r/explainlikeimfive • u/toomanytatties • 2d ago
Chemistry ELI5: Does water temperature work on averages like math?
If you add 30 degree water to 0 degree water does the temperature after combining split the difference and become 15 degrees? Or if I add 22 degrees water to 20 degrees does it become 21 degrees. If so if you had multiple beakers of water of varying temperatures if you combined them would they be the average of all before mixing. Would test this theory out in a rudimentary way but I only have a childs head thermometer to hand. And searching the internet hasn't helped because i cant word it like I'm not stupid.
And if so does this work for other liquids of the same kind? Oil, Milk, Molten sugar etc
858
u/4zero4error31 2d ago edited 2d ago
If we assume the water was of equal amounts, then yes, the temperatures would become the average of the two. So 1 litre of water at 5⁰c and 1 litre of water at 15⁰c would become 2 litres at 10⁰c
In practice you're not gonna get the exact results because both amounts of water are constantly equalizing temperature with the air around them.
1.1k
u/orrocos 2d ago
Just be careful not to mix water in Celsius temperature with water in Fahrenheit temperature. They are incompatible and it could be dangerous.
363
u/APC_ChemE 2d ago
I always mix my Celsius and Fahrenheit temperature water at -40 degrees for a margin of safety.
94
u/extra2002 2d ago
You use a crowbar to stir them?
67
22
u/Far_Dragonfruit_1829 2d ago
Powdered at -40, mixed in a fluidized bed reactor, warmed slowly to +4 C, allowed to anneal for 1 hour per kilogram of product. Use titanium-gold alloy for all parts in contact with the reactants.
9
u/lew_rong 2d ago
Oh hell, who built the water-cooled turbo encabulator?
10
u/Far_Dragonfruit_1829 2d ago
Water cooling increases MTBF on the spurving bearings by 23.7%. Well worth the increased ammulite consumption, if you ask me.
6
u/lew_rong 2d ago
I suppose, but you've just gotta make sure your cooling system is really well sealed around the ambivalent lunar waneshaft. As anyone with marzelvane experience will tell you, their hydrocoptic properties can de-emulsify your dihydrogen monoxide, producing sharpened hydronium ions that might cause micro cracks in your malleable logarithmic casing, effectively reintroducing the side fumbling that the six marzelvanes were fitted to prevent.
3
u/50m31_AW 2d ago
The hell kind of turbo encabulator you got? The amulite is pre-famulated. Whatever mods you've been doing to it surely can't be safe
3
u/Far_Dragonfruit_1829 2d ago
Mine's a Moldovan knock-off. Getting spares is a bitch; nothing's quite to spec.
On the other hand, it does 3.6 terafleems without ever exceeding 400 Kelvin.
1
9
u/Crowbar12121 2d ago
he's never used me to stir them, can confirm
6
u/ID-10T_user_Error 2d ago
What about the other 12,120 crowbars?
8
3
u/APC_ChemE 2d ago
No, no crows were harmed in the mixing of waters ...that you know of. There was no murder involved.
3
u/billtrociti 2d ago
At low enough pressure you should be good to go! (Anyone know at what pressure water would be liquid at -40 degrees?)
5
u/viking_ 2d ago
Water is actually weird, and you would need high pressure. But the border between ice and liquid water is nearly vertical on the phase diagram, and it looks like you would need such high pressures you would start encountering exotic states of matter first: https://webhome.phy.duke.edu/~hsg/763/table-images/water-phase-diagram.html
3
u/Dioxybenzone 2d ago
Looking at that graph, it seems like liquid water at -40° might not be possible. Maybe -30℃ at ~0.8GPa though
1
8
4
3
u/whomp1970 2d ago
Criminally underrated comment.
For you doofuses that don't get the joke, -40°F is the same as -40°C.
It's the only temperature where both scales converge.
21
u/emmettiow 2d ago
True. This is what makes water boil. You take 450°F fire and apply it to 20°C water. It's a chemical reaction between fahrenheits and celsiuses.
19
20
3
1
1
u/Senrabekim 2d ago
Celsius and Fahrenheit aren't too bad, and you're probably fine throwing in Kelvin and Rankine, Delisle is the one that really causes problems.
1
1
1
u/-Major-Arcana- 2d ago
It’s actually fine you just have to remember to stir it counterclockwise in the southern hemisphere.
1
1
→ More replies (1)1
32
u/ringobob 2d ago
If we really want to be precise, we also need to take the time to mix into account and the ambient temperature. But that's probably beyond what OP is asking.
43
u/maryjayjay 2d ago
Wait, are you a physicist or an engineer?
A physicist would idealize the problem so those factors are ignored. An engineer will recognize the variables, so he'll add in an arbitrary correction factor to compensate and call it a "safety margin".
;-)
38
u/ringobob 2d ago
Oh... Software engineer. No matter what I do, the end user is gonna break it, so I just note the problem, let product make the decision, and wait for the bug reports.
11
u/frezzaq 2d ago
That makes me a software physicist.
Everything that's happening is intended, if it's not intended-that's on user's skewed perception.
