Suppose I warm a glass of water up to 50°C in the microwave. In the other hand, I warm up exactly the same amount of water to the same temperature, but in the oven. In this situation I would have two identical glasses of water at the same temperature, but warmed up by different sources and heating processes (radiation and convection mainly). Now, if I hang the glasses to someone, can he/she distinguish which one was heated up with the oven and which one with the microwave?, can we tell the difference?.

EDIT: I understand that the person would likely feel a difference in the glasses temperature, but what happens if I pour the water into a new pair of glasses after I warm it up ?. Is there a way by which I can tell if the water was heated up mainly via radiation and not something else ?

  • $\begingroup$ @PabloNavarrete Regarding edit 1, what do you mean by the quality of both ovens being the same? The quality of a microwave oven has nothing to do with the interaction of microwaves with glass. They just generally don't. Regarding edit 2 I already answered in comments. There should be no difference. $\endgroup$
    – Bob D
    Commented Jul 7, 2019 at 21:29

2 Answers 2


The exact state of a system is defined by a set of parameters. In thermodynamics, we can prove the thermodynamic state of a closed system with the same mass of a pure compound (your glass of water) is defined by only two of the three parameters (T, p, and V). Unless one process changes the pure water to something else chemically while the other does not, the two systems of pure water at the same (T, p, V, and m) will be indistinguishable. This is especially true to a "casual observer".

You might ask whether one process (microwaves) can leave a different distinguishing signature about itself on the pure water that another (conventional oven) does not. Focusing only on microwaves, the fundamental interaction is generally to increase the rotational energy of molecules. Focusing on convectional ovens, the fundamental interaction is generally to increase the overall energy of the modes of translation, rotation, and vibration. In this case, translation is easiest to excite, followed by rotation, then vibration.

You might expect then to find that water heated by microwaves is distinguishable from water heated by a conventional oven using a tool that can tell which modes are more populated. In a process that uses microwaves, the rotational modes of the pure water should be more highly populated, or so you might suspect.

However, excitation of the rotational modes cause the molecule to bump into other neighboring molecules and transfer the energy. The energy transfer happens to create a thermal equilibrium throughout the water.

So, while the water in the microwave heats up ...

  • the water molecules are excited to higher rotational energies
  • the excited water molecules transfer the energy by collision to translation and vibrational modes of their neighbors
  • since translation is easier to excite than vibration, the rotational energy is transferred to vibrational energy

In the meantime, in the conventional oven, all modes get excited (translation first, then rotation, then vibration).

At the end, pure liquid water at a given temperature, pressure, volume, and mass is the same regardless of how you brought it to that state. The modes (translation, rotation, and vibration) are all normalized to the same population statistics. That indeed is the definition of an equilibrium temperature state for the pure water itself.

Quite a different picture will arise when you take an individual water molecule in a gas phase. You can excite that molecule thermally by having it bounce off of a hotter surface. The water molecule will move faster or rotate in the next highest state or vibrate in the next highest state. You can do this using only an infinitesimal amount of heat, but that will only cause translational excitation. You can excite an individual gas phase water molecule rotationally by hitting it with the right quantum level energy (frequency) of microwave radiation (determined as $\Delta E_{rot} = h\nu$). It can remain in that state until the energy transfers internally to other modes. This is internal relaxation of the excited state. Otherwise, the molecule will loose the excited energy when it collides with a system or molecule, especially one at a lower energy state. You can excite an individual gas phase water molecule vibrationally by hitting it with the right quantum level energy (frequency) of infrared radiation. Finally, you can excite an individual gas phase water molecule electronically by hitting it with the right quantum level energy (frequency) of ultra-violet radiation. In each case, you can usually detect the internal decay (relaxation) of the excited state (rotational, vibrational, or electronic) by looking for emission of radiation. By example, a fluorescent molecule is one that has been excited electronically and is relaxing by emitting visible radiation.

Excited states are not ground states. They are not the thermodynamic system of pure liquid water after it is heated. To capture the excited state of the water molecules, you have to isolate them. The minute an excited state molecule collides with another molecule, energy is transferred.

