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bacon
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I think that I’m doing this problem correctly, but the answer seems a bit unreasonable. Can someone else check my work?
A thermometer has a quartz body within which is sealed a total volume of 0.400cm^{3} of mercury. The stem contains a cylindrical hole with a bore diameter of 0.10mm. How far does the mercury column extend in the process of rising from 10deg C to 90deg C? Neglect any change in the volume of the quartz.
I first found the change in volume of the mercury.
\Delta V=\beta V_{0}\Delta T
\beta = 182x10^{-6}K^{-1} from a table in the book.
V_{0}=.400cm^{3}
\Delta T=90-10=80 deg
\Delta V=.005824cm^{3}=5.824mm^{3}
This change in volume will fill a portion of the slender cylindrical hole in the quartz.
The volume of a cylinder is V=L\pi \frac{D}{4}^{2} Where L is the length of the cylinder and D is its diameter.
L =\frac{4V}{\pi D^{2}}
V is the \Delta V calculated above and D is given as 0.10mm.
L=741mm
This seems like a very long thermometer.
Thanks in advance for any responses.
A thermometer has a quartz body within which is sealed a total volume of 0.400cm^{3} of mercury. The stem contains a cylindrical hole with a bore diameter of 0.10mm. How far does the mercury column extend in the process of rising from 10deg C to 90deg C? Neglect any change in the volume of the quartz.
I first found the change in volume of the mercury.
\Delta V=\beta V_{0}\Delta T
\beta = 182x10^{-6}K^{-1} from a table in the book.
V_{0}=.400cm^{3}
\Delta T=90-10=80 deg
\Delta V=.005824cm^{3}=5.824mm^{3}
This change in volume will fill a portion of the slender cylindrical hole in the quartz.
The volume of a cylinder is V=L\pi \frac{D}{4}^{2} Where L is the length of the cylinder and D is its diameter.
L =\frac{4V}{\pi D^{2}}
V is the \Delta V calculated above and D is given as 0.10mm.
L=741mm
This seems like a very long thermometer.
Thanks in advance for any responses.