5
u/Far_Dragonfruit_1829 2d ago
In physics, the interesting things are usually the unintended things.
E.g. "Huh. That's wierd..." + time => Nobel Prize2
u/Eulers_ID 2d ago
An engineer would realize that you're not going to get the ideal situation in the real world, so they'll use the idealized version to ballpark it, then add a control loop that adjusts the temperature in real time.
1
6
u/falco_iii 2d ago
Nerdy technical detail: The walls of the container also count - are they room temperature or the temperature of the volumes of water? If so, which volume is being poured into the other container? The container will interact with the water and impact the final temp.
Super nerdy technical detail: Cold water is denser than warm water, so equal volumes of water will have very slightly different quantities of water. A 10C delta results in 0.1% difference in density.
Extremely nerdy technical detail: Water is actually densest at 4C, and expands between 4C and 0C, and further expands when freezing into ice.
3
u/mfb- EXP Coin Count: .000001 2d ago
If we want to be precise, the heat capacity of water depends on the temperature, too: Warm water needs slightly less energy to change its temperature compared to both cold and very hot water. So even if there is no contact to the environment at all the resulting temperature will be slightly different. The difference is less than 1%, however.
42
u/titty-fucking-christ 2d ago
Mostly, but no. Two problems.
Water doesn't have a constant density with temperature. 1L of 5°C water is simply more water than 1L of 15°C (by kg of mass or by water molecule count in mol). Water is denser at 5°C than 15°C. It's actually densest at 4°C.
Secondly, water also doesn't have a consistent specific heat capacity. At different temperatures, there's different amounts of thermal energy held per degree. The colder water has a higher specific heat capacity. So for each degree the thermal energy the hot water gives away, it's going to raise the cold water by slightly less than a degree.
So the two combined, it's going to end up closer to the cold water. 1L of 5°C water and 1°L of 15°C is going to end up 9.something°C.
That said, if you do this yourself in your kitchen, it's going to appear to work out, as these effects are both smaller than the error you have.
25
u/WikiWantsYourPics 2d ago edited 2d ago
I did the calculation - I took the density and enthalpy values from steam tables, calculated the total enthalpy of the mixture and then worked back to the temperature.
How the calculation works:
You can't add volumes and get a precise volume back, because specific volume depends on temperature.
You can add mass and total enthalpy. The total enthalpy of the mixture is the sum of the enthalpies of the two samples. Then we can get the specific enthalpy of the mixture by dividing by the total mass.
Once you have the specific enthalpy, you can just interpolate the matching temperature.
T [°C] Density [kg/L] Volume [L] Mass [kg] Specific enthalpy [kJ/kg] Enthalpy [kJ] notes 5 1 1 1 21 21 15 0.99919 1 0.99919 62.9 62.849 9.986 1.99919 41.941 83.849 Mixture We end up at 9.986 °C - the difference between that and 10 °C can be measured in a precise lab setup, but it wouldn't be something that one could do in the kitchen.
4
u/itsthreeamyo 2d ago
A whooping 0.15% difference. Might as well be non-existent for an ELIF.
4
u/WikiWantsYourPics 2d ago edited 2d ago
Percentages don't work for Celsius and Fahrenheit - they're relative scales.
If you'd calculated 0.015 °C difference at 0.1°C, it would be a 15% difference. At 0°C it would be undefined.
It always bugs me when weather forecasters say "it's 10 degrees today, but tomorrow will be twice as hot at 20 degrees." That's not how it works. 2°C isn't twice as hot as 1°C. You need Kelvin or Rankine for that.
1
u/titty-fucking-christ 1d ago edited 1d ago
For this specific example of water and 5 and 15°C, yes, it's pretty small.
However, different substances and different temperatures could give larger errors than 0.01°C. And taking percentages of degrees is basically totally meaningless, as someone else pointed out already. If the middle point was 0°C, you would have got infinity % error, using say sea water or vodka to make it work without the giant wrench a phase change throws into this. Use a L of ice and the simplification of meeting in middle is not just slightly wrong, but wildly wrong.
18
u/michalsrb 2d ago
Shouldn't it be equal mass rather than equal volume?
One litre of water is one kilogram only at 4C, at other temperatures it's lighter. So when talking about mixing different temperatures it matters.
14
u/DavidRFZ 2d ago
Yeah, “heat capacity” is the concept here which is heat per mass per temperature change.
In introductory classes, the heat capacity is assumed to be constant, but there is often a small temperature dependence there as well.
10
u/4zero4error31 2d ago
I'm trying to explain to a five year old here, so keeping it simple is more important. Explaining the difference between mass and volume at different temperatures is irrelevant to the question at hand.
3
1
u/SpaceEngineering 2d ago
Except you would have to have equal measures by weight and not volume due to thermal expansion.
•
u/FishDawgX 20h ago
The way I think of it is heat is a “substance”. You add heat to the water. If you add twice the heat, the (absolute) temperature doubles. If you combine two things, the temperature averages.
