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# SolutionsSet10 - Homework Solutions Set 10 Physics 121 Spring 2008 Problem 1 After the system equilibrates its temperature will be Tf The final

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Homework Solutions Physics 121 Set 10 Spring 2008 - 2 - Problem 2 Since the system is insulated, there will be no exchange of heat with the environment. In order to melt all the ice, a total heat of LM ice is required. The maximum amount of heat that can be released by the water is the amount of heat that is released when the water cools to 0°C. This heat is equal to Q water = c water M water T f ! T water ( ) = ! c water M water T water (note: Q water is negative, indicating that heat is removed from the water). Since the heat released by the water is less than the heat required to melt all the ice, only a fraction of the ice will melt. The amount of ice that will melt, M ice-water , can be calculated easily by requiring that all heat released by the water when it cools to 0°C is used to melt ice: LM ice ! water ! c water M water T water = 0 The amount of ice that will melt is thus equal to M ice ! water = c water M water T water L The total amount of liquid water is thus equal to M liquid = M ice ! water + M water = c water M water T water L + M water
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## This note was uploaded on 10/20/2010 for the course PHY PHY 121 taught by Professor Wolfs during the Spring '08 term at Rochester.

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SolutionsSet10 - Homework Solutions Set 10 Physics 121 Spring 2008 Problem 1 After the system equilibrates its temperature will be Tf The final

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