Physics for Scientists and Engineers: Foundations and Connections
Physics for Scientists and Engineers: Foundations and Connections
1st Edition
ISBN: 9781133939146
Author: Katz, Debora M.
Publisher: Cengage Learning
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Question
Chapter 21, Problem 77PQ

(a)

To determine

The work done on the lead block in the process.

(a)

Expert Solution
Check Mark

Answer to Problem 77PQ

The work done on the lead block in the process is 0.279J_.

Explanation of Solution

Given that the mass of the lead block is 10.0kg, the temperature of the block is increased from 18.0°C to 55.0°C, the pressure remains constant at 1.00atm. The coefficient of linear expansion of lead is 28.0×106°C1.

Write the expression for the work done at constant pressure.

  W=PΔV                                                                                                        (I)

Here, W is the work done at constant pressure, P is the pressure, and ΔV is the change in volume of the block.

Write the expression for the change in the volume of lead block when its temperature changes.

  ΔV=βV(TfTi)                                                                                            (II)

Here, β is the coefficient of volume expansion, V is the initial volume of the block, Tf is the final temperature of the block, and Ti is the initial temperature of the block.

Write the expression for the coefficient of volume expansion in terms of the coefficient of linear expansion.

  β=3α                                                                                                                    (III)

Here, α is the coefficient of linear expansion of the lead block.

Use expression (III) in (II).

  ΔV=(3α)V(TfTi)                                                                                       (IV)

Write the expression for the volume of the lead block in terms of its mass and density.

  V=mρ                                                                                                              (V)

Here, m is the mass of the block, and ρ is the density of the block.

Use expression (V) in (IV).

  ΔV=mρ(3α)(TfTi)                                                                                 (VI)

Use expression (VI) in (I).

  W=P[mρ(3α)(TfTi)]                                                                             (VII)

Conclusion:

Substitute 1.00atm for P, 28.0×106°C1 for α, 10.0kg for m, 11.3×103kg/m3 for ρ, 55.0°C for Tf, and 18.0°C for Ti in equation (VII) to find W.

  W=(1.00atm×1.013×105N/m21.00atm)[3(28.0×106°C1)(10.0kg11.3×103kg/m3)(55.0°C18.0°C)]=0.279J

Therefore, the work done on the lead block when the temperature increases is 0.279J_.

(b)

To determine

The amount of energy added to the block by heat in the process.

(b)

Expert Solution
Check Mark

Answer to Problem 77PQ

The amount of energy added to the block by heat in the process is 47.4kJ_.

Explanation of Solution

Write the expression for the heat added up into the block while heating.

  Q=mc(TfTi)                                                                                           (VIII)

Here, Q is the energy or heat added to the block, and c is the specific heat capacity of lead.

Conclusion:

Substitute 128J/kg°C for c, 10.0kg for m, 55.0°C for Tf, and 18.0°C for Ti in equation (VIII) to find Q.

  Q=(128J/kg°C)(10.0kg)(55.0°C18.0°C)=0.474×102J=0.474×105J×1kJ1000J=47.4kJ

Therefore, the amount of energy added to the block by heat is 47.4kJ_.

(c)

To determine

The change in the internal energy of the block during the process.

(c)

Expert Solution
Check Mark

Answer to Problem 77PQ

The change in thermal energy of the block during the process is 47.4kJ_.

Explanation of Solution

The change in thermal energy of the block is equal to the sum of heat added up to the system and the work done.

Write the expression for the change in thermal energy of the block.

  ΔEth=Q+W                                                                                                (IX)

Here, ΔEth is the change in the thermal energy of block.

Conclusion:

Substitute 47.4kJ for Q, and 0.279J for W in equation (IX) to find ΔEth.

  ΔEth=47.4kJ0.279J×1kJ1000J=47.4kJ

Therefore, the change in thermal energy of the block during the process is 47.4kJ_.

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Chapter 21 Solutions

Physics for Scientists and Engineers: Foundations and Connections

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