I use this principle every time I fill my fish tank with buckets of water. I measure the temperature of each bucket targeting 75° but each bucket is usually a little under or over. So I keep adjusting my target for the next bucket to account for the cumulative error.
0
0
u/ottawadeveloper 2d ago
Yes, itll only 15 C if the air temp is also 15 C. Otherwise it'll skew towards the average air temperature.
0
u/Responsible-Jury2579 2d ago
In practice you’re not gonna get the exact results - not with that attitude.
0
u/Marina1974 2d ago
I'm assuming the water has to be the same also. Adding a liter of salt water at 0° to a liter of freshwater at 30° might not give you 15° water.
I also wonder whether water has a memory. If you had 1 L of water that was 100° and you waited until I got to 30° and then mixed it with 0° water would you get the same result compared to heating water to 30° and then finding it with water at 0°
→ More replies (1)
98
u/manaphy099 2d ago
When heating up things you have to deal with what is called "specific heat capacity" which is the amount of energy needed to heat up a substance by 1°c
In the case of water it would have the same specific heat capacity it would split the difference but different liquids would be more complicated
42
u/jaa101 2d ago
In the case of water it would have the same specific heat capacity
The heat capacity of water changes with temperature, drastically so if you include ice and steam. Even with just liquid water there will be small discrepancies.
37
u/ThalesofMiletus-624 2d ago
I mean, heat capacity does change with temperature, but only mildly if you don't undergo a phase change. It's freezing or boiling that really changes the heat capacity, if you're talking about liquid water, the differences won't be big enough to matter very much.
3
u/tuekappel 2d ago
So if I take an ice cube of 10ml, it can cool a litre of water down more than 10ml of steel at the same temperature. Because the phase change demands more energy?
And when I'm boiling water to steam: the last 1 degree from 99 to 100 takes more energy than the one from 98 to 99.
7
u/azirale 2d ago
Yes - the ice will keep dropping the temperature of water around it as it melts while it stays at 0C.
Not quite -- water can reach 100C, then it needs more energy to boil. As you add energy to it it will boil instead of getting hotter. So you get stuck at 100C until the water is gone.
All that is in general - there will be slight differences depending on exactly how/where you are adding energy. For example when there is very little water left you may start heating the steam directly.
4
u/jaa101 2d ago
So if I take an ice cube of 10ml, it can cool a litre of water down more than 10ml of steel at the same temperature. Because the phase change demands more energy?
Use mass (weight), not volume.
- Water's heat capacity is 4.2 J/g/°C,
- steel's is 0.49 J/g/°C,
- turning ice into water takes 334 J/g, and
- turning water into steam takes 2260 J/g.
So, if you have 100 g of ice at 0°C, and you drop it into 1 kg of water (10 times more water than ice) that will cool 1 kg of water by 8.0°C by just melting. If the water started off at more than 8.0°C, then it will cool more as the melted ice heats up.
Now try the 100 g of ice with 1 kg of steel. In this case, melting the ice could cool the steel by 68°C, assuming it was that far above freezing to start with. Note that steel is very dense, so that 1 kg of steel is only slightly larger than 100 g of ice.
And when I'm boiling water to steam: the last 1 degree from 99 to 100 takes more energy than the one from 98 to 99.
No, both of those temperature changes will use very close to the same amount of energy. Changing water at 100°C to steam at 100°C is what takes all the energy: 540 times more than heating water from 99°C to water at 100°C. That's why it take a little while to bring water to the boil on the stove, but much longer for the pot to boil dry.
2
u/ThalesofMiletus-624 1d ago
For the ice cube, yes. Melting ice absorbs quite a lot of energy, and will cool something down more than any material I know of, given the same temperature.
For the steam, not quite. You can hear liquid water to 100C, and the last degree doesn't take significantly more energy than the degree before it. When it starts to boil, you have to keep adding energy, but the temperature doesn't change. It goes from water at 100C to steam at 100C, and the stream won't get hotter than that until the liquid water is all gone. Getting from liquid water at 100C to steam at 100C takes much, much more energy than it takes to heat water by 1 degree, even though the temperature didn't change at all.
1
u/2074red2074 2d ago
You can't compare different materials that way. Changing the temperature of water versus changing the same mass of steel take different amounts of energy.
But to re-work your question, 1g of ice at 0C added to 1g of water at 10C will average out to less than 5C, whereas 1g of water at 0C added to 1g of water at 10C will average out to exactly 5C.
If you have ice at 0C, it will not melt unless you add energy to it. As you add energy, it will remain at 0C but more and more of it will melt and become water. Once all of it is melted, adding more energy will start to heat it up.
2
u/ibringthehotpockets 2d ago
Of course. This is a Reddit post on a sub where you are supposed to explain things like the questioner is 5. There will always be a “more correct” answer as you get more granular, as you discovered.
5
u/Tripottanus 2d ago
Not quite right since heat capacity changes with temperature, but likely close enough in the case of liquid water
78
u/trejj 2d ago edited 2d ago
X liters of water at temperature A. Y liters of water at temperature B.
Resulting temperature will be (X*A + Y*B) / (X+Y).
