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: How much energy is needed to convert 70.6 grams of ice at 0.00°C to water at 75.0°C?  specific heat (ice) = 2.10 J/g°Cspecific heat (water) = 4.18 J/g°Cheat of fusion = 333 J/gheat of vaporization = 2258 J/gA. 45.6 kJB. 2.53 kJC. 182 kJD. 22.1 kJE. 34.6 kJ

Problem

How much energy is needed to convert 70.6 grams of ice at 0.00°C to water at 75.0°C?  

specific heat (ice) = 2.10 J/g°C

specific heat (water) = 4.18 J/g°C

heat of fusion = 333 J/g

heat of vaporization = 2258 J/g

A. 45.6 kJ

B. 2.53 kJ

C. 182 kJ

D. 22.1 kJ

E. 34.6 kJ

Solution

We’re being asked to calculate the amount of energy required to convert 70.6 g of ice at 0.00˚C to water at 75˚C.


There are three heats involved in this problem:


1. q1 which is the heat in melting 70.6 g ice at 0.0°C.

2. q2 which is the heat in raising the temperature of  70.6 g of water from 0.00˚C to 75˚C.


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


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