Question

A hollow aluminum cylinder 23.0 cm deep has an internal capacity of 2.000 L at 24.0°C. It is completely filled with turpentine at 24.0°C. The turpentine and the aluminum cylinder are then slowly warmed together to 86.0°C. (The average linear expansion coefficient for aluminum is 24 ✕ 10⁻⁶°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 86.0°C? (Give your answer to at least four significant figures.)
L

(c) If the combination with this amount of turpentine is then cooled back to 24.0°C, how far below the cylinder's rim does the turpentine's surface recede?
cm

Answer 1

1-23cm -0.23m - 24 c v=2 litres - 2X10-3 md T2 = 86°C 2 = 24X10- B = q X10-4 ,- & Area = Volume = 2x10-3 lengh 0.23 8.6956410-3m² volume of liquid overflows => AV = V(B-34DDT - (2 x 10-3) (9x10-4 - 3624x108) (62) - 1.026724444 - 1.026 72x10 cm (1.626 x 10 cm3] [102.6 cm 3) Vrenang = V - DV = (2x10 3 - = (1-897L 1.02672410-4)