da Fonz

 

If you take one gallon boiling water (approximately 212 degrees, for the sake of argument) and mix it in a container with one gallon of water that is room temperature (74 degrees for the sake of argument), do you get two gallons of water that are 146 degrees?

Close. Because the specific heat of a substance varies, we cannot simply take the arithmetic average of the temperature. Also, the density of the water at 212 degrees is different from that of 74 degrees. I get 1.99 gallons at 141.7 degrees Fahrenheit. I used some very accurate property evaluation software, so this is more accurate than most folks need.

If so, then if you mix ANOTHER gallon of room temperature water, will you then have three gallons of water that are 110 degrees?
Donsgal

Here again, we have the complications of varying specific heat and density. I get a final volume of 2.987 gallons and a final temperature of 118.9 degrees Fahrenheit.

I can send you details, if you wish (preferably in MS Word format). PM me if you are interested.

By the way, I am not a physicist or a physics professor. I am a professor of mechanical engineering. I teach thermodynamics. So although I am not immune to making errors, this kind of calculation is standard for me.

River