Physics for Scientists and Engineers: Foundations and Connections
Physics for Scientists and Engineers: Foundations and Connections
1st Edition
ISBN: 9781133939146
Author: Katz, Debora M.
Publisher: Cengage Learning
bartleby

Videos

Question
Chapter 22, Problem 80PQ
To determine

The heat absorbed or released in the reaction.

Expert Solution & Answer
Check Mark

Answer to Problem 80PQ

The heat released in the reaction is 3.1×104J.

Explanation of Solution

Write the expression to calculate the entropy of reactant A.

  SA=0K300KcAdTT                                                                                                           (I)

Here, SA is the entropy of reactant A, cA is the molar specific heat capacity of A at constant pressure and T is the temperature.

Write the expression to calculate the entropy of reactant B.

  SB=0K300KcBdTT                                                                                                          (II)

Here, SB is the entropy of reactant A and cB is the molar specific heat capacity of B at constant pressure.

Write the expression to calculate the entropy of reactant D.

  SD=0K300KcDdTT                                                                                                        (III)

Here, SD is the entropy of reactant A, cD is the molar specific heat capacity of A at constant pressure.

Write the expression to calculate the entropy change for the reaction.

  Δs=SD(SA+SB)                                                                                               (IV)

Here, Δs is the change in entropy of the reaction.

Write the expression to calculate the heat.

  Q=TΔs                                                                                                                (V)

Here, Q is the amount of heat.

Conclusion:

Substitute 2T for cA in the above equation (I) to calculate SA.

  SA=0K300K(2T)dTT=20K300KdTT=2[2T]0K300K=69.3J/K

Substitute 4T for cB in the above equation (II) to calculate SB.

  SA=0K300K(4T)dTT=40K300KdTT=4[2T]0K300K=138.6J/K

Substitute 3T for cD in the above equation (III) to calculate SD.

  SA=0K300K(3T)dTT=30K300KdTT=3[2T]0K300K=103.9J/K

Substitute 69.3J/K for SA, 138.6J/K for SB and 103.9J/K for SD in the above equation (IV) to calculate Δs.

  Δs=103.9J/K(69.3J/K+138.6J/K)=104J/K

Substitute 104J/K for Δs and 300K for T in the equation (V) to calculate Q.

  Q=300K(104J/K)=3.1×104J

The value of Q is the negative thus, the heat will release the system after the reaction.

Therefore, the heat released in the reaction is 3.1×104J.

