Chemistry: The Molecular Nature of Matter and Change
Chemistry: The Molecular Nature of Matter and Change
8th Edition
ISBN: 9781259631757
Author: Martin Silberberg Dr., Patricia Amateis Professor
Publisher: McGraw-Hill Education
bartleby

Concept explainers

bartleby

Videos

Question
Book Icon
Chapter 12, Problem 12.142P

a.

Interpretation Introduction

Interpretation:

The time to heat the sample to its melting point is to be calculated.

Concept introduction:

Specific heat capacity (c) of a substance is the amount of heat needed to raise the temperature of 1g of a substance by 1K. The formula to calculate heat required is as follows:

  q=(mass)(c)(T2T1)        (1)

Here,

T2 is the final temperature.

T1 is the initial temperature.

q is the heat released or absorbed.

c is the specific heat capacity of the substance.

The formula to calculate heat at phase change is as follows:

  q=(mass)(ΔH)        (2)

Here,

ΔH is the enthalpy change.

a.

Expert Solution
Check Mark

Answer to Problem 12.142P

The time to heat the sample to its melting point is 1.1min.

Explanation of Solution

Substitute 25g for mass, 1.0J/g°C for c, 20.0°C for T2 and 40.0°C for T1 in the equation (1) to calculate the heat required to increase the temperature of sample from 40.0°C to 20.0°C.

  q=(25g)(1.0J/g°C)((20.0°C)(40.0°C))=500J

Constant rate of heating is 450J/min. The time to heat the sample to its melting point is calculated as follows:

  Time=(500J450J/min)=1.1111min1.1min.

Conclusion

The given sample reaches its melting point by heating for 1.1min.

(b)

Interpretation Introduction

Interpretation:

The time taken to melt the sample is to be calculated.

Concept introduction:

Specific heat capacity (c) of a substance is the amount of heat needed to raise the temperature of 1g of a substance by 1K. The formula to calculate heat required is as follows:

  q=(mass)(c)(T2T1)        (1)

Here,

T2 is the final temperature.

T1 is the initial temperature.

q is the heat released or absorbed.

c is the specific heat capacity of the substance.

The formula to calculate heat at phase change is as follows:

  q=(mass)(ΔH)        (2)

Here,

ΔH is the enthalpy change.

(b)

Expert Solution
Check Mark

Answer to Problem 12.142P

The time taken to melt the sample is 10min.

Explanation of Solution

Substitute 25g for mass and 180J/g for ΔHfus° in the equation (2) to change the phase of solid sample at 20.0°C to liquid sample at 20.0°C.

  q=(25g)(180J/g)=4500 J

Constant rate of heating is 450J/min. The time taken to melt the sample is calculated as follows:

  Time=(4500 J450J/min)=10min

Conclusion

The time taken to melt 25g sample is 10min.

(c)

Interpretation Introduction

Interpretation:

A curve of temperature vs. time for the entire heating process is to be calculated.

Concept introduction:

Specific heat capacity (c) of a substance is the amount of heat needed to raise the temperature of 1g of a substance by 1K. The formula to calculate heat required is as follows:

  q=(mass)(c)(T2T1)        (1)

Here,

T2 is the final temperature.

T1 is the initial temperature.

q is the heat released or absorbed.

c is the specific heat capacity of the substance.

The formula to calculate heat at phase change is as follows:

  q=(mass)(ΔH)        (2)

Here,

ΔH is the enthalpy change.

(c)

Expert Solution
Check Mark

Answer to Problem 12.142P

The curve of temperature vs. time for the entire heating process is as follows:

Chemistry: The Molecular Nature of Matter and Change, Chapter 12, Problem 12.142P , additional homework tip  1

Explanation of Solution

Substitute 25g for mass, 2.5J/g°C for c, 85°C for T2 and 20.0°C for T1 in the equation (1) to calculate the heat required to increase the temperature of the sample from 20.0°C to 85°C.

  q1=(25g)(2.5J/g°C)(85°C(20.0°C))=6562.5J

Constant rate of heating is 450J/min. The time to increase the temperature of the sample from 20.0°C to 85°C is calculated as follows:

  Time=(6562.5J450J/min)15min

Substitute 25g for mass and 500J/g for ΔHvap in the equation (2) to change the phase of the liquid sample at 85°C to gas sample at 85°C.

  q2=(25g)(500J/g)=12500J

Constant rate of heating is 450J/min. The time to change the phase of the liquid sample at 85°C to gas sample at 85°C is calculated as follows:

  Time=(12500J450J/min)28min

Substitute 25g for mass, 0.5J/g°C for c, 100°C for T2 and 85°C for T1 in the equation (1) to calculate the heat required to increase the temperature of the sample from 85°C to 100°C.

  q3=(25g)(0.5J/g°C)(100°C85°C)=187.5J

Constant rate of heating is 450J/min. The time to increase the temperature of the sample from 85°C to 100°C is calculated as follows:

  Time=(187.5J450J/min)0.4min.

