A hollow aluminum cylinder 17.0 cm deep has an internal capacity of 2.000 L at 1

Question

A hollow aluminum cylinder 17.0 cm deep has an internal capacity of 2.000 L at 16.0°C. It is completely filled with turpentine at 16.0°C. The turpentine and the aluminum cylinder are then slowly warmed together to 72.0°C. (The average linear expansion coefficient for aluminum is 2.4 10-5°C-1, and the average volume expansion coefficient for turpentine is 9.0 10-4°C-1.)
(a) How much turpentine overflows?
cm3

(b) What is the volume of turpentine remaining in the cylinder at 72.0°C? (Give your answer to four significant figures.)
WebAssign will check your answer for the correct number of significant figures. cm3

(c) If the combination with this amount of turpentine is then cooled back to 16.0°C, how far below the cylinder's rim does the turpentine's surface recede?
cm

Explanation / Answer

(a) Volume of the cylinder is

V = 2 L = 2000 cm^3

For isotropic materials, the coefficient of volume expansion V and linear expansion L are related as,

V = 3L

The new volume of the cylinder is,

V = V0(1 + 3*LT) = 2000*(1 + 3*2.4e-5*(72-16)) = 2008.1 cm^3

Volume of turpentine at 72 C is,

Vt = V0(1 + VT) = 2000*(1 + 9e-4*(72-16)) = 2100.8 cm^-3

The excess volume of turpentine which overflows is,

Vt - V = 2100.8 - 2008.1 = 92.7 cm^3 = 92.7 mL

(b) The turpentine remaining in the cylinder should equal the volume of the cylinder at 72 C,

i.e. V = 2008.1 cm^3

(c) Cooling this back to 16 C, the volume of the turpentine will now be,

Vf = V(1 + VT) = 2008.1*(1 + 9e-4*(16-72)) = 1906.9 cm^3

Volume of the turpentine inside the cylinder is r^2h

Since, r is the same as before,

V1/V2 = h1/h2

h2 = V2/V1*h1 = 1906.9/2000*17 = 16.2086 cm

Turpentine's surface recedes 17-16.2086 = 0.7914 cm below the rim.

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