Chapter 1.7, Problem 32P

Interpretation Introduction

Interpretation:

To find the missing data for different sources using the available data.

Concept introduction:

The only concept that is to be used here is that though the values are from different sources, the difference between the values two state points for any source remains constant.

To convert the unit of $\text{J}/\text{g}$ to $\frac{\text{ft}\cdot {\text{lb}}_{\text{f}}}{{\text{lb}}_{\text{m}}}$ follow the following step:

$\begin{array}{c}1\frac{\text{J}}{\text{g}}=\left(1\frac{\text{J}}{\text{g}}\right)\left(\frac{453.6\text{g}}{1{\text{lb}}_{\text{m}}}\right)\left(\frac{1\text{N}\cdot \text{m}}{\text{J}}\right)\left(\frac{\text{1}{\text{lb}}_{\text{f}}}{4.448\text{N}}\right)\left(\frac{1\text{ft}}{0.3048\text{m}}\right)\\ =334.6\frac{\text{ft}\cdot {\text{lb}}_{\text{f}}}{{\text{lb}}_{\text{m}}}\end{array}$

Explanation of Solution

The difference in specific internal energy for state A and B from source 2 is calculated.

${\widehat{U}}_B-{\widehat{U}}_A=20470\frac{\text{ft}\cdot {\text{lb}}_{\text{f}}}{{\text{lb}}_{\text{m}}}$

Here, specific internal energy of states A and B is ${\widehat{U}}_A\text{and}{\widehat{U}}_B$ respectively.

Substitute ${\widehat{U}}_B=-18.0\text{J}/\text{g}$, ${\widehat{U}}_B-{\widehat{U}}_A=20470\frac{\text{ft}\cdot {\text{lb}}_{\text{f}}}{{\text{lb}}_{\text{m}}}$ in source 1

$\begin{array}{l}-18.0\text{J}/\text{g}-{\widehat{U}}_A=20,470\frac{\text{ft}\cdot {\text{lb}}_{\text{f}}}{{\text{lb}}_{\text{m}}}\\ -18.0\text{J}/\text{g}-{\widehat{U}}_A=20,470\frac{\text{ft}\cdot {\text{lb}}_{\text{f}}}{{\text{lb}}_{\text{m}}}\times \frac{1}{334.6}\frac{\text{J}}{\text{g}\cdot \frac{\text{ft}\cdot {\text{lb}}_{\text{f}}}{{\text{lb}}_{\text{m}}}}\\ -{\widehat{U}}_A=20,470\times \frac{1}{334.6}\frac{\text{J}}{\text{g}}+18.0\frac{\text{J}}{\text{g}}\\ {\widehat{U}}_A=-79.2\frac{\text{J}}{\text{g}}\end{array}$

The value of specific internal energy for state A in source 1 is $-79.2\frac{\text{J}}{\text{g}}$

Similarly calculate ${\widehat{U}}_C-{\widehat{U}}_B$ from source 1.

Here, specific internal energy at state C is ${\widehat{U}}_C$.

Substitute ${\widehat{U}}_C=0$ and ${\widehat{U}}_B=-18.0\text{J}/\text{g}$.

$\begin{array}{l}{\widehat{U}}_C-{\widehat{U}}_B=0-\left(-18.0\text{J}/\text{g}\right)\\ {\widehat{U}}_C-{\widehat{U}}_B=18.0\text{J}/\text{g}\times 334.6\frac{\text{ft}\cdot {\text{lb}}_{\text{f}}}{{\text{lb}}_{\text{m}}}\\ {\widehat{U}}_C-{\widehat{U}}_B=6,022\frac{\text{ft}\cdot {\text{lb}}_{\text{f}}}{{\text{lb}}_{\text{m}}}\end{array}$

Using ${\widehat{U}}_C-{\widehat{U}}_B=6,022\frac{\text{ft}\cdot {\text{lb}}_{\text{f}}}{{\text{lb}}_{\text{m}}}$ and substituting ${\widehat{U}}_B=20,470\frac{\text{ft}\cdot {\text{lb}}_{\text{f}}}{{\text{lb}}_{\text{m}}}$ in source 2,

