   # What quantity of energy does it take to convert 0.500 kg ice at –20.°C to steam at 250.°C? Specific heat capacities: ice, 2.03 J/g • °C; liquid, 4.2 J/g . °C; steam, 2.0 J/g • °C; ∆ H vap = 40.7 kJ/mol; ∆ H fus = 6.02 kJ/mol. ### 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 9, Problem 99E
Textbook Problem
265 views

## What quantity of energy does it take to convert 0.500 kg ice at –20.°C to steam at 250.°C? Specific heat capacities: ice, 2.03 J/g • °C; liquid, 4.2 J/g . °C; steam, 2.0 J/g • °C; ∆Hvap = 40.7 kJ/mol; ∆Hfus = 6.02 kJ/mol.

Interpretation Introduction

Interpretation: The required amount of heat for water at -20°C to 250°C should be calculated.

Concept Introduction:

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

C = Heat absorbedTemperature change

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

Specific heat capacity =  Absorbed heat (J)/ Temperature change(c)×mass ofsubstance (g)......(1)

For the above equation heat is:

q = S×M×ΔT......(2)

q is heat J

M is mass of sample (g)

S is specific heat capacity (J/°C·g)

ΔT  is temperature change (°C)

### Explanation of Solution

Explanation

Record the given data:

Specific heat capacity of ice =   2.03 J/C.g

Specific heat capacity of steam = 2.0 J/C.g

Specific heat capacity of water 4.184 J/C.g

The  ΔΗvap =40.7 kJ/mol

The ΔΗfus = 6.02 kJ/mol

• The given specific heat capacities, ΔΗvap and ΔΗfus are recorded as shown above.

To calculate the required heat of ice -20°C to 0°C .

The reaction is,

H2O(s-20°C)H2O(s0°C)

q1=2.30Jg.°C×5.00×102×20°C=2.0×104J=20kJ

• The specific heat capacity of ice, temperature change and mass of ice are plugging in to equation 2 to give heat of ice -20°C to 0°C .
• The heat of ice -20°C to 0°C is 20kJ .

To calculate the required heat of ice 0°C to water at 0°C .

The reaction is,

H2O(sol0°C)H2O(lq0°C)

Molecular weight of water is 18.02g

q2=×5.00×102×1mol18.02g×6.02kJmol=167kJ

• The enthalpy of fusion and molecular weight of water are plugging in to equation  to give heat of ice 0°C to water at 0°C .
• The heat ice 0°C to water at 0°C is 167kJ .

To calculate the required heat of water at 0°C in to 100°C .

The reaction is,

H2O(lq0°C)H2O(1q100°C)

q3=4.2Jg.°C×5.00×102×100°C=2

### Still sussing out bartleby?

Check out a sample textbook solution.

See a sample solution

#### The Solution to Your Study Problems

Bartleby provides explanations to thousands of textbook problems written by our experts, many with advanced degrees!

Get Started

Find more solutions based on key concepts
Which of the following is not one of the ways fats are useful in foods? Fats contribute to the taste and smell ...

Nutrition: Concepts and Controversies - Standalone book (MindTap Course List)

Bacteria, Archaea, and Eukarya are three ________.

Biology: The Unity and Diversity of Life (MindTap Course List)

Example 10.6 Angular Acceleration of a Wheel A wheel of radius R mass M and moment of inertia I is mounted on a...

Physics for Scientists and Engineers, Technology Update (No access codes included)

What are possible fates (or end-states) for comets?

Foundations of Astronomy (MindTap Course List)

What technical innovation made surfing waves of this magnitude possible?

Oceanography: An Invitation To Marine Science, Loose-leaf Versin 