In our test there was a question that went like so:

Question 4 You have a glass of iced water on an unshaded picnic table and went for a walk for 30 minutes. When you return you noticed the glass has water on the outside of it.

a. In terms of heat transfer explain what has happened to the glass of water.

The majority of the class understood this question and answered it correctly.

b. Would there have been more or less water on the outside of the glass if the picnic table was in the shade? Explain.

This question caused a lot of controversy with the majority of the students (including myself) believing that the shade would have caused more water on the outside. Whereas the teacher and a few students thought that the sun would have caused more water on the outside.

The reasoning that the teacher provided was not very convincing and so we have come to this forum to ask what is the correct answer to part b and most of all WHY?

We are 16 -17 years of age if you need to know the level for the explanation.

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    $\begingroup$ The question is ambiguous. Was all the ice melted in the "sunny" glass upon your return? Was there still some ice in the "shady" glass upon your return? The answer is going to be somewhat dependent on how much ice you started with, the ambient temperature and relative humidity of the air around the glass, the time of day (was sunlight striking the glass at noon or dusk?), the surface area of the glass, the wind conditions (i.e., whether or not forced convection or natural convection was involved), etc. $\endgroup$ – David White Feb 22 '17 at 3:53
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    $\begingroup$ @DavidWhite Yes I think this is contributing to the why their is such a divide for the answer $\endgroup$ – class_question Feb 22 '17 at 4:03
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    $\begingroup$ as a side note, if you want to get a quality education you are going to need to get comfortable with people around you being wrong. perhaps that is the lesson your physics teacher is teaching you today. $\endgroup$ – james turner Feb 22 '17 at 18:43
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    $\begingroup$ @Dschoni You won't get condensation at all in that case. The fact of condensation proves that the air temperature exceeds the temperature of the glass. $\endgroup$ – dmckee Feb 22 '17 at 22:54
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    $\begingroup$ There is an easy way to settle the question. Obtain a glass of water, some ice, and a picnic table. Get out of the classroom and do some experiments! $\endgroup$ – Eric Lippert Feb 23 '17 at 0:01

10 Answers 10


Obviously it is the water vapor in the air that condensed onto the surface of the glass, and the ice inside the glass played the role of maintaining the glass-surface at a certain temperature. I am now going to simplify the problem to its essentials.

I don't know what material the glass had, but say it was made of some highly conducting material, so that we may assume that the temperature was uniform over the entire glass (outer) surface. Also since it is ice inside the glass, it is reasonable to assume that temperature inside the glass is constant over time, with heat flux (from ambient to glass) adjusting itself to maintain this temperature. Also we shall assume that steady state has been reached. Let us also assume that glass is closed at its top, although this is not a serious assumption and may be relaxed. We shall neglect effect of winds and consequent evaporation. We shall assume that water vapor pressure in the air remains constant.

So here's the simplified problem: given two closed hollow objects, both filled with ice so that temperature at inside wall of the object ($T_{inside}$) is constant, and one is kept in the shade while the other is in the sun, which will have greater rate of condensate formation (at its outer surface) after steady state has been reached?

Heat flux to glass in the shade is $Q_{shade}$, and that to the glass in the sun is $Q_{sun}$, with $Q_{shade}<Q_{sun}$. Let $T_{shade}$ and $T_{sun}$ be the temperature of the outer surface of the glass in the shade and in the sun respectively; both must of course be higher than $T_{inside}$. Then since $Q_{shade}<Q_{sun}$ we must have $T_{shade}<T_{sun}$ (why? hint: heat transfer through wall depends on temperature difference across the wall). Greater rate of condensate formation occurs on the cooler surface, which is that in the shade.

P.S. I will briefly explain why condensation rate must be higher for a colder surface. Condensation of water vapor on to a surface is a result of the fact that more number of water molecules are being deposited on the surface from air, than are being lost to air. The rate at which deposition takes place depends on water vapor pressure in air, which we have assumed to be constant and therefore is the same for both glasses. However the rate at which water molecules deposited on the glass are lost to air depends on the temperature of the surface, and lower the temperature lower is this net outward flux (see here).