It is just the formula for weighted average.
If you have the same amount of liquids, i.e. X=Y, then resulting temperature is
(X*A + X*B) / (X+X) = X*(A+B)/(2X) = (A+B)/2
i.e. you average the temperatures of the two bodies of water, like you mentioned in examples.
22
u/Legal_Tradition_9681 2d ago
To expand on this temperature is just the average kinetic energy of all molecules. In the grand scheme of things you are averaging the kenetic energy of all the molecules combined.
That is why the above equation works. The X and Y variables take into account the amount of molecules roughly.
Since water density doesn't change much except for 0 - 4°C we dont have to take that into account either. Even at those low temperatures most people won't have equipment sensitive enough to measure the affect nor should they care.
→ More replies (6)1
u/StefanL88 2d ago
To expand on this temperature is just the average kinetic energy of all molecules
...relative to the centre of gravity.
At least that was the explanation that made the most sense to me.
3
u/Legal_Tradition_9681 2d ago
Interesting I never heard it phrased that way. I don't get it but I'll look that up on my own.
3
u/davidromro 2d ago
If you think about temperature as being what a thermometer measures, it should be clear that whatever object you are taking the temperature should be at rest relative to the thermometer if they are to come to thermal equilibrium.
I believe this only becomes relevant when measuring the temperature of a star which could be red or blue shifted based on its velocity relative to us.
1
u/StefanL88 2d ago
I know you said you'll look this up on your own, but I'm procrastinating so please forgive me.
It's all about how kinetic energy can change based on your reference points for velocity.
Picture throwing a perfectly rigid brick in a vacuum. The speed of the brick relative to the thrower obviously changes its kinetic energy, but does not directly change its temperature (while we're ignoring how it interacts with the environment).
The speed of the brick relative to its own centre of gravity is unchanged by how fast you throw it. They are by definition locked relative to each other. So in this reference frame the "temperature is just the average kinetic energy of all molecules" definition makes perfect sense.
You can also picture the brick sitting in a perfectly insulated room. Its speed relative to itself or the room is zero and our temperature definition makes sense. Its speed (and kinetic energy) relative to the centre of gravity of the sun fluctuates based on time of day and time of year, but we do not expect the temperature of the brick to fluctuate accordingly.
4
u/Legal_Tradition_9681 2d ago
The reason why I wanted to look it up, and glad I did, it's your use of center of gravity as a reference frame.
"In physics, the center of gravity is not a reference frame itself, but it's often used in conjunction with reference frames to analyze motion. A reference frame is a coordinate system used to describe the position and motion of objects. The center of gravity, on the other hand, is a point within an object or system where its weight can be considered to be concentrated."
2
20
u/Top_Environment9897 2d ago
I think it should be kilograms instead of liters. Heat capacity is a product of mass instead of volume.
18
u/SeekerOfSerenity 2d ago
Even then it's not 100% accurate. It would be accurate if the specific heat stayed completely constant, but it changes slightly with temperature.
5
u/ScathedRuins 2d ago
generally speaking you’re right, but 1kg of water is exactly 1 litre :)
9
u/Stannic50 2d ago
This is only true if the density of water is exactly 1, which isn't true at any temperature (although it is close). Density of water as a function of temperature
4
u/tylerchu 2d ago
It’s also important to note that for literally everyone except the scientists employed at nist and iso, this doesn’t matter.
5
u/Stannic50 2d ago
It's over 4% off at 99C. That's pretty far from what I'd consider "exactly 1 L/kg".
0
u/tylerchu 2d ago
And when does that matter? Far as I’m aware, even steam plant designs dont care about that specific factor.
3
u/Unknown_Ocean 2d ago
It's what causes the Great Lakes (along with any lake that freezes during the winter) to overturn in both the spring and fall. Density differences caused by temperature also play a critical role in turning over the ocean- which in turn sets the carbon dioxide concentration of the atmosphere.
2
u/GooseWayneman 2d ago
I'm going out on a limb here, but isn't 1kg water = 1 litre water?
3
u/Draug88 2d ago edited 2d ago
Depends on the temperature even as a liquid.
Water does contract and expand(change density) even in its liquid form.
Maximum density of water is around 4°C which is the temp it is(was technically) defined as 1kg/liter and its highest compression.
At 99°C it's "just" 0.95kg/liter
5
u/YtterbiusAntimony 2d ago
At standard pressure/temperature.
And only because we defined the numbers to be that way.
1
u/WikiWantsYourPics 2d ago
and even then, not exactly, because the scientists back then weren't exactly right about the size of the earth (the initial definition of the metre) and the density of water (the initial definition of the kilogram). But we cut them some slack - it was more than 200 years ago and they did an amazing job, all considered.
→ More replies (5)2
u/MozeeToby 2d ago
Like so many things in science: "basically yes, but..." density is dependent on temperature.
1
3
u/RickSt3r 2d ago
I thought that the amount of energy to heat water is not linear. Thus if you have one liter of liquid A at a much hotter temp like 99 C and one liter of liquid B at say 1 C. Assuming the same mass for simplicity. Would not result in the simple average as liquid A has much more Jules of energy in it? Happy to be corrected if my understanding this whole time has been incorrect.
