Giancoli 7th Edition textbook cover
Giancoli's Physics: Principles with Applications, 7th Edition
15
The Laws of Thermodynamics
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15-1 and 15-2: First Law of Thermodynamics
15-3: Human Metabolism
15-5: Heat Engines
15-6: Refrigerators, Air Conditioners, Heat Pumps
15-7: Entropy
15-10: Statistical Interpretation
15-11: Energy Resources

Question by Giancoli, Douglas C., Physics: Principles with Applications, 7th Ed., ©2014, Reprinted by permission of Pearson Education Inc., New York.
Problem 41
Q

What is the change in entropy of 1.00 m31.00 \textrm{ m}^3 of water at 0C0 ^\circ \textrm{C} when it is frozen to ice at 0C0 ^\circ \textrm{C}?

A
1.22×106 J/K-1.22 \times 10^6 \textrm{ J/K}
Giancoli 7th Edition, Chapter 15, Problem 41 solution video poster
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VIDEO TRANSCRIPT

This is Giancoli answers with Mr. Dychko. The change in entropy of water as it changes from water to ice all at zero degrees Celsius is gonna be the heat that the water absorbs divided by its temperature. Now it's not absorbing any heat it's actually losing heat and so that means we have a negative sign there and it's going to lose the amount of heat due to the latent heat of fusion. So this is the Heat lost just as a result of the phase change from liquid into solid. So we have mass times leading into fusion so the mass is going to be its density times its volume and we're given the volume of one cubic meter times up by one times ten to the three kilograms per cubic meter. The negative is outside there because it's losing heat, and times by 3.33 times ten to the five joules per kilogram latent heat of fusion divided by zero degrees Celsius converted into Kelvin by adding 273 and you get about negative 1.22 times ten to the six Joules per Kelvin.

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