Ch.11 - Liquids, Solids & Intermolecular ForcesWorksheetSee all chapters
All Chapters
Ch.1 - Intro to General Chemistry
Ch.2 - Atoms & Elements
Ch.3 - Chemical Reactions
BONUS: Lab Techniques and Procedures
BONUS: Mathematical Operations and Functions
Ch.4 - Chemical Quantities & Aqueous Reactions
Ch.5 - Gases
Ch.6 - Thermochemistry
Ch.7 - Quantum Mechanics
Ch.8 - Periodic Properties of the Elements
Ch.9 - Bonding & Molecular Structure
Ch.10 - Molecular Shapes & Valence Bond Theory
Ch.11 - Liquids, Solids & Intermolecular Forces
Ch.12 - Solutions
Ch.13 - Chemical Kinetics
Ch.14 - Chemical Equilibrium
Ch.15 - Acid and Base Equilibrium
Ch.16 - Aqueous Equilibrium
Ch. 17 - Chemical Thermodynamics
Ch.18 - Electrochemistry
Ch.19 - Nuclear Chemistry
Ch.20 - Organic Chemistry
Ch.22 - Chemistry of the Nonmetals
Ch.23 - Transition Metals and Coordination Compounds

Solution: Based on the thermodynamic properties provided for water, determine the energy change when the temperature of 0.850 kg of water decreased from 123°C to 19.0°C.

Solution: Based on the thermodynamic properties provided for water, determine the energy change when the temperature of 0.850 kg of water decreased from 123°C to 19.0°C.

Problem

Based on the thermodynamic properties provided for water, determine the energy change when the temperature of 0.850 kg of water decreased from 123°C to 19.0°C.


Solution

We’re being asked to calculate the amount of energy released when 0.850 kg of water is cooled from 123 ˚C to 19.0 ˚C.


Note that cooling and freezing is an exothermic process, which means q is negative (–)


There are two heats involved in this problem:

1. q1 which is the heat in cooling 0.850 kg of H2O from 123 ˚C to 100 ˚C

2. q2 which is the heat in condensing 0.850 kg of H2O at 100 ˚C

2. q3 which is the heat in cooling 0.850 kg of H2O from 100 ˚C to 19.0 ˚C


We need to solve for each heat individually then add them together to get the final answer.


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