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    $\begingroup$ "I don't know what material the glass had" I'd say it was probably made out of... glass. $\endgroup$ – David Richerby Feb 22 '17 at 13:43
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    $\begingroup$ @DavidRicherby Which one? $\endgroup$ – Sabre Feb 22 '17 at 13:58
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    $\begingroup$ @Sabre Are there any that would fit the description of a "highly conducting material" given later in the same sentence? If not, it doesn't really matter. $\endgroup$ – David Richerby Feb 22 '17 at 14:02
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    $\begingroup$ You are assuming the vapor pressure of water is constant but are you also assuming it is uniform? If the air in the sun is at 100% relative humidity, then either the air in the shade is supersaturated or it has less vapor pressure. In practice I think it will be supersaturated, but if it did have a lower vapor pressure that would lead to a decrease in condensation and even cause more condensation in the sun in the limit of very conductive glass. $\endgroup$ – Brian Moths Feb 22 '17 at 15:53
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    $\begingroup$ @DavidRicherby You are partially right in saying that glass should be made of...well...glass, unless "glass" is used as a more generic word for a utensil used to drink water from, in which case a "glass" could be made of stainless steel or brass or whatever. You are not to be blamed for not knowing this of course, because this queer usage of word "glass" exists in my country and perhaps only here :-) $\endgroup$ – Deep Feb 23 '17 at 4:53

The best thing for your class to do would be to do an experiment. You can think about how best to do this, but a few suggestions are:

  • Use several glasses of water. For example, put five glasses in the sun and five in the shade.
  • Do your measurements on the glasses in the sun and glasses in the shade at the same time.

We can also try to predict the outcome with a rough argument. In my argument, I make the simplification that at the end of the 30 minutes there is still ice left in the glass. This is a useful simplification because it means that the water in the glasses will remain at a constant temperature throughout the experiment. I will also imagine a somewhat more precise description of the experiment. I'll say that we have a picnic table in the sun and an umbrella covers half of the table in shade.

Then, we can think about what the variables are that control the rate at which condensation forms on the glass. Some candidates are:

  • The temperature of the surrounding air
  • The temperature of water in the glass
  • The humidity of the surrounding air

We can notice that the temperature of water in the glass is the same in both cases, since they both have ice in them. The humidity won't differ much between sun and shade. This is because the water molecules in the air are moving around randomly, and this makes them tend to distribute themselves evenly. In fact, the air temperature in the sun and shade are also equal. The random collisions in the air tends to smooth out the distribution of heat in the air. Since we have chosen our two glasses to be near each other (on the same picnic table), the air temperature and humidity will be almost the same in the two cases.

We see that all three variables which control the rate of condensation forming are equal in the two cases. Thus I would predict that in the experiment you would not be able to notice a significant difference in the amount of condensation in the shaded and un-shaded glasses.

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    $\begingroup$ This answer ignores the effect of the temperature difference across the inner and outer surface of the glass, as discussed in "Deep"s answer. Considering that effect I agree with his response. I'd be interested to see some numbers put into Deeps argument though, I suspect the effect would be too small to notice by eye. $\endgroup$ – Alex May Feb 22 '17 at 6:48
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    $\begingroup$ +1 for suggesting an experiment. The basis of knowledge is observation. $\endgroup$ – Neal Feb 22 '17 at 15:05
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    $\begingroup$ I just did the experiment and the shady side (barely) had more condensed water. $\endgroup$ – bobuhito Feb 22 '17 at 20:15
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    $\begingroup$ I suspect that in the sun, the condensate has some slightly higher rate of re-evaporation due to absorbing additional energy from the sunlight. I'd be interested to know if you see a differential with a glass half in and half out of the sun... $\endgroup$ – Zimul8r Feb 23 '17 at 3:53

Consider the possibility that there is no "correct" answer.

Maybe the purpose of the question was just to get you to think about and discuss a physically interesting situation.

Condensation will depend on at least the following factors

  • Air humidity
  • Air temperature
  • Air circulation
  • Water temperature

It is possible for direct sun light to affect all of these factors and the effects on condensation will work in different directions. Which effect dominates may depend on several factors not specified in the question. Even if there is a "correct" answer in most situations, it may not be possible to confidently conclude this based on a simple physical argument.

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    $\begingroup$ +1 When complicated effects are taken into account it is hard to say what will happen. $\endgroup$ – Deep Feb 22 '17 at 11:25
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    $\begingroup$ While this would be a much better use of the question, it was apparently a test question and the teacher intended for there to be a "correct" answer. One can only hope the question will be excluded from the exam. $\endgroup$ – jpmc26 Feb 23 '17 at 0:16
  • $\begingroup$ @jpmc26 Oh, I missed the part about this being an exam question. That seems like a really poor choice by the teacher. $\endgroup$ – jkej Feb 23 '17 at 10:00

The answer is related to the first question.

When ice is kept in glass it will absorb heat from the water vapours which are present in atmosphere around it.

So water condensers around it.

Now the difference when it is kept in shade and sun is that when it is kept in sun the ice will absorb heat from the Sun rays falling on ice too. It will Avalon more amount of heat from sun than through water vapours and atmosphere.

