A hollow aluminum cylinder 18.0 cm deep has an internal capacity of 2.000 L at 2

Question

A hollow aluminum cylinder 18.0 cm deep has an internal capacity of 2.000 L at 23.0°C. It is completely filled with turpentine at 23.0°C. The turpentine and the aluminum cylinder are then slowly warmed together to 89.0°C. (The average linear expansion coefficient for aluminum is 2.4 10-5°C1, and the average volume expansion coefficient for turpentine is 9.0 10-4°C1.)
(a) How much turpentine overflows?

(b) What is the volume of turpentine remaining in the cylinder at 89.0°C? (Give you answer to four significant figures.)

(c) If the combination with this amount of turpentine is then cooled back to 23.0°C, how far below the cylinder's rim does the turpentine's surface recede?

Explanation / Answer

Given

Initial volume of the cylinder and turpentine Vo = 2.000 L = 2000 cm3

Initial temperature To = 23.0oC

Final temperature T = 89.0oC

Coefficient of liner expansion of aluminum = 2.4 x 10-5oC-1

Coefficient of volume expansion of turpentine = 9.0 x 10-4oC-1

Solution

a)

The new volume of the cylinder

VC = Vo[1+3(T-To)]

VC = 2000[1 + 3 x 2.4 x 10-5x (89.0-23.0)]

VC = 2009.504 cm3

The new volume of the turpentine

VT = Vo[1+(T-To)]

VT = 2000[1 + 9.0 x 10-4x (89.0-23.0)]

VT = 2118.8 cm3

since VT is greater than VC the turpentine will overflow. The volume of the excess turpentine

V = VT - VC

V = 2118.8 - 2009.504

V = 109.296 cm3

V = 109.3 cm3 (rounding off to 4 significant figures)

V = 0.1093 L

b)

Since the turpentine is overflowing there will be turpentine inside the container to its fullest fill.

Which means the full volume of the container is filled with turpentine

The volume of turpentine inside container = Vc

VC = 2009.504 cm3

VC = 2.009504 L

VC = 2.01 L

c)

In this scenario

The initial temperature is T = 89oC

The final temperature T’ = 23oC

Initial volume of turpentine VC = 2009.504 cm3+

Final volume of turpentine

VC’ = VC [1 + (T’ - T)]

VC’ = 2009.504 [1 + 9 x 10-4 x (23-89)]

VC’ = 1890.1394624 cm3

the cylinder will shrink back to its original volume Vo

VC’ /Vo = 1890.1394624/2000

r2h’/ r2h = 0.9450697312

h’/h = 0.9450697312

h’ = 0.9450697312 x h

h’ = 0.9450697312 x 20

h’ = 18.901394624 cm

the new height of turpentine inside the chamber is 18.9 cm

h = h – h’

h = 20 – 18.9

h = 1.1 cm

The turpentine is 1.1 cm below the cylinder’s rim

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