• :00Days
  • :00Hours
  • :00Mins
  • 00Seconds
A new era for learning is coming soonSign up for free
Log In Start studying!

Select your language

Suggested languages for you:
Answers without the blur. Sign up and see all textbooks for free! Illustration

4.28

Expert-verified
An Introduction to Thermal Physics
Found in: Page 137
An Introduction to Thermal Physics

An Introduction to Thermal Physics

Book edition 1st
Author(s) Daniel V. Schroeder
Pages 356 pages
ISBN 9780201380279

Answers without the blur.

Just sign up for free and you're in.

Illustration

Short Answer

Imagine that your dog has eaten the portion of Table 4.1 that gives entropy data; only the enthalpy data remains. Explain how you could reconstruct the missing portion of the table. Use your method to explicitly check a few of the entries for consistency. How much of Table 4.2 could you reconstruct if it were missing? Explain.

We can say that the missing portion of table 4.1 is reconstructed by applying the second law of thermodynamics and the table 4.2.

See the step by step solution

Step by Step Solution

Step1 : Given information

Given table 4.1

And Table 4.2

Step2: Explanation

First express the change in entropy for steam at zero temperature
ΔS1=QT......(1)Where, Q is the heat change (absorbed or lost) and T is the temperature.

We know that the enthalpy is described as the energy that is required to produce a substance at fixed pressure.Now write an expression of the change in entropy for water at given temperatureΔS2=ΔHT......(2)Where, ΔH - change in enthalpy

Substitute Q = 2501 kJ/kg and T =273 K in equation (1), we get
ΔS1=2501 kJ·kg-1(273 K)=9.156 kJ·kg-1·K-1This is the same value as in Table

Now, substitute ΔH=42 kJ·kg-1 and T=278 K in equation (2), we get

ΔS2=42 kJ.kg-1(278 K)=0.151 kJ·kg-1.K-1

This value is the same as in given table

Now write expression for the change in entropy for steam at given temperature
ΔS=ΔHT-nRΔPP......(3)Where, n is number of moles, R is gas constant, P is average pressure and ΔP is change in pressure.

Substitute ΔH=19 kJ·kg-1 , T=278K, n= 55.55, R= 8.314 J/mole K,ΔP = 0.006 and P=0.009 bar in expression (3), we get

ΔS=19 kJ·kg-1(278 K)-(55.55)(8.314 J/mole·K)(0.006bar)(0.009bar)=-0.239 kJ·kg-1·K-1

The entropy for steam at finite temperature is:

ΔS1+ΔS=9.156 kJ·kg-1·K-1+-0.239 kJ·kg-1·K-1=8.917 kJ·kg-1·K-1

This value is very close to the value in the table.

So we can say that the missing portion of table 4.1 is reconstructed by applying the second law of thermodynamics and the table 4.2.

Most popular questions for Physics Textbooks

Icon

Want to see more solutions like these?

Sign up for free to discover our expert answers
Get Started - It’s free

Recommended explanations on Physics Textbooks

94% of StudySmarter users get better grades.

Sign up for free
94% of StudySmarter users get better grades.