The curve of temperature vs. time for the entire heating process is as follows:

Chemistry: The Molecular Nature of Matter and Change, Chapter 12, Problem 12.142P , additional homework tip  2

Conclusion

The temperature versus time plot for the given sample represents the phase change.

Want to see more full solutions like this?

Subscribe now to access step-by-step solutions to millions of textbook problems written by subject matter experts!

Chapter 12 Solutions

Chemistry: The Molecular Nature of Matter and Change

Ch. 12.6 - For each of the following crystalline solids,...Ch. 12.6 - Prob. 12.6BFPCh. 12.6 - Prob. 12.7AFPCh. 12.6 - Iron crystallizes in a body-centered cubic...Ch. 12.6 - Prob. 12.8AFPCh. 12.6 - Prob. 12.8BFPCh. 12.6 - Prob. B12.1PCh. 12.6 - Prob. B12.2PCh. 12 - Prob. 12.1PCh. 12 - Prob. 12.2PCh. 12 - Prob. 12.3PCh. 12 - Prob. 12.4PCh. 12 - Prob. 12.5PCh. 12 - Prob. 12.6PCh. 12 - Prob. 12.7PCh. 12 - Name the phase change in each of these events: (a)...Ch. 12 - Prob. 12.9PCh. 12 - Many heat-sensitive and oxygen-sensitive solids,...Ch. 12 - Prob. 12.11PCh. 12 - Prob. 12.12PCh. 12 - Prob. 12.13PCh. 12 - Prob. 12.14PCh. 12 - Prob. 12.15PCh. 12 - Prob. 12.16PCh. 12 - Prob. 12.17PCh. 12 - Prob. 12.18PCh. 12 - From the data below, calculate the total heat (in...Ch. 12 - Prob. 12.20PCh. 12 - Prob. 12.21PCh. 12 - Prob. 12.22PCh. 12 - Prob. 12.23PCh. 12 - Prob. 12.24PCh. 12 - Prob. 12.25PCh. 12 - Prob. 12.26PCh. 12 - Prob. 12.27PCh. 12 - Prob. 12.28PCh. 12 - Prob. 12.29PCh. 12 - Prob. 12.30PCh. 12 - Use Figure 12.10 to answer the following: Carbon...Ch. 12 - Prob. 12.32PCh. 12 - Prob. 12.33PCh. 12 - Prob. 12.34PCh. 12 - Prob. 12.35PCh. 12 - Prob. 12.36PCh. 12 - Distinguish between polarizability and polarity....Ch. 12 - Prob. 12.38PCh. 12 - Prob. 12.39PCh. 12 - Prob. 12.40PCh. 12 - Prob. 12.41PCh. 12 - Prob. 12.42PCh. 12 - Prob. 12.43PCh. 12 - Prob. 12.44PCh. 12 - Prob. 12.45PCh. 12 - Prob. 12.46PCh. 12 - Prob. 12.47PCh. 12 - Prob. 12.48PCh. 12 - Prob. 12.49PCh. 12 - Which liquid in each pair has the lower vapor...Ch. 12 - Which substance has the lower boiling point?...Ch. 12 - Which substance has the higher boiling point?...Ch. 12 - Prob. 12.53PCh. 12 - Prob. 12.54PCh. 12 - Prob. 12.55PCh. 12 - Prob. 12.56PCh. 12 - Why does the antifreeze ingredient ethylene glycol...Ch. 12 - Prob. 12.58PCh. 12 - Prob. 12.59PCh. 12 - Why does an aqueous solution of ethanol (CH3CH2OH)...Ch. 12 - Prob. 12.61PCh. 12 - Prob. 12.62PCh. 12 - Prob. 12.63PCh. 12 - Prob. 12.64PCh. 12 - Prob. 12.65PCh. 12 - Prob. 12.66PCh. 12 - Prob. 12.67PCh. 12 - Prob. 12.68PCh. 12 - Prob. 12.69PCh. 12 - Prob. 12.70PCh. 12 - Prob. 12.71PCh. 12 - Prob. 12.72PCh. 12 - Prob. 12.73PCh. 12 - Prob. 12.74PCh. 12 - Prob. 12.75PCh. 12 - Prob. 12.76PCh. 12 - Prob. 12.77PCh. 12 - Prob. 12.78PCh. 12 - Prob. 12.79PCh. 12 - Prob. 12.80PCh. 12 - Prob. 12.81PCh. 12 - Prob. 12.82PCh. 12 - Prob. 12.83PCh. 12 - Prob. 12.84PCh. 12 - Besides the type of unit cell, what information is...Ch. 12 - What type of unit cell