$\begin{array}{c}{\widehat{U}}_C-{\widehat{U}}_B=6,022\frac{\text{ft}\cdot {\text{lb}}_{\text{f}}}{{\text{lb}}_{\text{m}}}\\ {\widehat{U}}_C-20,470\frac{\text{ft}\cdot {\text{lb}}_{\text{f}}}{{\text{lb}}_{\text{m}}}=6,022\frac{\text{ft}\cdot {\text{lb}}_{\text{f}}}{{\text{lb}}_{\text{m}}}\\ {\widehat{U}}_C=6,022\frac{\text{ft}\cdot {\text{lb}}_{\text{f}}}{{\text{lb}}_{\text{m}}}+20,470\frac{\text{ft}\cdot {\text{lb}}_{\text{f}}}{{\text{lb}}_{\text{m}}}\\ {\widehat{U}}_C=26,492\frac{\text{ft}\cdot {\text{lb}}_{\text{f}}}{{\text{lb}}_{\text{m}}}\end{array}$

The value of specific internal energy for state C in source 2 is $26,492\frac{\text{ft}\cdot {\text{lb}}_{\text{f}}}{{\text{lb}}_{\text{m}}}$

From source 2, calculate ${\widehat{U}}_D-{\widehat{U}}_C$.

Here, specific internal energy at state D is ${\widehat{U}}_D$.

Substitute ${\widehat{U}}_C=26,492\frac{\text{ft}\cdot {\text{lb}}_{\text{f}}}{{\text{lb}}_{\text{m}}}$ and ${\widehat{U}}_D=102,970\frac{\text{ft}\cdot {\text{lb}}_{\text{f}}}{{\text{lb}}_{\text{m}}}$.

$\begin{array}{c}{\widehat{U}}_D-{\widehat{U}}_C=102,970\frac{\text{ft}\cdot {\text{lb}}_{\text{f}}}{{\text{lb}}_{\text{m}}}-26,492\frac{\text{ft}\cdot {\text{lb}}_{\text{f}}}{{\text{lb}}_{\text{m}}}\\ {\widehat{U}}_D-{\widehat{U}}_C=76,478\frac{\text{ft}\cdot {\text{lb}}_{\text{f}}}{{\text{lb}}_{\text{m}}}\times \frac{1}{334.6}\frac{\text{J}}{\text{g}\cdot \frac{\text{ft}\cdot {\text{lb}}_{\text{f}}}{{\text{lb}}_{\text{m}}}}\\ {\widehat{U}}_D-{\widehat{U}}_C=228.6\text{J}/\text{g}\end{array}$

Using ${\widehat{U}}_D-{\widehat{U}}_C=228.6\text{J}/\text{g}$ in source 1 to calculate ${\widehat{U}}_D$.

Substitute ${\widehat{U}}_C=0$.

$\begin{array}{l}{\widehat{U}}_D-0=228.6\text{J}/\text{g}\\ {\widehat{U}}_D=228.6\text{J}/\text{g}\end{array}$

The value of specific internal energy for state D in source 1 is $228.6\text{J}/\text{g}$

We know that $\widehat{U}$ for state D relative to all three reference states. The unit for source 1 and source 3 is same, but $\widehat{U}$ is $228.6\text{J}/\text{g}$ higher on the reference state which is used in source 1. So, the values from source 1 at all states will be $228.6\text{J}/\text{g}$ which is higher than the values for source 3.

From source 3, calculate ${\widehat{U}}_E-{\widehat{U}}_D$, ${\widehat{U}}_F-{\widehat{U}}_D$ and ${\widehat{U}}_G-{\widehat{U}}_D$.

Here, specific internal energy at state D is ${\widehat{U}}_D$, state E is ${\widehat{U}}_E$, state F is ${\widehat{U}}_F$, state G is ${\widehat{U}}_G$.

