   Chapter 10, Problem 21P

Chapter
Section
Textbook Problem

A hollow aluminum cylinder 20.0 cm deep has an internal capacity of 2.000 L at 20.0°C. It is completely filled with turpentine at 20.0°C. The turpentine and the aluminum cylinder are then slowly warmed together to 80.0°C. (a) How much turpentine overflows? (b) What is the volume of the turpentine remaining in the cylinder at 80.0°C? (c) If the combination with this amount of turpentine is then cooled back to 20.0°C, how far below the cylinder’s rim does the turpentine’s surface recede?

(a)

To determine
The volume of turpentine that has overflowed.

Explanation

Given Info: The length of the hollow cylinder (l) is 20.0 cm. The capacity of the cylinder ( V0 ) is 2.000 L at T1=20.0οC . The temperature is changed to T2=80.0οC . The co-efficient of volume expansion of turpentine is βT=9.0×104/οC . The co-efficient of volume expansion of aluminum is βAl=72×106/οC .

Formula to calculate the volume of turpentine that has overflowed is,

Vof=ΔVTΔVAl       (I)

• ΔVT is the change in volume of turpentine.
• ΔVAl is the change in volume of Aluminum.

Formula to calculate the change in volume of turpentine is,

ΔVT=V0βT(T2T1)       (II)

Formula to calculate the change in volume of aluminum is,

ΔVAl=V0βAl(T2T1)       (III)

Substitute Equations (II) and (III) in (I).

Vof=V0(T2T1)(βTβAl)

Substitute 2

(b)

To determine
The volume of turpentine remaining.

(c)

To determine
The distance between the rim and the surface of turpentine (h).

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