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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? γ ~ ( t u r p ) = ( 9.00 × 10^ − 4 / ∘ C ) , γ ~ ( a l ) = ( 0.72 × 10^ − 4 / ∘ C )

A

0.11L

B

0.22L

C

0.14L

D

None of the Above

Text Solution

Verified by Experts

The correct Answer is:
A
Volume of overflow
`=Delta V_("turpentine")- Delta V_(Al)= gamma_("turpentine")V_(0) Delta theta- gamma_(Al) V_(0)Delta theta`
`=(gamma_("Turpentine")- 3 alpha_(Al))V_(0) Delta theta= (9.00 xx 10^(-4) -2.16 xx 10^(-4)) V_(0)Delta theta`
`=6.84 xx 10^(-4) xx 2 xx 80L = 0.11L`
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