Finding Increse in Volume Given Temp, Vol Expansion....

AI Thread Summary
The discussion revolves around calculating the increase in volume of mercury in a thermometer due to temperature change. The coefficient of volume expansion for mercury is given as 1.8 x 10^-4 per degree Celsius, and the temperature change is from 30 to 70 degrees Fahrenheit, which is equivalent to 22 degrees Celsius. The original volume of mercury in the bulb is assumed to be 0.2 milliliters, and the tube's diameter is 0.2 millimeters. Participants clarify that the increase in volume can be calculated using the formula DeltaV = BV(DeltaT) and that the height of mercury rise can be determined by converting the volume increase into height using the tube's cross-sectional area. The final calculation results in a mercury rise of approximately 2.52 centimeters in the tube.
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Homework Statement


I am supposed to find the increase in volume, or more specifically how much mercury will rise inside of a thermometer, given the following values. The volume expansion is 1.8 * 10-4/C(degree Celcius). The temperature increase from 30 to 70 degrees F (already converted the temp increase to 22 C). Finally, I am given that the bulb contains .2 millimeters (it actually does not say .2 millimeters of mercury but I assume that is what they are referring to) of mercury and that the tube has a diameter of .2 millimeters.

Homework Equations


Increase in Volume(DeltaV) = Volume Expansion * Original Volume(V) * The change in Temp(Delta T)
X = 1.8x10-4/C * ? * 22 C

Volume of a Cylinder = pieR2H

The Attempt at a Solution


I have so far been able to form the above, however, I am not sure how to proceed in finding the original volume.
I have the diameter so therefore I have the radius, however, I do not have the height and i do not understand how knowing that the bulb contains 0.2mm of mercury helps in finding the answer. I am tempted to try using the amount in the bulb as the height, or even just multiplying the 2 together, however I do not see how that helps me get any closer to finding the volume. Thanks in advance for any help and I feel like I am missing something very simple.

Original Problem Statement:
We want to construct a thermometer using mercuyr. As the mercury in the bulb is heated, it expands and moves up the thin capillary tube connected to the bulb. The symbol used for the coefficient of of volume expansion of a substance due to ta temperature increase is B(sigma). it is used in the following equation: DeltaV = BV(DeltaT)

here, DeltaV is the increase in volume, V is the original volume, and DeltaT is the temperature increase. The value of B is 1.8x10-4[1/degree Celsius]. If the bulb contains 0.2 milliliters and the tupe has a diameter of 0.2 milimeters, how much will the mercury rise in the tube in units of centimeters if we increse the temperature from 30 degrees F to 70 degrees F?
 
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Are you sure that the bulb information was not its volume in milliliters? Or perhaps cubic mm? One way or another you'll need to have the volume of the mercury in the bulb, since that's what's going to be expanding with the temperature change.

Can you share word for word the text of the problem dealing with the bulb and its contents?
 
We want to construct a thermometer using mercuyr. As the mercury in the bulb is heated, it expands and moves up the thin capillary tube connected to the bulb. The symbol used for the coefficient of of volume expansion of a substance due to ta temperature increase is B(sigma). it is used in the following equation: DeltaV = BV(DeltaT)

here, DeltaV is the increase in volume, V is the original volume, and DeltaT is the temperature increase. The value of B is 1.8x10-4[1/degree Celsius]. If the bulb contains 0.2 milliliters and the tupe has a diameter of 0.2 milimeters, how much will the mercury rise in the tube in units of centimeters if we increse the temperature from 30 degrees F to 70 degrees F?
 
gneill said:
Are you sure that the bulb information was not its volume in milliliters? Or perhaps cubic mm? One way or another you'll need to have the volume of the mercury in the bulb, since that's what's going to be expanding with the temperature change.

Can you share word for word the text of the problem dealing with the bulb and its contents?
I saw that i misread it for millimeters instead of millimeters, but even when I plug that into the eqaution and get .000792cm3 for the increase in volume I don't see how that is supposed to help me get how much the mercury rose. I took the cubed root and got .09205213cm, but that doesn't seem right at all.
 
Supposing that the bulb itself does not change dimensions (maybe it's made of pyrex glass), and only the mercury expands, where will the excess mercury go? What length of tubing will it occupy knowing its shape?
 
gneill said:
Supposing that the bulb itself does not change dimensions (maybe it's made of pyrex glass), and only the mercury expands, where will the excess mercury go? What length of tubing will it occupy knowing its shape?
It will shoot up the tube, but we don't know the length of the tubing, only its diameter. I suppose that possibly the increase in volume (.000792cm3) could represent the height since it is shooting through the tube, but we would have to get the unit in centimeters if it were to make any sense with the Volume of a cylinder. Then again, we are not concerned with the volume, just how much the mercury rises in the tube in units of centimeters. Since I got the increase in volume to be .000792cm3 I suppose that does represent the height once you take the cubed root of it. Since volume is LxWxH essentially. Wati... do i use V = pier^2H and plug in .000792cm^3 as the Volume, and the diameter as the radius to find the height? I think I do... yeah I think that makes sense, since the mercury is shooting into the portion of the tube where the diameter is known.
 
I think you're starting to put it together. Carry on...
 
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gneill said:
I think you're starting to put it together. Carry on...
Diameter was .2mm convert that to radius which is .1mm convert that to cm which is .01cm, square that which is .0001cmm.
then I take the volume over the radius^2*pie (.000792cm3/(pie)(.01cm2)) and get 2.52cm. Thanks a bunch for the help and also thanks for guiding me along, self-discovery is a beautiful thing. I love these Eureaka moments almost as much as I love the ones I have in my programming classes. Cheers much thanks mate.
 
You're very welcome.
 
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