We get,

$\begin{array}{c}{\widehat{U}}_E-{\widehat{U}}_D=\text{67}\text{.0}\frac{\text{J}}{\text{g}}\text{=22,420}\frac{\text{ft}{\text{.lb}}_{\text{f}}}{{\text{lb}}_{\text{m}}}\\ {\widehat{U}}_F-{\widehat{U}}_D=\text{136}\text{.0}\frac{\text{J}}{\text{g}}\text{=45,500}\frac{\text{ft}{\text{.lb}}_{\text{f}}}{{\text{lb}}_{\text{m}}}\\ {\widehat{U}}_G-{\widehat{U}}_D=\text{37}\text{.0}\frac{\text{J}}{\text{g}}\text{=12,380}\frac{\text{ft}{\text{.lb}}_{\text{f}}}{{\text{lb}}_{\text{m}}}\end{array}$

Substitute ${\widehat{U}}_D=102,970\frac{\text{ft}\cdot {\text{lb}}_{\text{f}}}{{\text{lb}}_{\text{m}}}$ in the above and get the values of ${\widehat{U}}_E$, ${\widehat{U}}_F$ and ${\widehat{U}}_G$ as,

$\begin{array}{l}{\widehat{U}}_E-{\widehat{U}}_D=22,420\frac{ft.l{b}_f}{l{b}_m}\\ {\widehat{U}}_E-102,970\frac{ft.l{b}_f}{l{b}_m}=22,420\frac{ft.l{b}_f}{l{b}_m}\\ {\widehat{U}}_E=125,390\frac{ft.l{b}_f}{l{b}_m}\end{array}$

The value of specific internal energy for state E in source 2 is $125,390\frac{ft.l{b}_f}{l{b}_m}$

$\begin{array}{l}{\widehat{U}}_F-{\widehat{U}}_D=45,500\frac{ft.l{b}_f}{l{b}_m}\\ {\widehat{U}}_F-102,970\frac{ft.l{b}_f}{l{b}_m}=45,500\frac{ft.l{b}_f}{l{b}_m}\\ {\widehat{U}}_F=148,470\frac{ft.l{b}_f}{l{b}_m}\end{array}$

The value of specific internal energy for state F in source 2 is $148,470\frac{ft.l{b}_f}{l{b}_m}$

$\begin{array}{l}{\widehat{U}}_G-{\widehat{U}}_D=12,380\frac{ft.l{b}_f}{l{b}_m}\\ {\widehat{U}}_G-102,970\frac{ft.l{b}_f}{l{b}_m}=12,380\frac{ft.l{b}_f}{l{b}_m}\\ {\widehat{U}}_G=115,350\frac{ft.l{b}_f}{l{b}_m}\end{array}$

The value of specific internal energy for state G in source 2 is $115,350\frac{ft.l{b}_f}{l{b}_m}$

Similarly from source 3, calculate ${\widehat{U}}_E-{\widehat{U}}_D$, ${\widehat{U}}_F-{\widehat{U}}_D$ and ${\widehat{U}}_G-{\widehat{U}}_D$.

Here, specific internal energy at state D is ${\widehat{U}}_D$, state E is ${\widehat{U}}_E$, state F is ${\widehat{U}}_F$, state G is ${\widehat{U}}_G$.

We get,

$\begin{array}{c}{\widehat{U}}_E-{\widehat{U}}_D=67.0\frac{J}{g}\\ {\widehat{U}}_F-{\widehat{U}}_D=136.0\frac{J}{g}\\ {\widehat{U}}_G-{\widehat{U}}_D=37.0\frac{J}{g}\end{array}$

Substitute ${\widehat{U}}_D=228.6\frac{J}{g}$ in the above and get the values of ${\widehat{U}}_E$, ${\widehat{U}}_F$ and ${\widehat{U}}_G$ as,

$\begin{array}{c}{\widehat{U}}_E-{\widehat{U}}_D=67.0\frac{J}{g}\\ {\widehat{U}}_E-228.6\frac{J}{g}=67.0\frac{J}{g}\\ {\widehat{U}}_E=295.6\frac{J}{g}\end{array}$

