A hollow aluminum cylinder 19.0 cm deep has an internal capacity of 2.000 L at 23.0°C. It is completely filled with turpentine at 23.0°C. The turpentine and the aluminum cylinder are then slowly warmed together to 91.0°C. (The average linear expansion coefficient for aluminum is 2.4 x10^-5/°C, and the average volume expansion coefficient for turpentine is 9.0 x10^-4/°C.)
(a) How much turpentine overflows?
in cm^3?
(b) What is the volume of turpentine remaining in the cylinder at 91.0°C? (Give you answer to four significant figures.)
in cm^3?
(c) If the combination with this amount of turpentine is then cooled back to 23.0°C, how far below the cylinder’s rim does the turpentine’s surface recede?
in cm?


  1. Answer:

    (a) 0.1134 L

    (b) 2.009 L

    (c) 0.07 cm from the top


    The volume of the aluminum container and the turpentine is 2.000 L at  23 °C.

    (a) When the entire system is heated to from 23 °C to 91 °C. The volume of the aluminum container is:

    Vf = Vi*(1 + 3α*ΔT)


    Vi is the initial volume, Vf the final volume, α is the linear expansion coefficient and ΔT is the temperature change.

    Vf = 2*[1 + 3*2.4 *10^-5*(91 – 23)] = 2.009 L

    The volume of the turpentine is

    Vf = Vi*(1 + β*ΔT)


    β is the  volume expansion coefficient

    Vf = 2*[1 + 9.0*10^-4*(91 – 23)] = 2.1224 L

    The overflow is:

    2.1224  – 2.009 = 0.1134 L

    (b) the volume of turpentine remaining in the cylinder at 91 °C is the same volume of the hollow aluminum cylinder, that is, 2.009 L

    (c) Cooling the system back to 23 °C means the aluminum container is back to 2.000 L. The  volume of turpentine is:

    Vf = 2.1224*[1 + 9.0*10^-4*(23-91)] = 1.9925 L

    The following proportion must be  satisfied:

    Volume of turpentine / Volume of container = deep turpentine / deep of container

    1.9925 / 2.000 = deep turpentine/19

    deep turpentine = (1.9925 / 2.000)*19 = 18.93 cm

    This is 19 – 18.93 = 0.07 cm from the top

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