a) Explanation Write the energy balance for the closed system. E in − E out = Δ E system (I) Here, total energy entering is E in , total energy leaving is E out , and change in total energy of a system is Δ E system . Since there is no energy entering and leaving, the value of net energy transfers E in − E out = 0 . Rearrange Equation (1) and substitute 0 for E in − E out . Δ U = 0 Δ U ice + Δ U w = 0 ( [ m c p ( 0 ° C − T 1 ) ] solid + m h i f + [ m c p ( T 2 − 0 ° C ) liquid ] ) ice + ( m c P ( T 2 − T 1 ) ) water = 0 (II) Here, change in internal energy of ice and water is Δ U Ice and Δ U w , mass of copper and water is m ice and m w , initial and final temperature of ice and water is T 1 and T 2 . Conclusion: Refer to Table A-3, “properties of common liquids, solids, and foods”, obtain the following properties of water and ice. c P w = 4.18 kJ / kg ⋅ °C c P i c e = 2.11 kJ / kg ⋅ °C T m e l t i n g = 0 ° C h i f = 333 b) To determine The exergy destroyed during the process.

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ISBN: 9781307434316

Author: CENGEL

Publisher: INTER MCG

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Chapter 8.8, Problem 111RP

To determine

The final equilibrium temperature.

To determine

The exergy destroyed during the process.

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Chapter 8.8, Problem 109RP

Chapter 8.8, Problem 112RP

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THERMODYNAMICS(SI UNITS,INTL.ED)EBOOK>I

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