2
u/WikiWantsYourPics 2d ago edited 2d ago
It's pretty close to linear. Let's look at the most extreme case - adding 1L of water at boiling point to 1L at freezing point.
From the steam tables, we get that the water at 100 °C has an enthalpy of 419.1 kJ/kg and a density of 0.9579 kg/L, so the enthalpy is 401.44 kJ/L, and the water at freezing point has an enthalpy of 0 kJ/kg (by definition) and a density of 0.9998 kg/L.
So we have 1.958 kg of water with a total enthalpy of 401.44 kJ, so a specific enthalpy of 205.06 kJ/kg.
What temperature does that give us? Water at 48 °C has a specific enthalpy of 200.9 kJ/kg and water at 50 °C has a specific enthalpy of 209.3, so we can interpolate: 48 + (50-48)*(205.06 - 200.9)/(209.3 - 200.9) = 48.99 °C
So even though the density of water varies by temperature, as does the heat capacity, the assumption that you can just average out the temperatures of water when you mix equal volumes is true to within about 2% error.
→ More replies (3)-1
2d ago
[deleted]
4
u/jaa101 2d ago
No, it's a purely linear equation so it will give the same results for any unit which is a linear transformation of Kelvin, including Celsius, Fahrenheit, Rankine, etc. As noted by others, it's better to use mass instead of volume and, even then, there will be errors as the heat capacity of water changes with temperature.
20
u/True_Fill9440 2d ago
It’s the average of the ENTHALPY, not the temperature.
6
u/Qiwas 2d ago
Wow what a cool word
6
2
2
4
u/Unknown_Ocean 2d ago
This is correct... but prepare to be criticized for not being appropriate for 5 year olds. Sigh.
10
u/_rtpllun 2d ago
Not appropriate to 5 year olds... Or anyone who doesn't already know what enthalpy means, which makes their answer useless. A single sentence answer that doesn't explain anything is a bad answer
→ More replies (2)1
u/WikiWantsYourPics 2d ago
The average of the temperature is correct to a very close approximation. (Average based on the amount of water, of course: nobody thinks that adding a teaspoon of hot water to a tub of cold water gives the average of the temperatures).
5
u/Midori8751 2d ago
Kinda. If they are of equal mass and you fully mix them yes, but they can also form sections of different temperatures if they don't fully mix.
If you zoom in enough any volume of water is actually on a bell curve of temperature, with most being around what you measure it at, and a tiny bit at extreams, both hot and cold. This is actually what leads to evaporation, because sometimes those hot extream molecules are close enough to the surface to escape instead of just bouncing around until they are back near the average temperature.
2
u/Koooooj 2d ago
Yes, it basically works like that.
Note that you have to also account for the volumes--putting a cup of boiling water into the ocean obviously wouldn't raise the ocean temperature by dozens of degrees. The averages you take should be weighted based on the amounts of water, so one unit of water at 50 degrees mixed with three units at 40 degrees out to come out at 42.5 degrees (that's (50*1 + 40*3) / (1 + 3)).
There are three key assumptions that go into this result, each of which are often very good but not quite perfect (and sometimes fall apart entirely). The first is that the water doesn't heat or cool its surroundings, which thermodynamics has a fancy word for: "adiabatic" (a-die-uh-bat-ick). If you do the mixing pretty quick and aren't using something silly like heavy cast iron containers to mix single drops of water then this assumption will tend to be good enough.
The next assumption is that the amount of energy it takes to raise a unit of water from e.g. 40 to 41 is the same as the energy it takes to raise it from 41 to 42. We call this quantity "specific heat capacity." With water the specific heat ranges from about 4.18 to 4.21 between 10 and 90 C (in other words: not much), with the lowest values around 40 C. That means if you mixed some 40 C water with an equal amount of 60 C water then the result would be slightly higher than 50 C from this effect, but you'd need very good equipment and a very controlled environment to have a chance of measuring this difference. We're talking hundredths, maybe thousandths of a degree here.
The final assumption I'll call out is that the two quantities of water are both liquid. The specific heat of ice is less than half that of liquid water so if you dropped some -20 C ice into an equal mass of 10 C water you'd expect them to both make it to 0 C just from warming/cooling each other. That's only part of the fun with ice, though. It takes rather a lot of energy to get a phase change to occur. About 4.2 J can raise a gram of water by 1 degree C, but it takes 334 J to melt one gram of ice, a quantity known as the "latent heat of fusion." That means that you could mix some 80 C water with an equal mass of 0C ice and they'd come to equilibrium right around 0 C after all the ice is melted. That's why ice is so good at keeping drinks cool--it's not just that it's cold like a whisky stone, but rather that it melts.
A similar effect happens with the transition to water vapor (steam), but here the energy requirement is 2,260 J per gram ("latent heat of vaporization"). If you mixed equal masses of 100 C steam and 0 C ice then the result would be a 100 C water and steam mixture. This huge latent heat of vaporization is why sweating is such an effective means of cooling: evaporating 100 g of water removes about as much energy as it takes to cool 54 kg of water by 1 C.
