A hollow aluminum cylinder 15.0 cm deep has an internal capacity of 2.000 L at 2

Question

A hollow aluminum cylinder 15.0 cm deep has an internal capacity of 2.000 L at 23.0 degree C. It is completely filled with turpentine at 23.0 degree C. The turpentine and the aluminum cylinder are then slowly warmed together to 91.0 degree C. (The average linear expansion coefficient for aluminum is 24 times 10^-6 degree C^-1, and the average volume expansion coefficient for turpentine is 9.0 times 10^-4 degree C^-1.) How much turpentine overflows? What is the volume of turpentine remaining in the cylinder at 91.0 degree C? (Give your answer to at least four significant figures.) If the combination with this amount of turpentine is then cooled back to 23.0 degree C, how far below the cylinder's rim does the turpentine's surface recede? cm

Explanation / Answer

a)

when the entire system is heated to 91deg the volume of the aluminum container is

vf = vi * ( 1 + 3 * alpha * deltaT)

vf = 2 * ( 1 + 3 * 24 * 10^-6 * 68deg )

vf = 2.009792 L

the volume of the turpentine is

vf = vi * ( 1 + beta * deltaT)

vf = 2 * ( 1 + 9 * 10^-4 * 68 )

vf = 2.1224 L

the overflow is = Vturpentine - Valuminium

= 2.1224 - 2.009792 = 0.1126 L

b)

the turpentine volume remaining is the same as the aluminium container volume which is 2.009792 L

c)

vf = vi * ( 1 + beta * deltaT)

vf = 2.009792* ( 1 + ( 9 * 10^-4 * -68))

vf = 1.886 L

this is the following fraction of the volume of the container

Vturpentine / Vcontainer = 1.886 / 2 = 0.943

it is also the following fraction of the 15 cm height of the container

h(turpentine) = 0.943 * 15 = 14.145 cm

form the top is = 15 - 14.145 = 0.855 cm

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