Is all water created equally?

January 11, 2007 1:25 PM

PhysicsFilter: A room (at room temp) has two identical cups of water in it. The water in one cup is +10 degrees over room temp, the other is -10 degrees. Left alone, will they reach room temp at the same time?

1a) If the initial temp difference between the two cups was more dramatic, say, a cup of frozen water and a cup of boiling water (both equidistant from room temp) does this change the answer?

1b) If the equilibrium temp is moved to a much hotter room, or much colder room, does this change the answer?


(And no, this isn't homework, I'm just curious)

posted by 777 to Science & Nature (25 answers total)

Newton's Law of Cooling says that the rate of temperature change is proportional to the difference in temperature between two bodies. I'm not physicist, and I realize that this "law" is a simplification, but I take it to mean that neither (a) nor (b) will change the answer of "yes, they will reach room temperature at the same time."
posted by Khalad at 1:32 PM on January 11, 2007

I'm not proofreader, either.
posted by Khalad at 1:32 PM on January 11, 2007

Assuming identical geometry of the samples and complete isolation, I believe the answer is yes. If 1a involves a state change, however (ice -> water), then the answer may be no if the heat transfer coefficient for ice and air is different from the heat transfer coefficient for water and air.
posted by Krrrlson at 1:38 PM on January 11, 2007

No, the warm cup will loose heat from evaporation. Evaporation doesn't help the cold cup heat up. In fact it will actually cool off from evaporation. This is what allows those cloth water bags to cool water even when it's 40C outside.
posted by Mitheral at 1:38 PM on January 11, 2007

Mitheral, the cold cup could pick up heat from condensation of water vapor in the room.
posted by Good Brain at 1:42 PM on January 11, 2007

Convective and conductive heat transfer coefficients depend on the absolute temperatures of the fluids involved (as viscosity is a function of temperature), so yes there will be a slight difference in the rates of temperature change of the two cups. I'll leave it to someone still in mechanical engineering school to look up which one will be faster.
posted by cardboard at 1:44 PM on January 11, 2007

Newton's Law of Cooling is a simplification. The energy content is a function of temperature and pressure. In the same room, equal pressure can be assumed without loss of generality, but the difference between room temperature thermal energy content and each state (corresponding to 10*C difference in either direction) is not necessarily equal. Indeed the modes of heat transfer will be different.

Also, the rate of heat exchange by free convection is dependent on temperature (of the medium, in this case ambient air) as well, via the Rayleigh number (which utilizes temperature dependent quantities such as viscosity and density) and in the immediate vicinity of the cups, this could be different as the colder cup cools the air.

I know from a heat transfer course I took this past term that free convection to a cold plate is half as fast as from a warm plate (or thereabouts, depending on the correlation you use) because of the flow geometries involved.

For all practical purposes with a 10*C difference, the disparity in equilibration time is probably negligible, but over large temperature differences, Newton's law of cooling does not accurately describe matters because the specific heat varies (i.e. in the equation Q=m*c*deltaT, the c term is itself a function of T)

I hope that helps. Email me if you want more info or references.
posted by KevCed at 1:45 PM on January 11, 2007

To clarify: the cold cup will warm up but not as fast as say a bottle of cold water at the same temp. This is because despite a gain temperature from radiation and conduction from the surfaces around the cup the cold cup will also be losing heat by evaporation.

On preview:
Good Brain writes "he cold cup could pick up heat from condensation of water vapor in the room"

Possibly, if the cup was below the dew point.
posted by Mitheral at 1:46 PM on January 11, 2007

Cardboard: Haha, I'm still in school for mechanical engineering. It would be impossible to look up which is faster without a metric butt-load more detail on geometry and conditions. Only then can the monkey work (plug and chug) of cranking through the correlations begin.
posted by KevCed at 1:49 PM on January 11, 2007

I'm with Mithereal, the evaporation could significantly change things. However, these things are actually much more complex than just evaporation and convection. Which freezes faster, warm water or cool water? Sometimes it is the warm water. This is called the Mpemba Effect, and apparently it is still not entirely understood.
posted by caddis at 1:58 PM on January 11, 2007

So the answer is that, yes, in general they'll reach room temperature at the same time, especially if the water is in a closed container so as to remove the evaporative cooling factor and many of the other complexities introduced above.
posted by vacapinta at 2:20 PM on January 11, 2007

I thought hot water froze faster because when water freezes, the crystal structure is less dense than the liquid is (which is what makes it so wierd). When the liquid is hot, the molecules are less dense, and moving more freely, which allows them to arrange into the ice lattice easier than when they're cold, moving slow, and packed too closely. At least that's what they told me in college.
posted by Green Eyed Monster at 3:16 PM on January 11, 2007

As regards the extreme case where the water is frozen, there is what is known as "heat of crystallization", an amount of energy associated with the transition from solid to liquid, which doesn't manifest as a temperature change. Depending on the substance involved, that transition may yield energy or consume it. In the case of water, energy is consumed by the melting process.