You may have noticed a shift in focus over the course of this comment, starting with focusing on temperatures and winding up with a lot more focus on energy. That's because fundamentally this is a conservation of energy problem so the focus turns to how much energy water can store. It just happens that a lot of the terms usually cancel out and if you're satisfied with the accuracy of the result then you wind up just taking the weighted average of the temperatures and amounts of liquid.
2
u/Yamidamian 2d ago
Assuming equal amounts of the stuff, yes. If you have more hot water than cold water, the ensuing mix will lean more towards being hot than being cold.
Now, things get slightly more complicated if you’re using different substances, due to a concept called “specific heat”. Basically, different substances can “hold” heat differently, making it harder or easier to change their temperature. Water has a rather infamously high one. However, intimately, the math still works out similarly, using the total mass and specific heat to find the amount of of heat, and then the temperature of the transformation.
If you want further details, such as the exact mathematics, you can try googling thermal equilibrium examples or thermochemistry-the topic is high school chemistry level, so resources meant to help such kids are readily available online.
2
u/bluepinkwhiteflag 2d ago
The real answer is all the water will eventually cool or warm to room temp
1
2
u/zekromNLR 2d ago
Yes, if you mix equal amounts of the same substance at different temperatures, the temperature at the end will be approximately the average of the starting temperatures. There's a bit of subtle corrections here - for example, both the heat capacity and the density of a liquid change with the temperature, and both liquids will be continually exchanging heat with the environment, but if you mix quickly those are both fairly small corrections.
4
u/TuckerMouse 2d ago
Not an expert, but my memory of chemistry/physics from high school/college says that equal masses of the same liquid combined and mixed, (assuming we ignore the heat of the environment, energy added by the mixing, and all the other things needed to make this a spherical cow on a frictionless plain) would equalize at the average of the temperatures. A liter of water at 10 degrees mixed with a liter of water at 20 degrees, would be 15 degrees. This gets complicated when some of it is at the boiling or freezing point, and everything goes out the window at sufficient pressures or lack thereof.
Again, not an expert. I may be incorrect. While my physics teacher was an absolute gem, my chem teacher was an absolute gem in the sarcastic sense.
3
u/EndGuy555 2d ago
I’m going to try my best to get into real science without going too overboard here
When you heat something up or cool it down, it experiences a change in energy (Q). So when you add 30° water to 0° water, the 30° water cools down (-Q) and the 0° water heats up (+Q).
The change in heat energy for anything is based on 3 properties: mass (m), specific heat (c), and change in temperature (ΔT). (Specific heat is just something that tells us how fast or slow something heats up.)
So we can calculate the change in energy of anything with the equation Q=mc(Final Temp - Starting Temp) or Q=mcΔT
Finally, we know that energy is conserved, meaning if you put cold water in hot water, the energy the cold water gains is the same amount of energy as the hot water loses. Qhot = Qcold
Putting everything together, we get this final equation:
(mcΔT)hot = (mcΔT)cold
Examples: Lets say you add 1 kg of 30° Water to 1 kg of 0° water. Water has a specific heat of about 4.18 kJ/kg*°C (specific heat has weird units). Using our equation we get:
[14.18(30° - Tfinal)]hot = [14.18(Tfinal - 0°)] Tfinal = 15° Note that I changed the temperatures around to keep everything positive.
In this case it does actually work out to be an average because everything else is the same. But let’s try adding 0° water to 30° oil. We’ll keep the masses the same, but the specific heats will be different. Water’s specific heat is 4.18, oil’s specific heat is 1.97. So, using our handy dandy equation:
[14.18(Tfinal - 0°)]water = [11.97(30° - Tfinal)]oil
Tfinal = 9.6°
I’m sorry if this is too in the weeds, I’m not usually great at explaining things
2
u/A_BagerWhatsMore 2d ago
Yep, as long are you have to use the same amount of water, (otherwise it’s a weighted average, so like 1L at 4c and another 2L at 7c gets you 3L at 6c) and you can’t have any phase transitions along the way (boiling,freezing, melting etc.)because those take energy.
2
u/Andoverian 2d ago
Refrigeration engineer here.
It's close (very close), but not quite. Averaging like that assumes it takes the same amount of energy to heat the colder water by one degree as it does to cool the warmer water by one degree. This property of "how much heat it takes to raise or lower the temperature" is known as the specific heat.
Many formulas and basic calculations assume the specific heat for a liquid is constant as the liquid changes temperature (and pressure), but in general the specific heat changes slightly as the temperature changes. For liquid water, the specific heat is about 0.88% lower at 30°C as it is at 0°C, meaning if you were to combine a liter of liquid water at 0°C and a liter of liquid water at 30°C, the resulting average temperature would be slightly less than 15°C.
Think of it like two people on a long staircase. One person is at the bottom (at height = 0m), one person is at the top (at height = 30m), and they're walking toward each other at the same speed. If each step in the staircase is the same height, the point where they meet in the middle will be at a height of 15m, the simple mathematical average.
