# How does the change in temperature of water in a tube depend on the angle that the tube makes with the gravitational field/the horizontal?

Let's imagine a tube that is open at one end and is filled with water. A constant source of heat is applied to a specific spot somewhere at the middle of the tube, for a example a preheated soldering iron of constant temperature. Let's say you measure the temperature of water at the upper end of the tube over a specific amount of time when the tube is parallel to the gravitational force, or 90 degrees to the horizontal. If you keep all the conditions constant, including the amount of time for heating, but change the angle so that it is 80 degrees to the horizontal, then 70, 60 and so on... You will find that the increase in the temperature becomes less as the angle decreases. But what would the graph of the increase in temperature vs sine of the angle look like? And what is the theory behind it? I undertstand that there are convection currents of natural convection caused by a buoyant force, which is caused by the difference in density and which will be counteracted by a normal force of reaction as the the tube becomes tilted. However, I would like to understand more of the physics involved in order to model the phenomenon.

The following graphs are based on primary data (collected by me). The tube used was a burette, the mark of 17 ml was in between the mark of 1 ml and the source of heat, which was at the mark of 25 ml. I realized that the wire of the thermometer at the mark of 17 ml could pose a confounding variable as it was curved, so it was not used in the fourth measurement session.

• Did you factor in the heat conduction of the glass tube? Dec 11 '20 at 20:26
• @SolarMike No, I did not. The processing of the data only involved calculating the sine (by dividing the hight between the two ends of the tube by the length of the tube) and the increase in temperature (final temperature minus initial temperature). I would appreciate it if you refer to any literature relevant to the question. Dec 12 '20 at 6:55