A hollow aluminum cylinder 15.5 cm deep has an internal capacity of 2.000 L at 2

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       

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