Ch.11 - Liquids, Solids & Intermolecular ForcesWorksheetSee all chapters
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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: From the data below, calculate the total heat (in J) needed to convert 12.0 g of ice at -5.0 °C to liquid water at 0.5 °C:ΔHfus= 6.02 kJ/mol, cliquid = 4.18 J/g °C, csolid = 2.09 J/g °CA. –2.2 x 103 J

Problem

From the data below, calculate the total heat (in J) needed to convert 12.0 g of ice at -5.0 °C to liquid water at 0.5 °C:

ΔHfus= 6.02 kJ/mol, cliquid = 4.18 J/g °C, csolid = 2.09 J/g °C

A. –2.2 x 103 J

B. 2.1 x 102 J

C. –2.1 x 102 J

D. 4.2 x 103 J

E. 2.2 x 103 J

Solution

We’re being asked to calculate the amount of heat needed to convert 12.0 g of ice at -5.0˚C to liquid water at 0.5°C. There are three heats involved in this problem:

Step 1. q1 which is the heat in raising the temperature of 12.0 g of ice from 5.0˚C to 0˚C

Step 2. q2 which is the heat in melting 12.0 g of ice at 0˚C

Step 3. q3 which is the heat in raising the temperature of 12.0 g of water from 0˚C to 0.5˚C
Step 4. Calculate Qtotal.


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


Step 1. The temperature changes from  –5.0˚C to 0˚C

Because of this, the equation we’ll use is:

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