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Assuming standard atmospheric pressure, Does the water vapour created from boiling water have the temp of 100 °C? What about water vapour created by evaporation at room temperature?

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If you have an equilibrium state then the answer is "yes" the temperature of the liquid and the vapour will be the same.
In both cases you have a large volume of system to keep at the same temperature.
When you boil some water in a kettle the vapour soon loses heat to the cooler surrounding air and starts to condense back to a liquid.
That is the reason that you can see steam.

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  • $\begingroup$ Thanks. Why would the temp be same at equilibrium - given that temp is directly proportional to kinetic energy of molecules and vapor molecules have higher kinetic energy than liquid molecules. $\endgroup$
    – user13107
    Aug 1, 2017 at 7:36
  • $\begingroup$ In the equilibrium state the temperature of liquid and vapour is the same. If that were not the case then there would be a net transfer of heat from one state to the other. $\endgroup$
    – Farcher
    Aug 1, 2017 at 7:58
  • $\begingroup$ Ok. do you mean to say average kinetic energy of water vapour molecules is the same as that of liquid molecules, it is in equilibrium with? $\endgroup$
    – user13107
    Aug 1, 2017 at 8:06
  • $\begingroup$ @user13107 Temperature is not kinetic energy. Certainly not in water with all the hydrogens having a lot of zero-point motion. $\endgroup$
    – user137289
    Aug 1, 2017 at 9:00

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