A typical mercury thermometer is made up of a thin, cylindrical capillary tube with a diameter of 0.0040 cm, and the spherical bulb with a diameter of 0.25 cm. If we DON'T Neglect the expansion of the glasas, find the change in height of the mercury column for a temperature change of 30 degree. In the solution, it was given that the thermal volume expansion of mercury = 18 * 10^{-5} and the thermal volume expansion of glass = 2 x 10^{-5} 3. The attempt at a solution I have attempted to solve the problem, however, I don't really know which volume expansion value should I use in the problem. I have setup: [tex]\Delta[/tex] V = V_{expansion} (V_{0} + [tex]\Delta[/tex]A) * [tex]\Delta[/tex]h = V_{0} + V_{0} * volume expansion value * [tex]\Delta[/tex] Temperature [tex]\Pi[/tex] *(0.0040 cm/2)^{2} + [tex]\Pi[/tex] *(0.0040 cm/2)^{2} * ( thermal expansion of "what material???" * (30 degree) * ([tex]\Delta[/tex] h =( ([tex]\frac{4}{3}[/tex] * [tex]\Pi[/tex] * (0.25cm/2)^{3} ) + ([tex]\frac{4}{3}[/tex] * [tex]\Pi[/tex] * (0.25cm/2)^{3} * (volume expansion of "which material" ) * (30 degree) I have tried to solve with both thermal expansion number, however, I have got 650 cm and 631cm. The solution of the problem is [tex]\Delta[/tex] h = 3.1cm... I wonder where did I do wrong? Whcih volume of expansion should I choose?
Firstly I assume the volume expansion for glass and mercury are given in the correct units. [here using cm to be consistent with the other units in the question] You are on the right track... You start by calculating the volume expansion of the mercury and subtract from it the expansion due to the glass. This gives the effective expansion of the mercury into the tube. The volume expansion is found by multiplying the original volume by the rise in temperature and the expansion coefficient. Equate this volume expansion to the volume of a length L in the tube. Volume in tube is A x L where L is what you want to calculate. A is the cross section area.