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 22 Solutions

Physics for Scientists and Engineers: Foundations and Connections

Ch. 22 - Prob. 6PQCh. 22 - An engine with an efficiency of 0.36 can supply a...Ch. 22 - Prob. 8PQCh. 22 - Prob. 9PQCh. 22 - Prob. 10PQCh. 22 - Prob. 11PQCh. 22 - Prob. 12PQCh. 22 - Prob. 13PQCh. 22 - Prob. 14PQCh. 22 - Prob. 15PQCh. 22 - Prob. 16PQCh. 22 - Prob. 17PQCh. 22 - Prob. 18PQCh. 22 - Prob. 19PQCh. 22 - Prob. 20PQCh. 22 - Prob. 21PQCh. 22 - In 1816, Robert Stirling, a Scottish minister,...Ch. 22 - Prob. 23PQCh. 22 - Prob. 24PQCh. 22 - Prob. 25PQCh. 22 - Prob. 26PQCh. 22 - Prob. 27PQCh. 22 - Prob. 28PQCh. 22 - Prob. 29PQCh. 22 - Prob. 30PQCh. 22 - Prob. 31PQCh. 22 - Prob. 32PQCh. 22 - Prob. 33PQCh. 22 - Prob. 34PQCh. 22 - Prob. 35PQCh. 22 - Estimate the change in entropy of the Universe if...Ch. 22 - Prob. 37PQCh. 22 - Prob. 38PQCh. 22 - Prob. 39PQCh. 22 - Prob. 40PQCh. 22 - Prob. 41PQCh. 22 - Prob. 42PQCh. 22 - Prob. 43PQCh. 22 - Prob. 44PQCh. 22 - Prob. 45PQCh. 22 - Prob. 46PQCh. 22 - Prob. 47PQCh. 22 - Prob. 48PQCh. 22 - Prob. 49PQCh. 22 - Prob. 50PQCh. 22 - Prob. 51PQCh. 22 - Prob. 52PQCh. 22 - Prob. 53PQCh. 22 - Prob. 54PQCh. 22 - Prob. 55PQCh. 22 - Prob. 56PQCh. 22 - What is the entropy of a freshly shuffled deck of...Ch. 22 - Prob. 58PQCh. 22 - Prob. 59PQCh. 22 - Prob. 60PQCh. 22 - Prob. 61PQCh. 22 - Prob. 62PQCh. 22 - Prob. 63PQCh. 22 - Prob. 64PQCh. 22 - Prob. 65PQCh. 22 - Prob. 66PQCh. 22 - Prob. 67PQCh. 22 - Prob. 68PQCh. 22 - Prob. 69PQCh. 22 - Prob. 70PQCh. 22 - A system consisting of 10.0 g of water at a...Ch. 22 - Prob. 72PQCh. 22 - Figure P22.73 illustrates the cycle ABCA for a...Ch. 22 - Prob. 74PQCh. 22 - Prob. 75PQCh. 22 - Prob. 76PQCh. 22 - Prob. 77PQCh. 22 - Prob. 78PQCh. 22 - Prob. 79PQCh. 22 - Prob. 80PQCh. 22 - Prob. 81PQ
Knowledge Booster
Physics
Learn more about
Need a deep-dive on the concept behind this application? Look no further. Learn more about this topic, physics and related others by exploring similar questions and additional content below.
Similar questions
  • Of the following, which is not a statement of the second law of thermodynamics? (a) No heat engine operating in a cycle can absorb energy from a reservoir and use it entirely to do work, (b) No real engine operating between two energy reservoirs can be more efficient than a Carnot engine operating between the same two reservoirs, (c) When a system undergoes a change in state, the change in the internal energy of the system is the sum of the energy transferred to the system by heat and the work done on the system, (d) The entropy of the Universe increases in all natural processes, (e) Energy will not spontaneously transfer by heat from a cold object to a hot object.
    Consider cyclic processes completely characterized by each of the following net energy inputs and outputs. In each case, the energy transfers listed are the only ones occurring. Classify each process as (a) possible, (b) impossible according to the first law of thermodynamics, (c) impossible according to the second law of thermodynamics, or (d) impossible according to both the first and second laws. (i) Input is 5 J of work, and output is 4 J of work. (ii) Input is 5 J of work, and output is 5 J of energy transferred by heat. (iii) Input is 5 J of energy transferred by electrical transmission, and output is 6 J of work. (iv) Input is 5 J of energy transferred by heat, and output is 5 J of energy transferred by heat. (v) Input is 5 J of energy transferred by heat, and output is 5 J of work. (vi) Input is 5 J of energy transferred by heat, and output is 3 J of work plus 2 J of energy transferred by heat.
    An ideal gas is taken from an initial temperature Ti to a higher final temperature Tf along two different reversible paths. Path A is at constant pressure, and path B is at constant volume. What is the relation between the entropy changes of the gas for these paths? (a) SA SB (b) SA = SB (c) SA SB
    Recommended textbooks for you
  • Physics for Scientists and Engineers: Foundations...
    Physics
    ISBN:9781133939146
    Author:Katz, Debora M.
    Publisher:Cengage Learning
    Principles of Physics: A Calculus-Based Text
    Physics
    ISBN:9781133104261
    Author:Raymond A. Serway, John W. Jewett
    Publisher:Cengage Learning
    Physics for Scientists and Engineers, Technology ...
    Physics
    ISBN:9781305116399
    Author:Raymond A. Serway, John W. Jewett
    Publisher:Cengage Learning
  • Physics for Scientists and Engineers
    Physics
    ISBN:9781337553278
    Author:Raymond A. Serway, John W. Jewett
    Publisher:Cengage Learning
    Physics for Scientists and Engineers with Modern ...
    Physics
    ISBN:9781337553292
    Author:Raymond A. Serway, John W. Jewett
    Publisher:Cengage Learning
    University Physics Volume 2
    Physics
    ISBN:9781938168161
    Author:OpenStax
    Publisher:OpenStax
  • Physics for Scientists and Engineers: Foundations...
    Physics
    ISBN:9781133939146
    Author:Katz, Debora M.
    Publisher:Cengage Learning
    Principles of Physics: A Calculus-Based Text
    Physics
    ISBN:9781133104261
    Author:Raymond A. Serway, John W. Jewett
    Publisher:Cengage Learning
    Physics for Scientists and Engineers, Technology ...
    Physics
    ISBN:9781305116399
    Author:Raymond A. Serway, John W. Jewett
    Publisher:Cengage Learning
    Physics for Scientists and Engineers
    Physics
    ISBN:9781337553278
    Author:Raymond A. Serway, John W. Jewett
    Publisher:Cengage Learning
    Physics for Scientists and Engineers with Modern ...
    Physics
    ISBN:9781337553292
    Author:Raymond A. Serway, John W. Jewett
    Publisher:Cengage Learning
    University Physics Volume 2
    Physics
    ISBN:9781938168161
    Author:OpenStax
    Publisher:OpenStax
    The Second Law of Thermodynamics: Heat Flow, Entropy, and Microstates; Author: Professor Dave Explains;https://www.youtube.com/watch?v=MrwW4w2nAMc;License: Standard YouTube License, CC-BY