QuestionJuly 16, 2025

Consider the melting of ice at 10^circ C and 1 atm. For this process, q_(sys)gt 0 What are the signs of Delta S_(sys),Delta S_(surr), and Delta S_(univ) Delta Ssysgt 0,Delta Ssurrlt 0,Delta Sunivgt 0 Delta Ssysgt 0,Delta Ssurrlt 0,Delta Sunivlt 0 Delta Ssyslt 0,Delta Ssurrgt 0,Delta Sunivgt 0 Delta Ssysgt 0,Delta Ssurrgt 0,Delta Sunivgt 0

Consider the melting of ice at 10^circ C and 1 atm. For this process, q_(sys)gt 0 What are the signs of Delta S_(sys),Delta S_(surr), and Delta S_(univ) Delta Ssysgt 0,Delta Ssurrlt 0,Delta Sunivgt 0 Delta Ssysgt 0,Delta Ssurrlt 0,Delta Sunivlt 0 Delta Ssyslt 0,Delta Ssurrgt 0,Delta Sunivgt 0 Delta Ssysgt 0,Delta Ssurrgt 0,Delta Sunivgt 0
Consider the melting of ice at 10^circ C and 1 atm. For this process, q_(sys)gt 0 What are the signs of Delta S_(sys),Delta S_(surr), and Delta S_(univ) Delta Ssysgt 0,Delta Ssurrlt 0,Delta Sunivgt 0 Delta Ssysgt 0,Delta Ssurrlt 0,Delta Sunivlt 0 Delta Ssyslt 0,Delta Ssurrgt 0,Delta Sunivgt 0 Delta Ssysgt 0,Delta Ssurrgt 0,Delta Sunivgt 0

Solution
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Answer

\Delta S_{sys} > 0, \Delta S_{surr} 0 Explanation 1. Determine \Delta S_{sys} Melting is an endothermic process, so heat is absorbed by the system. Therefore, \Delta S_{sys} > 0. 2. Determine \Delta S_{surr} Since the system absorbs heat, the surroundings lose heat. Thus, \Delta S_{surr} 0.

Explanation

1. Determine $\Delta S_{sys}$<br /> Melting is an endothermic process, so heat is absorbed by the system. Therefore, $\Delta S_{sys} > 0$.<br /><br />2. Determine $\Delta S_{surr}$<br /> Since the system absorbs heat, the surroundings lose heat. Thus, $\Delta S_{surr} < 0$.<br /><br />3. Determine $\Delta S_{univ}$<br /> For a spontaneous process at constant temperature and pressure, $\Delta S_{univ} = \Delta S_{sys} + \Delta S_{surr} > 0$.
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