Hi Willy, you are quite right about there being so much thermodynamics going on. It is indeed very difficult to frame these questions really completely.
When you take the thermometer out of the water, it is wet, and the water starts to evaporate. But remember your description of the problem. The air and water are at the same temperature!
So why does the water evaporate? And yes, when it evaporates, it does cool the sheath.
We should tie up one loose end in the problem description, the air humidity was not defined. I would expect that the air humidity in the room was somewhat below 100%, probably less than 50%, so the water vapour pressure in the air was somewhat below the saturation vapour pressure. Consequently, some of the water molecules with something above the average energy level will escape the water surface, leaving the remaining water a little cooler. The thermometer will then provide some heat to the now cooler water, and the air will supply heat to return the whole to the original average temperature. We could carry this on to conclude that the room will end a little cooler, or the system maintaining the room temperature will supply some extra heat. And of course the bath is also evaporating a little if the water vapour content of the air near the surface is less than the equilibrium vapour pressure for the temperature. If the air is very still, the humidity builds near the water surface and the evaporation stops. However air movement near the surface removes those freshly evaporated water molecules and so the process continues further. But you might disturb this still air by entering the room to conduct the experiment.
The description can go on, but the important points are first that truly reaching thermal equilibrium, when there is no further heat transfer is a slow process, and difficult to achieve. In practical terms we are normally dealing with close enough rather than exact.
Second, the thermometer is only measuring the temperature of the thermometer, and we need to place the thermometer in close thermal contact with the object or fluid we are measuring, and give them time to reach the same temperature before we take the reading.
Thirdly, trying to answer this question also involves understanding of evaporation of fluids and the associated heat transfer.
Things operating in space involve a whole extra level of thermodynamics, but it is getting very far from my experience. Also, like the errors in the assumptions of conservation of mass, not likely to be of much practical use to most of us as model engine builders. So I will use that as an excuse to call the thread back to topic!
So time for me to wish you and yours a very happy and safe new year. Let’s hope it is a bit easier than this year for all.
Thank you for all the interesting questions that have kept my mind active throughout the year. I hope I have been able to help you and others who have been reading the thread to understand a little more thermodynamics.
The 9pm fireworks have just happened in Sydney so the kids can go to bed. The midnight ones will be a little shorter than usual, and people are being encouraged to watch from home. The Melbourne one have been cancelled to avoid tempting people to gather in large numbers to see the spectacle. A very different New Year’s Eve for most this year. Time for us all to imagine alternative ways to enjoy seeing the new year in.
MJM460
