A hollow aluminum cylinder 21.0 cm deep has an internal capacity of 2.000 L at 2

Question

A hollow aluminum cylinder 21.0 cm deep has an internal capacity of 2.000 L at 22.0^degreeC. It is completely filled with turpentine at 22.0^degreeC. The turpentine and the aluminum cylinder are then slowly warmed together to 76.0^degreeC. (The average linear expansion coefficient for aluminum is 24 times 10-6^degreeC-1, and the average volume expansion coefficient for turpentine is 9.0 times 10-4^degreeC-1.) How much turpentine overflows? What is the volume of turpentine remaining in the cylinder at 76.0^degreeC? (Give your answer to at least four significant figures.) If the combination with this amount of turpentine is then cooled back to 22.0^degreeC, how far below the cylinder's rim does the turpentine's surface recede?

Explanation / Answer

1) We assume, before heating, that the aluminum cylinder and the turpentine are both at temperature 22.0 C.

A. overflow = Vturpentine - Valuminium = 2*(1 + 9*10^-4*54) - 2*(1 + 3*24*10^-6*54) = 0.0894 L

B. The turpentine volume remaining is the same as the aluminum container volume which is 2.0077 L.

C. cooling turpentine then Vt = 2.0077*(1 + 9*10^-4*(-54)) = 1.9101 L

fraction of the volume of the container. = 1.9101/2 = 0.95505

hturpentine = 0.95505*21 = 20.05605 cm

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