EBK PHYSICS FOR SCIENTISTS AND ENGINEER
EBK PHYSICS FOR SCIENTISTS AND ENGINEER
16th Edition
ISBN: 8220100546716
Author: Katz
Publisher: CENGAGE L
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Chapter 22, Problem 71PQ

A system consisting of 10.0 g of water at a temperature of 20.0°C is converted into ice at –10.0°C at constant atmospheric pressure. Calculate the total change in entropy of the system. Assume the specific heat of water is 4.19 × 103 J/(kg · K), the specific heat of ice is 2.10 × 103 J/(kg · K), and the latent heat of fusion is 3.33 × 105 J/kg.

Expert Solution & Answer
Check Mark
To determine

The total change in entropy of the system.

Answer to Problem 71PQ

The total change in entropy of the system is 15.9J/K.

Explanation of Solution

Write the expression to calculate the entropy change when 293K or 20°C water convert to 273K or 0°C.

  Δs=mcpln(TfTi)                                                                                                       (I)

Here, Δs is the change in entropy, m is the mass of the water, cp is the specific heat capacity of the water, Tf is the final temperature and Ti is the initial temperature.

Write the expression to calculate the entropy change for the transformation of phase from ice to water.

  Δs1=mLfT                                                                                                              (II)

Here, Δs1 is the entropy change of the phase transformation, T is the temperature of the ice and Lf is the latent heat of fusion of the ice.

Write the expression to calculate the change in entropy of the ice to decrease the temperature from 273K or 0°C to 263K or 10.0°C.

  Δs2=mcpln(TfTi)                                                                                                 (III)

Here, Δs2 is the change in entropy for the ice, cp is the specific heat of ice, Tf is the final temperature and Ti is the initial temperature.

Write the expression to calculate the entropy change for the whole process.

  Δstot=Δs+Δs1+Δs2                                                                                              (IV)

Here, Δstot is the total entropy change of the whole process.

Conclusion:

Substitute 10.0g for m, 4.19×103J/kgK for, cp, 293K for Ti and 273K for Tf in the above equation to calculate Δs.

  Δs=(10.0g)(103kg1g)(4.19×103J/kgK)ln(273K293K)=2.96J/K

Substitute 10.0g for m, 3.33×105J/kg for Lf and 273K for T in the equation (II) to calculate Δs1.

  Δs1=(10.0g(103kg1g))(3.33×105J/kg)273K=12.2J/K

Substitute 10.0g for m, 2.10×103J/kgK for cp, 263K for Tf and 273K for Ti in the above equation (III) to calculate . Δs2

  Δs2=(10.0g(103kg1g))(2.10×103J/kgK)ln(263K273K)=0.78J/K

Substitute 2.96J/K for Δs, 12.2J/K for Δs1 and 0.78J/K for Δs2 in the above equation to calculate Δstot.

  Δstot=2.96J/K+(12.2J/K)+(0.78J/K)=15.9J/K

Therefore, the total change in entropy of the system is 15.9J/K.

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Chapter 22 Solutions

EBK PHYSICS FOR SCIENTISTS AND ENGINEER

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