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The mercury in a household glass-tube thermometer has a volume of 500mm^(3)(=5.0xx10^(-7)m^(3)) at T=19^(@)C. The hollow column within which the merucry can rise or fall has a cross-sectional area of 0.1mm^(2)(=1.0xx10^(-7)m^(2)). Ignoring the volume expansion of the glass, how much will the mercury rise in the thermometer when its temperature is 39^(@)C ? (The coefficient of volume expansion of mercury is 1.8xx10^(-4)//""^(@)C.) |
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Answer» SOLUTION :First let's figure out by how much the volume of the MERCURY increases. `DeltaV=betaV_(0)DeltaT=(1.8xx10^(-4))/(""^(@)C)(5.0xx10^(-7)m^(3))(39^(@)C-19^(@)C)=1.8xx10^(-9)m^(3)` Now, since volume=cross-sectional area `xx`height, the CHANGE in height of the mercury column will be `Deltah=(DeltaV)/(A)=(1.8xx10^(-9)m^(3))/(1.0xx10^(-7)m^(2))=1.8xx10^(-2)m=1.8m` |
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