While in shade as sun rays are not falling more heat will be absorbed from atmosphere.

So in shade more water will get condensed around glass than in sun because in sun ice has a better or another source to absorb heat.

So water will be more in shade.

  • $\begingroup$ Thanks for the answer but our physics teacher is holding his ground $\endgroup$ – class_question Feb 22 '17 at 3:37
  • $\begingroup$ Can we not say that without shade ice absorbs same amount of heat from atmosphere but also absorbs extra heat from sun? This will cause more ice to melt, but water condensed will be same. $\endgroup$ – Ketan Chaware Feb 22 '17 at 3:45
  • $\begingroup$ Heat will be absorbed till the ice melts and it's temperature equals to that of its surrounding or atmosphere. $\endgroup$ – ATHARVA Feb 22 '17 at 4:00
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    $\begingroup$ What is "Avalon"? @class_question: Anyway the sun will also increase the rate of evaporation of the condensed water, further reducing the amount that remains on the exterior of the glass. $\endgroup$ – user21820 Feb 22 '17 at 12:43
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    $\begingroup$ "Our physics teacher is holding his ground" — Has your physics teacher taught you about what Galileo did to Aristotle? Time for an experiment! $\endgroup$ – 200_success Feb 23 '17 at 22:51

You will see that your assumptions about the experiment can make a difference. At the end, I give you a twist that can make all of the difference:


Let's make some simplifying assumptions, and explain why we these assumptions make sense:

First let's address the macroscopic conditions. By this, I mean the conditions that exist uniformly over most of the space in which this experiment takes place:

  1. Regarding the glass of iced water, let's assume it contains ice for the duration of the experiment. (The reason is that this is apparently a given - you have a glass of iced water.)

  2. We can also assume that the glass of iced water is at a constant temperature for the duration of the experiment. The reason is that as ice melts, it absorbs heat without changing temperature. This is what happens when a material undergoes a phase change. (solid to liquid or liquid to gas, for example). So, with the first assumption (there is always ice present), then the temperature will remain constant. (Phase Transition of Water)

  3. For the experiment in the sun and the experiment in the shade, all other conditions are identical. (We might assume an open space, since a the glass is on a picnic table.) So, for example, air temperature (see below) and pressure are identical. If it's windy at all, then this is the same in both cases. (Let's assume it's not windy, but you can take this further and ask yourself how that would alter the conclusion, if it does.)

  4. The air temperature is the same in the shade as it is in the sun. I know what you are thinking, but you have to understand that air temperature is a measure of the kinetic energy of the molecules in the air. (Why Official Temperatures are Always Measured in the Shade) Since the molecules are actually travelling very fast (at the speed of sound, on average), it doesn't take long for them to move around between a shaded area and an unshaded area. (If you question that, think about how quickly an odor can travel, even in "still" air. If you're still not sure, try an experiment with something that you can easily smell.)

  5. The level of humidity is the same in the shade as it is in the sun. This is the same as saying that for a given volume of air, the fraction of molecules that are water molecules is constant regardless of whether they are in shade. The humidity depends on the air temperature and the kinetic energy of the air molecules to keep water from condensing out. Also, water molecules in the air travel at speeds similar to all other air molecules.

Now, let's look at some microscopic conditions, specifically at the surface of the glass:

  1. (Convection) heat transfer will occur at the surface of the glass.

    • Air molecules, including the evaporated water (the humidity) will lose energy when they encounter the cold surface of the glass. This is called convection heat transfer (convection), which is the transfer of kinetic energy from the gas to the cold glass. Enough water molecules can lose energy to come to the point of condensation, producing the water on the side of the glass. (Other types of molecules in the air will not condense so readily at the temperatures we are talking about here.)

    • Water molecules on the side of the glass will gain energy when they are hit by fast moving air molecules. This can be sufficient to cause evaporation.

This is a continuous process, but obviously if you see condensation happening, then the rate of condensation must be greater than the rate of evaporation!

  1. Radiation heat transfer will occur at the surface of the glass. Radiation from the sun impinging on a thermometer heats the thermometer. As explained earlier, this is why we measure air temperature in the shade. It can also transfer enough heat to cause some of the water molecules on the side of a glass to evaporate. This would be in addition to evaporation due to convection.


We have assumed that apart from the sun/shade difference, all conditions are equal between the two experiments:

  • The glass will keep a constant temperature so long as there is ice, because heat added from the environment only goes to changing the ice from solid to liquid.

  • The effects of air temperature has been explained as being equal, because air temperature is a measure of the kinetic motion of the air molecules, and all of the molecules move freely and quickly between the shade and the sun.

