Ch.6 - Thermochemistry WorksheetSee 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: What is the change in internal energy (in J) of a system that absorbs 0.615 kJ of heat from its surroundings and has 0.247 kcal of work done on it?

Solution: What is the change in internal energy (in J) of a system that absorbs 0.615 kJ of heat from its surroundings and has 0.247 kcal of work done on it?

Problem

What is the change in internal energy (in J) of a system that absorbs 0.615 kJ of heat from its surroundings and has 0.247 kcal of work done on it?

Solution

This problem related to the first law of thermodynamics.

Recall that:

  • Heat absorbed by the system adds to the total energy of the system, therefore its sign is positive.
  • Work is also a form of energy. Work done on the system also adds to the total energy of the system, therefore its sign in positive.


We will need the following equation to solve this problem:

<math xmlns="http://www.w3.org/1998/Math/MathML"><mo>&#x2206;</mo><mi>E</mi><mo>&#xA0;</mo><mo>=</mo><mo>&#xA0;</mo><mi>q</mi><mo>&#xA0;</mo><mo>+</mo><mo>&#xA0;</mo><mi>w</mi><mo>&#xA0;</mo><mo>-</mo><mo>-</mo><mo>-</mo><mo>-</mo><mfenced><mn>1</mn></mfenced></math>


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