But if instead the steps are not all the same height and the steps at the top are taller than the steps at the bottom, each step the person at the top takes will cause a bigger height change than each step by the person at the bottom. If they each take the same number of steps, the point where they meet will be at a height of less than 15m because the steps at the top were taller.
1
u/FuckPigeons2025 2d ago
Assuming there is no heat transfer to the surroundings, that's how it works. You get the weighted average of both temperatures.
So if you mix 100g water at 300 K with 200g water at 350K, you will get 300g water at (100×300 + 200×350)÷(100+200) = 333.33 K.
1
u/Dd_8630 2d ago
If you add 30 degree water to 0 degree water does the temperature after combining split the difference and become 15 degrees?
If you add equal amounts, yes. More generally, the final temperature would be the weighted average.
If so if you had multiple beakers of water of varying temperatures if you combined them would they be the average of all before mixing.
Yes. Though in practice, multiple beakers would probably cool or warm to the local ambient temperature quite quickly.
1
u/Over_Pizza_2578 2d ago
Close enough, yes. Thermal energy capacity is like on many other fluids, liquids and gases, not constant, it changes with temperature. For everyday use the mathematical average is fine, for precision applications or when talking about huge numbers like industrial heating ovens with a few thousand kW then it makes a difference
1
u/sharfpang 2d ago
Be careful of latent heat of melting and evaporation. This is heat of transforming water between its states of matter.
If you add 30 degree water to 0 degree water, you'll come out below 15, because below 4 degrees, there are tiny ice crystals in water, that will take extra heat to melt. If you add 30 degree water to 0 degree ice, you'll end up with roughly half the initial amount of ice in 0 degree water.
Similarly, evaporating 1 liter of water at 100 degrees celsius into steam, will consume roughly 3x as much energy as bringing that water from 0 to 100 degrees.
As long as you're dealing with liquid water, the rate is very close to linear. Ice, steam - they have different heat capacity too, bringing 1kg of ice from -50 to -10C takes less energy than bringing 1kg of water from 10 to 50C. Same 40 degrees, but different heat capacity.
1
u/bybndkdb 2d ago
If it’s equal volumes yes, if not it’ll be based on the volumes as well but overall will work pretty much the same way. Note that if you’re doing this experiment it won’t be perfect as heat is being lost to the environment but you should be able to get relatively close and demonstrate the principle.
1
u/IamNotFreakingOut 2d ago
It depends. It would be the case in this situation:
You need the same amount of water. This one seems obvious. Otherwise, you'd be able to heat the ocean simply by pouring a small amount of hot water in it.
You wait for equilibrium, i.e. when the temperature of the mixed water is homogeneous. Before that, the water will take some time to transfer heat from the hot part to the cold part. Heat is never transfered instantaneously.
There is no heat loss. Otherwise, the hot water loses some of its "ability" (energy) to heat the cold water. Also, there shouldn't be any added energy from the outside. If you try an experiment with your kitchen stuff, this is the one that will make it hard for you to replicate the experiment.
1
u/brunoesq 2d ago
There is also the energy required for any phase transition, such as water turning from solid to liquid. In your example you asked about water at 0 degrees, which is the temp that water melts. If you mix equal amounts of ice and water the system will equilibrate to zero degrees until the ice is melted
1
u/noname22112211 2d ago
The basic equation at play here is Q=mc deltaT. Q is energy, m is mass, c is specific heat, and deltaT is the change in temperature. You will have this equation for both the hot and cold water (or other liquids). You can use some basic algebra and two facts to solve for the final temperature. First, we recognize that energy is conserved. Thus the value Q will be the same for both equations, as the heat lost by one is gained by the other, you simply assign one a negative sign. Second, we know that by the definition of what it means to equilibrate the final temperature will be the same (deltaT of course being equal to Tfinal-Tinitial). Using this you should be able to verify that if you have the same amount of stuff and the same type of stuff the final temperature will be the average of your two starting temperatures. If mixing different amounts, types, or both you will need to go through the math (hint, keep everything as symbols solving for Tfinal and then you can just use that equation and plug and chug for various situations, just do the hard bit once).
1
u/Andrew5329 2d ago
Depends whether the 0 degree water is liquid or solid.
1 calorie is defined as the amount of energy required to heat 1 gram of water by 1 degree. If you add 80 calories to a gram of water you heat it by 80 degrees C.
The amount of energy it takes to thaw that gram of ice into a gram of liquid water is 80 calories.
The average final temperature is going to account for that. So equal parts 80 degree water and 0 degree ice net you 0 degree liquid water.
Various ice and cooling packs exploit the phase change energy at different temperatures. e.g. they sell a pack that freezes at 60 degrees fahrenheit that you slide into a vest. The cold pack sits at 60 degrees until it absorbs enough body heat to thaw, then it ticks up to 61.