  • Conditions at the surface of the glass can result in phase changes. This can cause condensation and evaporation all at the same time. Energy transfer mechanisms control the net results.

  • The experiment done in the sunlight has one additional energy source influencing the condensation and evaporation process. Radiation can increase evaporation.

We have to conclude that the net of evaporation vs. condensation differs between the two experiments, and there will be less water on the side of the glass when the experiment is done in the sunlight.

Now whether this will be a noticeable amount is a different matter. But under the right experimental conditions this can be measured with precision.

Now Remove One Assumption

I wouldn't put it past my high school physics teacher to do something like this, but lets assume that the "iced water" hasn't reached equilibrium when the experiment starts. That means that as ice melts, the released liquid water will be at a colder temperature than the rest of the liquid.

What this means is that some of the heat energy from the water must go to warming this released water.

In the shaded experiment, we expect the ice to take longer to melt. (Less radiant energy on the glass!).

Heat transfer between the air and the glass will be at a rate that is proportional to their temperature difference (everything else being equal). This will result in less condensation (and more evaporation) while the glass is at a temperature above equilibrium.

We would expect the glass in the sunlit experiment to reach equilibrium sooner than the one in the shaded experiment.

Now you have a case where the evaporation from radiation may be offset by this thermal equilibration process of the iced water.

If you were to integrate the net condensation over the 30 minute time of the experiment, you would see that the sunlit glass might actually produce more condensation.

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    $\begingroup$ Yes! The teacher may not realize that the air temperature and humidity are equal in the sun and shade. The "Warm air holds more water" argument does not apply here. Thanks also for pointing out that this is an equilibrium process whereby condensation is offset by evaporation for a net effect. Lastly, thanks for pointing out that there is an equilibrium process in the melting ice. My college chemistry lab demonstrated that the assumption of the water-ice mixture being isothermal was often flawed. In practice the ice blocks convective mixing and it's only an isotherm if you actively stir it. $\endgroup$ – Kengineer Feb 23 '17 at 19:55

Condensation on the glass is all to do with whether or not the the air near the surface of the glass is at the dew point (temperature).
The dew point depends on the humidity of the air.
At the dew point the water vapour in the air will condense to form water liquid.

What is interesting is that the rate of condensation is not just affected by the temperature of the air it also depends on the rate at which heat can be abstracted from the condensing water vapour in the air.

When the Sun is shining on the glass the radiant heat from the Sun is absorbed by the glass and the condensed water and so raises the outside temperature of the glass and condensed water.
This in turn increases the rate of transfer of heat through the glass and together with the radiant heat absorbed by the ice/water in the glass increases the rate at which ice melts in the glass.
The volume of air around the glass which is at or below the dew point is probably greater when in the shade as the outside of the glass is at a lower temperature and this could result in a greater rate of condensation of water vapour.

It seems to me that a glass in a shady spot will result in more water vapour being condensed in a given time.

  • $\begingroup$ The dew point is not pressure dependent. $\endgroup$ – jkej Feb 22 '17 at 11:09
  • $\begingroup$ @jkej Change made as suggested. $\endgroup$ – Farcher Feb 22 '17 at 11:35

There are a lot of long answers here that probably go way beyond the model that a high-school physics teacher is going to use with his students.

You don't detail the teachers or your reasoning but I'd guess the teacher's answer rests on one simple fact: warm air can hold more water than cold air.

The table in the sun (by this line of thinking) is warmer than the table in the shade but the temperature of the glass is the same in both cases. Therefore there is more moisture in the hot air that will condense out.

Is this really what would happen?

There are a lot of complicating factors here. Would the air really have more evaporated moisture in the sun than in the shade? It doesn't seem likely to me at least not in near proximity. How does the solar radiation change things? Would it cause some of the water to evaporate? Very difficult to model especially without more details.

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    $\begingroup$ Yes, warm air can "hold" more water. But indeed, that doesn't mean it HAS more water. Air in one location will basically contain the same actual amount of water regardless of patches of shade or sun spots. The only possible ways that could be untrue was 1 - ABSOLUTELY no mixing at all (perfectly calm)... plus a significant source of water for the sun to evaporate into the warmer air or 2 - if the air were highly saturated already (unlikely in daytime), plus the ground cold, to cause dew). Even in those two cases, it would take a timescale of days to bring small % change in moisture quantity. $\endgroup$ – JeopardyTempest Feb 23 '17 at 8:39
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    $\begingroup$ @JeopardyTempest Agreed. Again, I'm just guessing at what the teacher is thinking here but it's the only thing that comes to mind that would be valid (in theory) at the high-school level of instruction. $\endgroup$ – JimmyJames Feb 23 '17 at 17:21

To answer this, some assumptions have to be made. I'll assume the following:

  1. There was enough ice in the glass so that all the ice didn't melt. This means the inside of the glass was at essentially constant temperature close to the freezing point of water the whole time.

