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I have two samples of water – 'A' and 'B'. My thermometer measured them to be 90 °C. They both have different volumes. So, the kinetic energy of the particles of sample A and B is same or different? Through this question, I want to ask whether temperature measurement results depend upon how many particles hit the bulb of my thermometer or it is independent of this? If it is independent of this, then how thermometer works?

Please provide me the simplest answer as I am in class 9 only.

As another example, let me consider a gas which has the density of 2 mol/22.4 L. Its temperature recorded by my thermometer is 60 °C. Then its density is reduced to about 10 atoms/22.4 litre. Now, its temperature is the same?

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  • $\begingroup$ The unit of temperature is Joule all right. The ambient temperature in my place is about $6.13\cdot10^{-21}\;\rm J$; what's yours? $\endgroup$ – Ivan Neretin Sep 6 at 12:32
  • $\begingroup$ Now to the first part. You seem to know that the temperature has something to do with average kinetic energy. If the temperature of two samples is the same, then so is their average kinetic energy; what's unclear about that? $\endgroup$ – Ivan Neretin Sep 6 at 12:55
  • $\begingroup$ Okay. Let me consider a gas which has the density of 2 mol/22.4L. Its temperature recorded by my thermometer is 60°C. Then its density is reduced to about 10 atoms/22.4 litre. Now, its temperature is same?? $\endgroup$ – user36956 Sep 6 at 12:58
  • $\begingroup$ I have the above doubt that temp. should vary in both these forms of gases because in first case, many gas molecules were hitting the bulb of my thermometer and when density was reduced, only few molecules hitted the bulb which should have reduced the recording. $\endgroup$ – user36956 Sep 6 at 13:05
  • $\begingroup$ @user36956 Temperature is the statistical parameter, implying particles are in frequent contact and mutually in equilibrium. that does not apply to 10 atoms in 22.4 L. 1 particle does not have temperature, nor do 10 particles. $\endgroup$ – Poutnik Sep 6 at 13:07
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I have two samples of water – 'A' and 'B'. My thermometer measured them to be 90 °C. They both have different volumes. So, the kinetic energy of the particles of sample A and B is same or different?

The total kinetic energy is different. The average kinetic energy per particle is the same.

Through this question, I want to ask whether temperature measurement results depend upon how many particles hit the bulb of my thermometer or it is independent of this?

In your specific example, the same number of particles hit the bulb because it just depends on the area of the bulb, as long as it is covered in water. Not only is water hitting the bulb, the atoms in the bulb are also hitting the water. After they have done this for a while, water and bulb are at the same temperature.

If it is independent of this, then how thermometer works? As another example, let me consider a gas which has the density of 2 mol/22.4 L. Its temperature recorded by my thermometer is 60 °C. Then its density is reduced to about 10 atoms/22.4 litre. Now, its temperature is the same?

For the example with the two gases at the same temperature but at different pressure (i.e. particle density), the situation is a bit different, but the thermometer still works. There are a different number of "hits" in this case, but if you wait long enough, the gas and the bulb will be at the same temperature.

Please provide me the simplest answer as I am in class 9 only.

I tried, but if the question addresses a complicated phenomenon, the answer can't be super-simple.

[From the comments:] So, if i have 10,000000 particles in 1 L and 10 particles in 1L, temperature is going to be same or different?? (I mean the readings)

Ideally, there should be more sample than thermometer. This means that no matter what the temperature of the thermometer was to start, it will have the temperature of the sample afterwards. With only 10 particles in the sample, the 10 particles will take on the temperature of the thermometer (and the enclosure), so you will be unable to measure the initial temperature of the sample.

[My own question] If the sample keeps hitting the thermometer, why doesn't it get hotter and hotter?

As I said above, the thermometer also keeps hitting the sample. There is some net transfer of energy while the temperatures are different (from hot to cold, warming up the colder and cooling down the hotter). Once the temperatures are the same, sample and thermometer keep "hitting each other", but the temperature of both stays constant (if the entire setup is insulated against the surrounding).

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  • $\begingroup$ One more question- why my thermometer measures only the avg. translational kinetic energy of particles, but not rotational or vibrational one? $\endgroup$ – user36956 Sep 7 at 17:17
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The short answer:

If 2 samples of water of the same temperature is considered,the average kinetic energy of water molecules of samples A nad B are the same. But, within the same water sample, water molecules have different kinetic energy with a particular statistical distribution.

Even if I had set by some magic wand the same energy for all water molecules at this moment, the energy would be redistributed back in nanoseconds to their favourite distribution.

The longer answer:

Temperature is not average kinetic energy of particles, but it is a measure of average energy per "degree of particle" freedom.

As the degree of freedom - imagine yourself it is a way the particle can move.

Imagine an oxygene molecule, consisting of 2 atoms.

It can independently move along 3 perpendicular axis x,y z -> 3 degrees of freedom. Its chemical bond can vibrate - 1 degree of freedom. It can rotate around 2 perpendicular axis - can vibrate - 2 degrees of freedom.

Each degree of freedom has average energy proportional to temperature(*).

The proportionality constant in $E = k. T$ is the Boltzmann constant about $\pu{1.38 J/K}$.

If 2 macroscopic objects with the same temperature are in contact, their average energy per a degree of freedom is the same. Their particles exchange energy, but as the average is the same, the net exchange of energy is zero and objects stay at the same temperature.

(*) - It is not fully true if we consider quantum effects of quantization. Molecules are allowed to have just particular discrete values of rotational and vibrational energy. Translational energy does not have this quantization ( at least not in steps measureable by curent methods ).

(**) - Not fully true, but for not it is.

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  • $\begingroup$ Didn't get anything at all! I expected some easy basic answer. $\endgroup$ – user36956 Sep 6 at 13:02
  • $\begingroup$ It was easy, basic answer. :) Try this article on the Simple version of wikipedia. simple.wikipedia.org/wiki/Temperature Also, try some textbooks or online courses. $\endgroup$ – Poutnik Sep 6 at 13:04
  • $\begingroup$ Hmm... Tried but it didn't answer my question. $\endgroup$ – user36956 Sep 6 at 13:07
  • $\begingroup$ @user36956 I have updated the short answer. $\endgroup$ – Poutnik Sep 6 at 13:34

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