  • $\begingroup$ So the normalization would occur before any feasible detection in differences could be made? $\endgroup$
    – user234190
    Commented Jul 7, 2019 at 22:07
  • $\begingroup$ Like say you had a probe of some sort ready to go as soon as the temperature was reached and the heat source turned off. Could one look at the probe and determine how the water was heated? And if so, for how long? $\endgroup$
    – user234190
    Commented Jul 7, 2019 at 22:13
  • $\begingroup$ @user47014 I have expanded my discussion to include gas phase states to answer to your additional questions. $\endgroup$ Commented Jul 7, 2019 at 23:41

Suppose I warm a glass of water up to 50°C in the microwave. In the other hand, I warm up exactly the same amount of water to the same temperature, but in the oven.

First you should be aware that heat is energy transfer due to temperature difference. A microwave oven does not increase the temperature by means of a heat transfer. Technically, is is the work that the alternating electric field does on the water molecule dipole giving it rotational kinetic energy which, in turn, results in the increase in water temperature. The conventional oven, on the other hand, raises the temperature of the water and the glass by heat transfer. The difference can be important in answering your question.

The microwave energy is absorbed primarily by the water. Depending on the composition of the glass, microwave generally pass thru it and not significantly raise the temperature of the glass directly. The temperature of the glass increases due to heat transfer from the water to the glass.

The oven, on the other hand, does heat the water. It also directly heats the glass at the same time so that glass will feel hotter coming out of the conventional oven.

One difficulty with your experiment is that the microwave oven will likely raise the temperature of the water to 50 C faster than the conventional oven, so you wouldn't necessarily be able to hand both glasses to the person at the same time. That would make the comparison more difficult.

But let's say you had a way to know exactly when the temperature of the water reached 50C in the microwave oven. You immediately take it out and immediately hand it to the person and ask how hot it feels.

Next you do the same with the glass of water from the conventional oven and get their reaction.

Depending on the thickness of the glass, it is likely the person will say the glass (that's what they touch) feels hotter coming from the conventional oven than from the microwave oven. That is one way they might know which oven it came from. The thicker the glass the greater will be the difference since it takes more time for heat to transfer from the water to the glass the thicker it is.

The important difference is the conventional oven directly heats the glass along with the water. The microwave oven heats the water and not the glass. Heat needs to transfer from the water to the glass to raise the glass temperature.

Because I'm wondering that if, even at the same temperature, you could say if they were heated up by different sources by looking at some property or condition of the water, even microscopically. My question is probably more about if there exists some "print" in the water that tell me that microwave radiation was mainly there in the heating process.

You would have no way of knowing that the water temperature was raised due to the microwave oven or conventional oven. There is no "print" to tell you the 50 C water came from a microwave oven or from a conventional oven.

Now different microscopic properties of the water are in fact involved when interacting with the microwave oven vs the conventional oven. Microwave electromagnetic radiation rotates the water dipoles giving them rotational kinetic energy that results in raising the temperature of the water to 50 C. For the conventional oven, infrared electromagnetic waves give the water molecules vibrational kinetic energy that raises the temperature to 50 C.

But there's no "print" that tells you it was rotational kinetic energy or vibrational kinetic energy that was responsible for raising the temperature.

Hope this helps.

  • $\begingroup$ I understand that the person would feel some temperature difference in the glass itself, but what happens if I pour the water into a new pair of glasses, so the person can't feel that difference ? $\endgroup$ Commented Jul 7, 2019 at 21:04
  • $\begingroup$ @PabloNavarrete Water at 50C is water at 50C regardless of whether it was due to a microwave oven or conventional oven so it would make no difference if you poured the water into new identical glass at the same initial temperature. My point of the answer was to show that you have a way to distinguish whether the original glass of water was microwaved or conventionally heated based on the difference in glass temperature. I thought that's what you wanted to know. $\endgroup$
    – Bob D
    Commented Jul 7, 2019 at 21:08
  • $\begingroup$ Then different heating processes would make no difference in the water itself ? (not taking the glass into account) $\endgroup$ Commented Jul 7, 2019 at 21:10
  • $\begingroup$ @PabloNavarrete That is correct. But why would you think there would be a difference between 50 C water from a microwave oven and 50 C water from a conventional oven? $\endgroup$
    – Bob D
    Commented Jul 7, 2019 at 21:10
  • $\begingroup$ @Bob D I think he is more asking something like if the emission spectra from water varies depending on the source that heated it. But then the next question may be, if the spectra do vary, for how long are the emission spectra different? $\endgroup$
    – user234190
    Commented Jul 7, 2019 at 21:21

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