  2. The glass is transparent, and therefore gets very little thermal power directly from the sun. In any case, the inside was kept at the freezing point of water since the ice didn't all melt.

  3. The sun and shade options are very localized. A few meters either way would change sun or shade. Therefore, the air around the glass is essentially the same temperature and humidity whether the glass is in the sun or the shade. This also means that whatever wind there is or isn't is the same in both cases.

With all these conditions, there should be no difference in condensation between the glass in the sun and the shade. You have a glass object held at a fixed temperature with the same air at the same temperature and humidity gently swirling around it.

So, the answer comes down to subtle effects that make some of the assumptions not quite true. Since such detail was not provided in the question, the correct answer is You can't tell from the question.

For example, assumption #2 won't be exactly true. Even a transparent glass with clear water and ice inside will absorb a little power from the sun. Most of that will go into heating the water, which just causes more ice to melt over the same time. That still keeps the water at the same temperature until the ice is exhausted.

However, the tiny amount of heating on the outside of the glass will cause a slight temperature rise there, despite the inside of the glass being at a fixed temperature. That slight rise would slightly decrease the amount of condensation.

But as a counterpoint, there could be both more and less condensation in the sun due to assumption #3 not being quite true. Suppose the picnic table is wood and a little damp. The sun heating it would cause more humidity right above the table where the glass is, leading to more condensation. If the table is dry, then the sun heating it would cause a little warmer air right above the table, leading to less condensation.

Again, there is no clear way to answer sun/shade question without more details about the conditions, since the difference in condensation depends on subtle effects that depend on the details.


Sometimes a mental experiment, making the conditions more extreme, can be illuminating. I can think of two:

  1. Have the glass be a cup instead, made from a more insulating material; just so that there is barely some borderline condensation happening when the cup is in the shade under an umbrella. Imagine the material black to increase light absorption.

    Now remove the umbrella. I bet the cup surface exposed to the sun will heat up (assuming the sun shines at an angle) and condensation will stop, at least on the side the sun shines on. The effect is the same with the mostly transparent glass, just weaker.

  2. Take the original glass setup. But do not expose it to the sun but instead to a 100kW spot light, as is used on movie sets, effectively increasing the light intensity by an order of magnitude or two. I bet the glass's surface and the condensing water (which absorbs infrared well) will warm up somewhat, leading to less net condensation.

As an aside, I understand why one could think that there is more condensation in the sun. It is a common experience that on warm days chilled beverages "sweat" a lot, while on cool days the effect is much milder, if there is any at all. Houstonians need coasters; Siberians, not so much. But how about San Antonians? People in Pecos?

The condensation depends mostly on the relative humidity of the air at the surface of the glass.

The relative humidity (can the air absorb any more water?), in turn, depends on the absolute humidity (grams of water/grams of air) and the temperature (because warm air can hold much more water). The air in warm climates like Houston can carry a lot of water, which will all condense on your cold glass. The air in cold climates is much drier, in absolute terms — the famous extreme is the air at the south pole.

Because it will be warmer in the sun one could be tempted to think that more water will condense, like it would in warm Houston. But the air in the shade and in the sun is the same air holding the same amount of absolute water. It will behave identically under identical conditions on the glass surface. If at all, the warmer air and the sun may warm up the glass's surface, leading to less condensation, and the warm air in the sun (which is relatively drier, because it can hold more water) will accelerate evaporation, shifting the dynamic equilibrium away from condensation.


I noticed that one commenter said that the question is inherently ambiguous concerning whether you count condensation that has rolled off the side of the glass.

Question: Would there have been more or less water on the outside of the glass if the picnic table was in the shade?

Indeed, if I were to interpret it in a particularly amusing way, in the shade less water would evaporate from inside the glass into the air, and hence there would be less water on the outside of the glass.

That aside, another commenter also mentioned that you can't expect the ice to remain unmelted if you leave it in the sun for half an hour, and it is very likely that after the ice has melted the sun will dry up the outer side of the glass by evaporation. In the shade it is easy to imagine the glass of iced water remaining iced when you come back, so there will still be water condensing on the side of the glass.

Therefore I don't see much need for complicated reasoning here.

  • $\begingroup$ Unless your iced water is more like a glass full of ice rather than a glass of water with a few ice cubes in it... $\endgroup$ – user21820 Feb 24 '17 at 9:42

protected by Qmechanic Feb 22 '17 at 7:53

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