does each metal use in its...Ch. 12 - What is the number of atoms per unit cell for each...Ch. 12 - Calcium crystallizes in a cubic closest packed...Ch. 12 - Chromium adopts the body-centered cubic unit cell...Ch. 12 - Prob. 12.90PCh. 12 - Prob. 12.91PCh. 12 - Prob. 12.92PCh. 12 - Prob. 12.93PCh. 12 - Prob. 12.94PCh. 12 - Prob. 12.95PCh. 12 - Prob. 12.96PCh. 12 - Prob. 12.97PCh. 12 - Prob. 12.98PCh. 12 - Prob. 12.99PCh. 12 - Prob. 12.100PCh. 12 - Prob. 12.101PCh. 12 - Prob. 12.102PCh. 12 - Prob. 12.103PCh. 12 - Polonium, the Period 6 member of Group 6A(16), is...Ch. 12 - Prob. 12.105PCh. 12 - Prob. 12.106PCh. 12 - Prob. 12.107PCh. 12 - Prob. 12.108PCh. 12 - Prob. 12.109PCh. 12 - Prob. 12.110PCh. 12 - Prob. 12.111PCh. 12 - Prob. 12.112PCh. 12 - Prob. 12.113PCh. 12 - Prob. 12.114PCh. 12 - Prob. 12.115PCh. 12 - Prob. 12.116PCh. 12 - Prob. 12.117PCh. 12 - Prob. 12.118PCh. 12 - Prob. 12.119PCh. 12 - Prob. 12.120PCh. 12 - Prob. 12.121PCh. 12 - Prob. 12.122PCh. 12 - Prob. 12.123PCh. 12 - Prob. 12.124PCh. 12 - Prob. 12.125PCh. 12 - Prob. 12.126PCh. 12 - Bismuth is used to calibrate instruments employed...Ch. 12 - Prob. 12.128PCh. 12 - Prob. 12.129PCh. 12 - Prob. 12.130PCh. 12 - Prob. 12.131PCh. 12 - Prob. 12.132PCh. 12 - Prob. 12.133PCh. 12 - Prob. 12.134PCh. 12 - Prob. 12.135PCh. 12 - Prob. 12.136PCh. 12 - Prob. 12.137PCh. 12 - Prob. 12.138PCh. 12 - Prob. 12.139PCh. 12 - Prob. 12.140PCh. 12 - Prob. 12.141PCh. 12 - Prob. 12.142PCh. 12 - Prob. 12.143PCh. 12 - Prob. 12.144PCh. 12 - Prob. 12.145PCh. 12 - The crystal structure of sodium is based on the...Ch. 12 - Prob. 12.147PCh. 12 - One way of purifying gaseous H2 is to pass it...
Knowledge Booster
Background pattern image
Chemistry
Learn more about
Need a deep-dive on the concept behind this application? Look no further. Learn more about this topic, chemistry and related others by exploring similar questions and additional content below.
Recommended textbooks for you
Text book image
Chemistry
Chemistry
ISBN:9781305957404
Author:Steven S. Zumdahl, Susan A. Zumdahl, Donald J. DeCoste
Publisher:Cengage Learning
Text book image
Chemistry
Chemistry
ISBN:9781259911156
Author:Raymond Chang Dr., Jason Overby Professor
Publisher:McGraw-Hill Education
Text book image
Principles of Instrumental Analysis
Chemistry
ISBN:9781305577213
Author:Douglas A. Skoog, F. James Holler, Stanley R. Crouch
Publisher:Cengage Learning
Text book image
Organic Chemistry
Chemistry
ISBN:9780078021558
Author:Janice Gorzynski Smith Dr.
Publisher:McGraw-Hill Education
Text book image
Chemistry: Principles and Reactions
Chemistry
ISBN:9781305079373
Author:William L. Masterton, Cecile N. Hurley
Publisher:Cengage Learning
Text book image
Elementary Principles of Chemical Processes, Bind...
Chemistry
ISBN:9781118431221
Author:Richard M. Felder, Ronald W. Rousseau, Lisa G. Bullard
Publisher:WILEY
Types of Matter: Elements, Compounds and Mixtures; Author: Professor Dave Explains;https://www.youtube.com/watch?v=dggHWvFJ8Xs;License: Standard YouTube License, CC-BY