   # You have a l.00-mole sample of water at −30.°C and you heat it until you have gaseous water at 140.°C. Calculate q for the entire process. Use the following data. Specific heat capacity of ice = 2.03 J/ ° C ⋅ g Specific heat capacity of water = 4.18 J/ ° C ⋅ g Specific heat capacity of steam = 2.02 J/ ° C ⋅ g H 2 O ( s ) → H 2 O ( l ) Δ H fusion = 6.02 KJ/mol ( at 0 ° C ) H 2 O ( l ) → H 2 O ( g ) Δ H vaporization = 40.7 KJ/mol ( at 100 . ° C ) ### Chemistry: An Atoms First Approach

2nd Edition
Steven S. Zumdahl + 1 other
Publisher: Cengage Learning
ISBN: 9781305079243

#### Solutions

Chapter
Section ### Chemistry: An Atoms First Approach

2nd Edition
Steven S. Zumdahl + 1 other
Publisher: Cengage Learning
ISBN: 9781305079243
Chapter 7, Problem 131CP
Textbook Problem
80 views

## You have a l.00-mole sample of water at −30.°C and you heat it until you have gaseous water at 140.°C. Calculate q for the entire process. Use the following data. Specific heat capacity of ice = 2.03   J/ ° C ⋅ g Specific heat capacity of water = 4.18   J/ ° C ⋅ g Specific heat capacity of steam = 2.02   J/ ° C ⋅ g H 2 O ( s ) → H 2 O ( l ) Δ H fusion = 6.02   KJ/mol   ( at 0 ° C ) H 2 O ( l ) → H 2 O ( g ) Δ H vaporization = 40.7   KJ/mol   ( at 100 . ° C )

Interpretation Introduction

Interpretation: The amount of heat required to convert 1.00 mol of H2O(s) at -30°C to H2O(g) at 140°C should be determined.

Concept Introduction:

The heat capacity C is defined as the relation of heat absorbed to the temperature change. It can be given by,

C = Heat absorbedTemperature change......(1)

Require heat for an one gram of substance raise to its temperature by one degree Celsius is called specific heat capacity.

Absorbed heat (J)=Specific heat capacity×Temperature change(c)×mass ofsubstance (g)......(2)

For the above equation heat is:

S = q×M×T.....(1)

q is heat (J)

M is mass of sample (g)

S is specific heat capacity (J/°C·g.)

T is temperature change (C)

For the process no heat loss to the surroundings means then the heat is

(absorbed)-q×M×ΔT=-q×M×ΔT(released)......(3)'

### Explanation of Solution

Explanation

Record the given data:

Specific heat capacity of ice = 2.03J/°C· g

Specific heat capacity of water = 4.18 J/°C · g

Specific heat capacity of steam = 2.02J/°C· g

Sample water is l.00 mole

Initial temperature is - 30°C

Final temperature is - 140°C.

H2O(s) H2O(l)ΔΗfusion=6.02kJ/mol(at0°C)H2O(l)H2O(g)ΔΗvaporization=40.7kJ/mol(at100°C)

To calculate the required to heat H2O(s) from -30°C to 0°C.

q(1)=2.30J°C.g×18.2g×(0-(-30))°C=1.1×103J

• The given values are plugging in equation (1) to give the required to heat H2O(s) from -30°C to 0°C.
• Required heat for vaporization of 1 mole of water is 1.1×103J.

To calculate the required heat (q) to convert 1 mol H2O(s) at 0°C  into 1 mol H2O(l) at 0°C.

q(1)=1.00mol×6.02×103J/mol=6.02×103J

• The given enthalpy of fusion is multiplied by the mole of water to give required heat (q) to convert 1 mol H2O(s) at 0°C into H2O(l) at 0°C.
• Required heat (q) to convert 1 mol H2O(s) at 0°C  into H2O(l) at 0°C is 6.02×103J.

To calculate the required to heat H2O(l) from 0°C to 100°C.

q(3)=4.18J°C.g×18.2g×(100-0)°C=7

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