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A hollow aluminium cylinder 20.0 cm deep...

A hollow aluminium cylinder 20.0 cm deep has an internal capacity of 2.000 L at `20.0^@` C. It is completely filled with tupentine and then slowly warmed to `80.0^@`C a. How much turpentine overflows? B. If the cylindre is then cooled back to `20.0^@`C, how far below the cylinder's rim dows the tupentine's surface recede?

Text Solution

Verified by Experts

We guess that vertical cubic centimetres of turpentine will overflow, and that the liquid level will drop about a centimetre. We will use the definition of the volume expansion coefficient. Both the liquid and the container wxpand. We will need to reason carefully about original, intermediate and final volumes of each.
When the temperature is increased from `20.0^@`C to `80.0^@`C both the cylinder and the turpentine in volume by `DeltaV=gammaVDeltaT`
a. The overflow is `V_(over)=Delta_(turp)-V_(A1)`

`V_(over)=(gammaVDeltaT)_(turp)-(gammaVDeltaT)_(A1)=V_iDeltaT(gamma_(tarp)-3alpha_(A1))`
`V_("over")=(2.000L)(60.0^@C)(9.00xx10^(-4)//^(@)C-0.720xx10^(-4)//^(@)C)`
`=0.099 4L`
After warming, the whole volume of the turpentine is
`V' 2000cm^(3)+(9.00xx10^(-4)//^(@)C)(2000cm^(3))(60.0^@C)=2108cm^(3)`
The fraction lost is `(99.4cm^3)/(2108cm^3)=4.71xx10^-2`
This also is the fraction of the cylinder that will be empty after cooling. Therefore, change in level
`Deltah=(4.71xx10^-2)(20.0cm)=0.943cm`
The change in volume of the container in not negligible, but is `8%` of the change in volume of the turpentine.
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