What that means is that ice has to absorb some heat from its environment in order to melt, without the absorbed energy manifesting as a significant temperature change. Ice at 0 degrees C is a lower energy state than water at 0 degrees C.
posted by Steven C. Den Beste at 3:25 PM on January 11, 2007

If you look at things from the point of view of entropy, or losing heat, then the hot water loses its heat to the air, and the air loses its heat to the cold water. This doesn't seem symmetrical to me. Suppose the hot water loses its heat to the air more quickly and efficiently than the air loses its heat to the cold water, and it raises the room temperature ever so slightly above the original median in the process. In this case, until final equilibrium is reached, the room temperature is closer to the temperature of the hot water than the cold water. So the hot water reaches equilibrium with the air first (at slightly warmer than the original median), then both cool (with tiny fluctuating versions of the original process) to meet the rising temperature of the cold water. Then you have final equilibrium, which, by definition, is simultaneous.
posted by weapons-grade pandemonium at 3:25 PM on January 11, 2007

They told you wrong, GEM. Hot water has to become cold water before it freezes, with all the attendant properties of cold water, including increased density. It can't go from hot water to ice without becoming cold water in the process (barring any very unusual conditions such as very strong electric fields).
posted by DevilsAdvocate at 3:29 PM on January 11, 2007

Just in passing, there's a similar energy hurdle involved in the transition from liquid to gas. Water at 100 degrees C is a lower energy state than steam at 100 degrees C. Collectively, those two are known as "latent heat", heat which consumed or produced during phase changes in matter.

According to a chart on that page, water consumes 335 joules per gram when it melts, and 2272 joules per gram when it vaporizes.

(For some reason Wikipedia thinks that "heat of crystallization" is called "heat of fusion". Ah, well... And I remembered wrong: the transition from solid to liquid and the transition from liquid to gas always consume energy.)
posted by Steven C. Den Beste at 3:30 PM on January 11, 2007

Man, I could have sworn that I heard that... I wonder how fast the transformation of grey matter to oatmeal is after graduation?
posted by Green Eyed Monster at 4:27 PM on January 11, 2007

Even if they appear to reach equilibrium at the same rate at a macro level, quantum fluctuations in the atomic structure of the individual hydrogen and oxygen atoms will cause them to not arrive at room temperature at the same time.

If you measure to sufficient precision, no two things ever happen at the same time. Similarly, if you round enough, everything happens simultaneously.
posted by Caviar at 9:02 PM on January 11, 2007

GEM - hot water freezing more quickly is a common myth/old wives tale. It gets debunked pretty regularly, but keeps popping up again. So you're probably remembering correctly, but were the victim of some bad information to begin with.
posted by chrisamiller at 9:09 PM on January 11, 2007

The way hot water cools down is the warmer molecules, which move faster (that's why they're warmer), either evaporate or bounce off the cup/air, transferring some of their momentum to the air/cup. The way cool water warms up in the air is the faster air molecules bounce off the cup/water and transfer some of their momentum. But air is much less dense than water, so the latter happens much less often. So it takes more time for the water to warm up than to cool down. Or at least, that's how I would explain it when I taught chemistry.
posted by raf at 9:22 PM on January 11, 2007

Rate of heat conduction to/from the cup will be the same probably, depending on convective patterns within the cup, which will depend on the shape of the cup. Cold cup has rising sides and falling core while warm cup has falling sides, rising core and that's assuming non-turbulent flow. Same effect applies to convective heat transfer to/from the air outside the cup: hot cup has rising air on its walls, cold cup has falling air on the walls. The shape of stuff around the cup could favour one direction over the other.

If the hot water is hot because it was boiled, it may have reduced gas content and increased salt content, changing its specific heat and thermal conductivity to the cup.

The heat gained/lost due to evaporation & condensation will depend on the relative humidity of the air; any change in this could affect the heat transfer in favour of one cup or the other.

Finally, the density (and therefore thermal conductivity) of air at the boundary layer just outside the cup will change due to the heating/cooling effects of the cups. This also interacts with the convective behaviour of the air around the cup, which comes down to the physical shape of the environment around the cups.

In summary: could be anything, depending on the circumstances. The problem is underspecified. My guess though is that the times will be so close to the same you're not going to care; all the differences between the two cases are pretty minor. If one cup is frozen or actually boiling, the answer is pretty much guaranteed to be different rates of thermal transfer.

As for GEM's "hot water freezes faster"; it might be true that boiled but not necessarily very hot water freezes faster due to the reduced gas content. Dunno about that.
posted by polyglot at 10:15 PM on January 11, 2007

I'm not even clear enough about what "temperature" is to know whether , if you pour the cup of hot water into the cup of cold water (let's say they're at +30 and +10 degrees, so both liquid), the result will be +20 degree water. I just don't understand well-enough what temperature is a measurement of to understand whether it would change in such a linear fashion. Maybe this should be a separate AskMe question, or maybe I should just go and read the relevant chapter of a physics textbook. But I think understanding this is probably a prerequisite to predicting what happens in your experiment.
posted by louigi at 1:50 AM on January 12, 2007

louigi: temperature is a measurement of the velocity of particles in a medium. It is proportional to the thermal energy density (kJ per kg) and inversely proportional to the specific heat. For water, specific heat is 4.187 kJ/(kg.K), i.e. it takes 4.187 kJ to heat a 1kg of water by 1 Kelvin.

The important thing is that it is a linear relationship. So if you have equal masses - the volumes will differ since density decreases with temperature - of water at two different temperatures and mix them perfectly, the resulting temperature will be the mean of the two input temperatures.
posted by polyglot at 3:32 AM on January 12, 2007

hot water freezing more quickly is a common myth/old wives tale

Really?
posted by caddis at 6:58 AM on January 12, 2007

i remember is sitting in lab with a styrofoam coffee cup with ice and water in it, and poking a thermometer through the foil on top of the cup to measure when the ice melted and calculating the value of the heat of fusion for water. at the time i also remember thinking the 'experiment' was silly, and the coffee cup was a pretty crappy device to make the water/ice into a closed system...and the whole thing was pointless because when i got a job as a chemist, i would never be doing this anyway... and guess what? i haven't. not once since i was through with that lab. learning the theory behind it was good, but this was one instance when learning the equations was actually more exciting than the related lab experiment.
posted by Green Eyed Monster at 9:02 AM on January 12, 2007