Physics: Principles with Applications
Physics: Principles with Applications
6th Edition
ISBN: 9780130606204
Author: Douglas C. Giancoli
Publisher: Prentice Hall
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Question
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Chapter 15, Problem 35P
To determine

To find: the change in entropy

Expert Solution & Answer
Check Mark

Answer to Problem 35P

The change in entropy is 1.5×103J/K

Explanation of Solution

Given:

  250 gram of steam condensed to water T=100°C

Formula used:

The expression of change in entropy,

  ΔS=dSAnd, dS=δqT

Then,

  ΔS=δqT=QT

Here Q is heat energy and T is temperature.

Heat energy for evaporation

  Q=mLvap

Here m is mass and Lvap is latent heat of evaporation.

Then change in entropy is,

  ΔS=mLveT

Calculation:

Now substitute 250g for m,22.6×105J/kg for Lvap and 100C for T in entropy change equation.

  ΔS=(250g)(22.6×105J/kg)100C=(250g)(103kg1g)(22.6×105J/kg)(100+273)K=(250×103kg)(22.6×105J/kg)(373)K=1514.75J/K

Conclusion:

Hence the change in entropy is 1.5×103J/K

Chapter 15 Solutions

Physics: Principles with Applications

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