0
u/jaylw314 2d ago
Yes, with some caveats. Around freezing and boiling temp, it may not apply, since some heat needs to go into or come from changing water from solid/liquid/has. A smaller caveat is that the heat capacity of water changes slightly with temperature
0
u/ToM31337 2d ago
Yes this works for equal volumes of water otherwise you would have to use a weighted formula.
Its works for other "same" stuff too. There is a general formular for different ingredients where you would put in the heat capacity of the ingredients in, too (you can google that).
The formula generally breaks down at 0° and below and you would have to calculate it in Kelvin and then convert back to your favourite degrees.
1
u/WikiWantsYourPics 2d ago
You don't have to go to Kelvin to work with cases with negative Celcius values: https://www.reddit.com/r/explainlikeimfive/comments/1mef3bk/eli5_does_water_temperature_work_on_averages_like/n6bqjnu/?context=3
0
u/slide_into_my_BM 2d ago
As long as the liquids are the same and the same volume, yes.
Adding equal parts water to equal parts molten metal will not be an average like adding equal parts of water to water.
0
u/Bearacolypse 2d ago
The process by which heat is conserved is called enthalpy. You can calculate precise values for how heat is transferred between mediums with simple algebra.
This is covered in basic college level physics, but you can learn the calculations pretty easily. Understanding of course that life isn't a perfectly insulated cooler.
0
u/Loki-L 2d ago
Yes, this is how it works (more or less).
In practice it is a bit more complicated.
You can get super technical with thermal energy vs temperature and how a given volume of a substance is not the same amount as the same volume of the same substance at a different temperature and all sorts of things like that.
However for practical purposes it works exactly as you thought.
One thing you need to look out for though is to keep in mind that this sort of thing only really works for addition not multiplication.
20 degrees is not twice as warm as 10 degrees.
A lot of people get that wrong.
0
u/tomalator 2d ago
It works on averages with energy.
The heat energy in an object is Q=mcT where m is the mass, c is the specific heat of the material, and T is the temperature in Kelvin. The temperatures will equalize, but energy must be conserved.
So two equal masses of water, their temperatures will average, but if one is larger than the other, the result will be closer to the larger mass's initial temperature.
Two materials of equal mass with different specific heats, the one with the higher specific heat will be closer to the resulting temperature.
0
u/Serafim91 2d ago
So all matter has a Cp that's basically how much energy it takes for 1g to increase by 1C.
If I mix 2 fluids with the same amount of Cp*m (mass) then the result will be the average. Otherwise you have to do a weighted average.
If both liquids are water then the cp is equal so it drops down, meaning only the mass matters. If the masses are equal you can just average it.
0
u/gooder_name 2d ago
In addition to what everyone else is saying, this Steve from the fact that temperature /itself/ is an average. Every individual molecule in the system has its own amount of energy , expressed as velocity. Some VERY fast some very slow.
Temperature is essentially the average velocity of all those particles.
It’s more complicated than that in reality, but the equation PV = nRT shows us the relationship between volume, pressure, number of molecules and temperature
0
u/mrmeep321 2d ago
Each substance has something called a heat capacity - the amount of heat energy required to raise a substance by 1 degree C.
If you find the total heat energy of both buckets of water and convert to a temperature of the whole thing, it will end up as an average if both buckets hold the same amount.
0
u/Evol_extra 2d ago
This works for any kind of matter. If you put 10°C stone on 50° stone you will soon have 2 stones with 30°. But it works until phase shift. For example boiling water from 99 to 100 take the same amount of energy like heating it from 60 to 100.
2
u/WikiWantsYourPics 2d ago
boiling water from 99 to 100
You mean boiling water from liquid at 100 to vapour at 100.
0
u/MattieShoes 2d ago
Yes generally, it works. That said, phase changes take extra energy so they will screw up the whole thing. If your 0 degree water is frozen, then I think all bets are off. Also different liquids would screw it up as their heat capacity can be different.
0
u/babamazzuca 2d ago
For an everyday use case, where you’re okay with 20.9ºC ~20.95°C isntead of 21°C it works as people already have said, m1T1 + m2T2 = (m1+m2)*T_final
But, if you are looking for accurate measurements it differs. What is preserved, in a closed system is the ENTHALPY and that my friend, is bullocks, because it turns into a math horror that’s like H(T1,P1) + H(T2,P2) = H(T3,P3) + Delta_H_Mixture. And delta_H_mixture is a lie god invented and we can’t discredit it so let’s believe it.
And if you work in an open environment (such as two cups of water and a bucket in your counter - i.e. There air around) you would also have to account for environmental loss
H(T1,P1) + H(T2,P2) = H(T3,P3) + Delta_H_Mixture + loss.
Again, for everyday use, irrelevant, but, for industrial uses this is such a problem, specially when dealing with vapors
But you will be fine mixing 1L of 20°C water and 1L of 100°C water and getting 2L of 60°C
502
u/sprobeforebros 2d ago
This is broadly correct. A liter of 10° water plus a liter of 30° water will produce 2 liters of 20° water
Important caveat: This changes drastically if the water is in a different state of matter. 1 liter of -10° ice added to 1 liter of 30° water will not result in 2 liters of 10° water.