Question
A hollow aluminum cylinder 15.5 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 76.0°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? __86.1 cm3
(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
Explanation / Answer
Given data: The depth of the aluminum cylinder is 15.5 cm The volume of the cylinder,
V = 2.000 L = 2000 cm3 The initial temperature,
T1 = 24oC The final temperature,
T2 = 76oC The average linear expansion coefficient for aluminum,
Al = 2.4x10-5°C1 The average volume expansion coefficient for turpentine,
= 9.0x10-4°C1 --------------------------------------------------------------------------------------- Solution: a) From the volume expansion, we have
V = VT-
V(3
)
T = [
-3
]
VT = [(9.0x10-4°C1) - (3)(2.4x10-5°C1)](2000 cm3)(76oC-24oC) = 86.112 cm3
= 86.1 cm3 ------------------------------------------------------------------------------------------- b) The volume of the turpentine can be calculated as
VT =
VT = (9.0x10-4°C1) (2000 cm3)(76oC-24oC) = 93.6 cm3 = (9.0x10-4°C1) (2000 cm3)(76oC-24oC) = 93.6 cm3 The remaining volume of the turpentine can be calculated as
V' = 2000 cm3 + 93.6 cm3 - 86.1 cm3
= 2007.5 cm3 = 2007.5 cm3 ------------------------------------------------------------------------------------------- c) The new volume for the turpentine can be calculated as
V = 2000 cm3 + (9.0x10-4°C1)(2000 cm3)(76oC-24oC) = 2093.6 cm3 Therefore, the fractional lost of the volume is 86.1 cm3/2093 cm3 = 4.1125x10-2 The above fracton of the cylinder's depth is empty upon cooling, (4.1125x10-2)(15.5 cm) =
0.637 cm = 0.64 cm = 0.64 cm