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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Chapter 21, Problem 84PQ

(a)

To determine

The amount of heat that need to flow into Zan to completely melt his body.

(a)

Expert Solution
Check Mark

Answer to Problem 84PQ

The amount of heat that need to flow into Zan to completely melt his body is 2.33×107J_.

Explanation of Solution

Zan is initially existing as ice at temperature 10.00°C (263.15K). The final state is water at 0°C (273.15K). The mass of the body is 65.88kg, the specific heat of ice is 2100J/kgK, and the latent heat of fusion of ice is 3.33×105J/kg.

In order Zan to completely melt his body, first the 10.00°C ice has to become 0°C ice, then the 0°C ice has to become 0°C water.

Write the expression for the heat corresponding to temperature change.

  Q1=mc(TfTi)                                                                                                     (I)

Here, Q1 is the heat, m is the mass, c is the specific heat, Tf is the final temperature, and Ti is the initial temperature.

Write the expression for the heat required for phase transition from solid to liquid.

  Q2=mLF                                                                                                               (II)

Here, Q2 is the heat required for phase transition, and LF is the latent heat of fusion.

Write the expression for the total heat absorbed by the body in the process.

  Qtot=Q1+Q2                                                                                                        (III)

Conclusion:

Substitute 65.88kg for m, 2100J/kgK for c, 273.15K for Tf and 263.15K for Ti in equation (I) to find Q1.

  Q1=(65.88kg)(2100J/kgK)(273.15K263.15K)=1.38×106J

Substitute 65.88kg for m, 3.33×105J/kg for LF in equation (II) to find Q2.

  Q2=(65.88kg)(3.33×105J/kg)=2.19×107J

Substitute 1.38×106J for Q1, and 2.19×107J for Q2 in equation (III) to find Qtot.

  Qtot=1.38×106J+2.19×107J=2.33×107J

Therefore, the amount of heat that need to flow into Zan to completely melt his body is 2.33×107J_.

(b)

To determine

The amount of additional heat required to make the water form of Zan’s body to become vapor at 100°C.

(b)

Expert Solution
Check Mark

Answer to Problem 84PQ

The amount of additional heat required to make the water form of Zan’s body to start vaporizing at 100°C is 2.76×107J_.

Explanation of Solution

Initial temperature of water is 0°C (273.15K), its mass is 65.88kg, and the final temperature is 100.0°C (373.15K).

Write the expression for the heat corresponding to the temperature change.

  Q=mc(TfTi)

Conclusion:

Substitute 65.88kg for m, 2100J/kgK for c, 373.15K for Tf and 273.15K for Ti in the above equation to find Q.

  Q=(65.88kg)(2100J/kgK)(373.15K273.15K)=2.76×107J

Therefore, the amount of additional heat required to make the water form of Zan’s body to start vaporizing at 100°C is 2.76×107J_.

(c)

To determine

The amount of additional heat required to make the 100°C water form of Zan’s body to vapor at 100°C.

(c)

Expert Solution
Check Mark

Answer to Problem 84PQ

The amount of additional heat required to make the 100°C water form of Zan’s body to vapor at 100°C is 1.486×108J_.

Explanation of Solution

The mass of water is 65.88kg, it undergoes phase transition from liquid to vapor at 100.0°C.

Write the expression for the heat corresponding to the phase transition from liquid to vapor.

  Q=mLV

Here, LV is the latent heat of vaporization (for water LV=2.256×106J/kg).

Conclusion:

Substitute 65.88kg for m, and 2.256×106J/kg for LV in the above equation to find Q.

  Q=(65.88kg)2.256×106J/kg=1.486×108J

Therefore, the amount of additional heat required to make the 100°C water form of Zan’s body to vapor at 100°C is 1.486×108J_.

(d)

To determine

The average kinetic energy of one of water vapor molecule that make up Zan.

(d)

Expert Solution
Check Mark

Answer to Problem 84PQ

The average kinetic energy of one of water vapor molecule that make up Zan is 6.88×1021_.

Explanation of Solution

The temperature of the water vapor is 100.0°C (373.15K).

Write the expression for the average kinetic energy of one molecule of water vapor.

  Kav=32kBT

Here, Kav is the average kinetic energy, kB is the Boltzmann constant, and T is the temperature.

Conclusion:

Substitute 1.23×1023J/K for kB, and 373.15K for T in the above equation to find Kav.

  Kav=32(1.23×1023J/K)(373.15K)=6.88×1021J

Therefore, the average kinetic energy of one of water vapor molecule that make up Zan is 6.88×1021_.

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Chapter 21 Solutions

Physics for Scientists and Engineers: Foundations and Connections

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