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: Dry ice is solid carbon dioxide. Instead of melting, solid carbon dioxide sublimes according to the equation: CO2(s) → CO2(g). When dry ice is added to warm water, heat from the water causes the dry i

Solution: Dry ice is solid carbon dioxide. Instead of melting, solid carbon dioxide sublimes according to the equation: CO2(s) → CO2(g). When dry ice is added to warm water, heat from the water causes the dry i

Problem

Dry ice is solid carbon dioxide. Instead of melting, solid carbon dioxide sublimes according to the equation: 

CO2(s) → CO2(g). 

When dry ice is added to warm water, heat from the water causes the dry ice to sublime more quickly. The evaporating carbon dioxide produces a dense fog often used to create special effects. In a simple dry ice fog machine, dry ice is added to warm water in a Styrofoam cooler. The dry ice produces fog until it evaporates away, or until the water gets too cold to sublime the dry ice quickly enough. Suppose that a small Styrofoam cooler holds 15.0 L of water heated to 86˚C.

Calculate the mass of dry ice that should be added to the water so that the dry ice completely sublimes away when the water reaches 23˚C. Assume no heat loss to the surroundings.

Solution

The first thing that we have to focus on is that there is no heat loss to the surroundings. So, all the heat lost by water is gained by carbon dioxide for sublimation. We can write:

<math xmlns="http://www.w3.org/1998/Math/MathML"><msub><mi>q</mi><mrow><mi>C</mi><msub><mi>O</mi><mn>2</mn></msub></mrow></msub><mo>&#xA0;</mo><mo>=</mo><mo>&#xA0;</mo><mo>-</mo><msub><mi>q</mi><mrow><msub><mi>H</mi><mn>2</mn></msub><mi>O</mi></mrow></msub><mo>&#xA0;</mo><mo>-</mo><mo>-</mo><mo>-</mo><mo>-</mo><mfenced><mn>1</mn></mfenced></math>

qCO2 is positive as CO2 gains heat in the process, while qH2O is negative because water is losing heat.


The temperature of water changes from 86 oC to 23 oC. Change of temperature is given by:

<math xmlns="http://www.w3.org/1998/Math/MathML"><mo mathvariant="italic">&#x2206;</mo><mi>T</mi><mo mathvariant="italic">&#xA0;</mo><mo mathvariant="italic">=</mo><mo mathvariant="italic">&#xA0;</mo><msub><mi>T</mi><mrow><mi mathvariant="italic">f</mi><mi mathvariant="italic">i</mi><mi mathvariant="italic">n</mi><mi mathvariant="italic">a</mi><mi mathvariant="italic">l</mi></mrow></msub><mo mathvariant="italic">&#xA0;</mo><mo mathvariant="italic">-</mo><mpadded lspace="-1px"><mo mathvariant="italic">&#xA0;</mo><msub><mi mathvariant="italic">T</mi><mrow><mi mathvariant="italic">i</mi><mi mathvariant="italic">n</mi><mi mathvariant="italic">i</mi><mi mathvariant="italic">t</mi><mi mathvariant="italic">i</mi><mi mathvariant="italic">a</mi><mi mathvariant="italic">l</mi></mrow></msub></mpadded></math>


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