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Bashyboy
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Homework Statement
When the temperature of liquid mercury increases by one degree Celsius (or one kelvin), its volume increases by one part in 550,000. The fractional increase in volume per unit change in temperature (when the pressure is held fixed) is called the thermal expansion coefficient, [itex]\beta[/itex]:
[itex]\displaystyle \beta \equiv \frac{\Delta V/V}{\Delta T}[/itex]
(where V is volume, T is temperature, and A signifies a change, which in this case should really be infinitesimal if 0 is to be well defined). So for mercury, (3 = 1/550,000 K"1 = 1.81 x 10~4 K_1. (The exact value varies with temperature, but between 0°C and 200° C the variation is less than 1%.)
(a) Get a mercury thermometer, estimate the size of the bulb at the bottom,
and then estimate what the inside diameter of the tube has to be in order for
the thermometer to work as required. Assume that the thermal expansion
of the glass is negligible.
Homework Equations
The Attempt at a Solution
Suppose at the temperature [itex]T_0[/itex] all of the mercury is in the bulb of the thermometer; and suppose the radius of the bulb is [itex]r_0[/itex], from which we can get the volume of the bulb, assuming it to be spherical, and consequently the volume of the mercury (the space it occupies). Suppose, by some external means, the temperature of the mercury increases from [itex]T_0[/itex] to [itex]T[/itex], [itex]\Delta T = T - T_0[/itex]. Due to the fact that things expand when their temperature changes, the mercury will begin to diffuse upwards into the cylinder portion of the thermometer, but some will remain in the bulb.
Now, suppose that when the mercury reaches the temperature [itex]T[/itex], the mercury occupies all of the volume that the thermometer can afford. So, when [itex]T_0[/itex], [itex]V_0 = \frac{4}{3} \pi r^3_0[/itex]; and when the temperature changes to [itex]T[/itex], [itex]V = \underbrace{V_0}_{Initial~volume} + \underbrace{l_0 \pi r_1^2}_{Volume~ due~ to~ expansion}[/itex], where [itex]l_0[/itex] is the length of the thermometer, and [itex]r_1[/itex] is the radius of the cylinder, the variable of which we are solving for.
The change in volume: [itex]\Delta V = (V_0 + l_0 \pi r^2_1) - V_0 = l_0 \pi r^2_1[/itex]
Rearranging the thermal expansion coefficient, and substituting,
[itex]\Delta V = \beta V_0 (T-T_0)[/itex]
[itex]l_0 \pi r^2_1 = \beta V_0 (T-T_0) [/itex]
[itex]\displaystyle r_1 = \sqrt{\frac{\beta V_0 (T-T_0)}{l_0 \pi}}[/itex]
At this point I would make assumptions regarding the values of the radius of the bulb and the length of the thermometer, as per the directions of the problem statement
If I understand this question properly, then if one was to build a thermometer, such that at temperature [itex]T_0[/itex], the mercury occupied the volume of the bulb. And because of these conditions, these would determine what radius of the of the cylinder portion of the thermometer has to be. These seems to make some sense, but I would appreciate if someone could logically explain these, to clarify this issue. Another thing I am dealing with is, notice the formula that I derived: the radius of the cylinder does not depend on the [itex]T[/itex] or [itex]T_0[/itex] individually, but on the change in temperature. Why is that so, why does it not depend on the beginning and ending temperature, aside from it being mathematically true? I am having a little difficulty grasping that.
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