A hollow aluminum cylinder 25.0 cm deep has an internal capacity of 2.000 L at 2

Question

A hollow aluminum cylinder 25.0 cm deep has an internal capacity of 2.000 L at 21.0 degree C. It is completely filled with turpentine at 21.0 degree C. The turpentine and the aluminum cylinder are then slowly warmed together to 85.0 degree C. (The average linear expansion coefficient for aluminum is 2.4 x 10-5 degree C_1, and the average volume expansion coefficient for turpentine is 9.0 x 10-4 degree C_1.) How much turpentine overflows? cm3 What is the volume of turpentine remaining in the cylinder at 85.0 degree C? (Give your answer to four significant figures.) cm3 If the combination with this amount of turpentine is then cooled back to 21.0 degree C, how far below the cylinder's rim does the turpentine's surface recede? cm

Explanation / Answer

Following data are given: Volume of turpentine=2 L

Volume ofcanister=2L

Coefficient of volume expansion of turpentine = 9*10^-4

Coefficient of linear expansion of canister = 2.4*10^-5

Coefficient of volume expansion of canister = 3*Coefficient of linear expansion of canister

= 7.2*10^-5

Change in temperature = 85 - 21 = 64 C

a)

Change in volume =InitialVolume*Change in temperature*Coefficient of Expansion

Final volume ofcanister = 2*7.2*10^-5*64 + 2 = 2.009216 L

Final volume ofTurpentine = 2*9*10^-4*64 + 2 = 2.1152 L

Volume of turpentine that overflowed = 2.1152 - 2.009216 =0.105984 L

b)

Volume of turpentine in canister = volume of canister = 2.0092 L

c)

This volume of turpentine at 21 C = 2.0092 - 2.0092*64*9*10^-4 = 1.89347 L

Therefore in 2 L canister height of turpentine = 1.89347*25/2 = 23.668 cm

Therefore turpentine recedes by 25 - 23.668 = 1.332 cm

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