Another problem on Thermal Expansion

AI Thread Summary
The discussion revolves around calculating the overflow of mercury from a beaker due to thermal expansion when the temperature rises from -10 to 30 degrees Celsius. Key points include the need to determine the expansion of the beaker, lead sphere, and mercury using their respective coefficients of thermal expansion. Participants emphasize calculating the initial volume of the beaker by adding the volume of mercury to the volume of the lead sphere. There is a debate on whether the expansion of the beaker will prevent overflow, with suggestions to focus on the actual volume changes. Ultimately, the problem requires precise calculations to ascertain how much mercury will overflow.
andrew410
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A beaker made of ordinary glass contains a lead sphere of diameter 4 cm firmly attached to its bottom. At a uniform temperature of -10 degrees Celsius, the beaker is filled to the brim with 118 cm^3 of mercury, which completely covers the sphere. How much mercury overflows from the beaker if the temperature is raised to 30 degrees Celsius?

Things I know so far:
average linear expansion coefficient of glass: 9*10^-6
average linear expansion coefficient of lead: 29*10^-6
average volume expansion coefficient of mercury: 1.82*10^-4
\Delta V=\beta V_{i}\Delta T

Wouldn't nothing spill because when the beaker's temperature rises then the beaker expands more.
If I'm wrong then could someone lead me into the right direction?
Thanks in advance! :)
 
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andrew410 said:
Wouldn't nothing spill because when the beaker's temperature rises then the beaker expands more.
Don't guess, figure it out! How much does the beaker expand? The lead sphere? So... how much volume is left available in the beaker? How much does the mercury expand? So... how much must overflow?

Hint: Given the linear expansion coefficient, how can you find the volume expansion coefficient?
 
How would you get the height of the beaker? I assume that the diameter of the beaker is approximately 4 cm because the sphere is firmly attached to the bottom of the beaker. Or is this the wrong way to get the initial volume of the beaker?
 
You have the initial volume of mercury and the dimensions of the sphere. That's all you need to find the volume of the beaker.
 
So the volume of the beaker is volume of mercury + volume of the sphere?
 
andrew410 said:
So the volume of the beaker is volume of mercury + volume of the sphere?
Right. At least initially.
 
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