Physics for Scientists and Engineers With Modern Physics
Physics for Scientists and Engineers With Modern Physics
9th Edition
ISBN: 9781133953982
Author: SERWAY, Raymond A./
Publisher: Cengage Learning
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Chapter 19, Problem 21P

(a)

To determine

The amount of turpentine overflows.

(a)

Expert Solution
Check Mark

Answer to Problem 21P

The amount of turpentine overflows is 99.4cm3_.

Explanation of Solution

Write the equation of volume that overflows.

  (ΔV)=(β3α)Vi(T2T1)                                                                              (I)

Here, (ΔV) is the amount of overflow liquid, β is the volume expansion coefficient, α is the linear expansion coefficient, Vi is the initial volume, T1 is the initial temperature and T2 is the final temperature.

Conclusion:

Substitute, 9.00×104/°C for β, 24.0×106/°C for α, 2.000L for Vi, 20.0°C for T1 and 80.0°C for T2 in equation (I) to find (ΔV).

  (ΔV)=([9.00×104/°C]3[24.0×106/°C])(2.000L)(80.0°C20.0°C)=([9.00×104/°C]3[24.0×106/°C])(2.000L)(60.0°C)=99.4L=99.4cm3

Thus, the amount of turpentine overflows is 99.4cm3_.

(b)

To determine

The remaining volume of the turpentine in the cylinder.

(b)

Expert Solution
Check Mark

Answer to Problem 21P

The remaining volume of the turpentine in the cylinder is 2.01L_.

Explanation of Solution

Write the equation of remaining volume of the liquid in the cylinder.

  Vr=Vi+3αVi(T2T1)                                                                                   (II)

Here, Vr is the remaining volume of liquid.

Conclusion:

Substitute, 24.0×106/°C for α, 2.000L for Vi, 20.0°C for T1 and 80.0°C for T2 in equation (I) to find Vr.

  Vr=(2.000L)+3(24.0×106/°C)(2.000L)(80.0°C20.0°C)=(2.000L)+3(24.0×106/°C)(2.000L)(60.0°C)=2.01L

Thus, the remaining volume of the turpentine in the cylinder is 2.01L_.

(c)

To determine

The distance at which the cylinder’s rim does the turpentine’s surface recede.

(c)

Expert Solution
Check Mark

Answer to Problem 21P

The distance at which the cylinder’s rim does the turpentine’s surface recede is 0.998cm_.

Explanation of Solution

Write the equation for the distance.

  d=ph                                                                                      (III)

Here, d is the distance, p is the percentage of the empty cylinder and h is the depth of the cylinder.

Write the expression for the percentage.

  p=(ViV'Vi)×100%                                                                              (IV)

Here, V' is the volume of the liquid in the cylinder after it cools back to 20°C.

Write the expression for the volume of the liquid in the cylinder after it cools back to 20°C.

  V'=V(1+β(T2T1))                                                                            (V)

Rewrite the expression for the distance from equation (III) by using (IV) and (V).

  d=[(Vi[V(1+β(T2T1))]Vi)×100%]h                                   (VI)

Conclusion:

Substitute, 9.00×104/°C for β, 2.01L for V, 2.000L for Vi, 80.0°C for T1, 20.0cm for h and 20.0°C for T2 in equation (I) to find (ΔV).

  d=[((2.000L)[(2.01L)(1+(9.00×104/°C)(20.0°C80.0°C))](2.000L))×100%](20.0cm)=(4.99%)(20.0cm)=0.998cm

Therefore, the distance at which the cylinder’s rim does the turpentine’s surface recede is 0.998cm_.

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Chapter 19 Solutions

Physics for Scientists and Engineers With Modern Physics

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