The value of specific internal energy for state E in source 1 is $295.6\frac{J}{g}$

$\begin{array}{c}{\widehat{U}}_F-{\widehat{U}}_D=136.0\frac{J}{g}\\ {\widehat{U}}_F-228.6\frac{J}{g}=136.0\frac{J}{g}\\ {\widehat{U}}_F=364.6\frac{J}{g}\end{array}$

The value of specific internal energy for state F in source 1 is $364.6\frac{J}{g}$

$\begin{array}{c}{\widehat{U}}_G-{\widehat{U}}_D=37.0\frac{J}{g}\\ {\widehat{U}}_G-228.6\frac{J}{g}=37.0\frac{J}{g}\\ {\widehat{U}}_G=265.6\frac{J}{g}\end{array}$

The value of specific internal energy for state G in source 1 is $265.6\frac{J}{g}$

From source 1, calculate ${\widehat{U}}_D-{\widehat{U}}_C$,

Here, specific internal energy at state D is ${\widehat{U}}_D$, state C is ${\widehat{U}}_C$

$\begin{array}{c}{\widehat{U}}_D-{\widehat{U}}_C=228.6\frac{J}{g}-0\frac{J}{g}\\ =228.6\frac{J}{g}\end{array}$

From source 3, ${\widehat{U}}_D=0$ hence ${\widehat{U}}_C$ can be calculated as follows

$\begin{array}{c}{\widehat{U}}_D-{\widehat{U}}_C=228.6\frac{J}{g}\\ 0\frac{J}{g}-{\widehat{U}}_C=228.6\frac{J}{g}\\ {\widehat{U}}_C=-228.6\frac{J}{g}\end{array}$

The value of specific internal energy for state C in source 3 is $-228.6\frac{J}{g}$

From source 1, calculate ${\widehat{U}}_C-{\widehat{U}}_B$,

Here, specific internal energy at state B is ${\widehat{U}}_B$, state C is ${\widehat{U}}_C$

$\begin{array}{c}{\widehat{U}}_C-{\widehat{U}}_B=0\frac{J}{g}-\left(-18.0\frac{J}{g}\right)\\ =18.0\frac{J}{g}\end{array}$

From source 3, ${\widehat{U}}_C=-228.6\frac{J}{g}$ hence ${\widehat{U}}_B$ can be calculated as follows

$\begin{array}{c}{\widehat{U}}_C-{\widehat{U}}_B=18.0\frac{J}{g}\\ -228.6\frac{J}{g}-{\widehat{U}}_B=18.0\frac{J}{g}\\ {\widehat{U}}_B=-246.6\frac{J}{g}\end{array}$

The value of specific internal energy for state B in source 3 is $-246.6\frac{J}{g}$

From source 1, calculate ${\widehat{U}}_C-{\widehat{U}}_A$,

Here, specific internal energy at state A is ${\widehat{U}}_A$, state C is ${\widehat{U}}_C$

$\begin{array}{c}{\widehat{U}}_C-{\widehat{U}}_A=0\frac{J}{g}-\left(-79.2\frac{J}{g}\right)\\ =79.2\frac{J}{g}\end{array}$

From source 3, ${\widehat{U}}_C=-228.6\frac{J}{g}$ hence ${\widehat{U}}_A$ can be calculated as follows

$\begin{array}{c}{\widehat{U}}_C-{\widehat{U}}_A=79.2\frac{J}{g}\\ -228.6\frac{J}{g}-{\widehat{U}}_A=79.2\frac{J}{g}\\ {\widehat{U}}_A=-307.8\frac{J}{g}\end{array}$

The value of specific internal energy for state A in source 3 is $-307.8\frac{J}{g}$

Conclusion:

The correct values of $\widehat{U}$ for all seven states (A-G) from all three references are shown in Table.

State	Phase	$T\left(°\text{C}\right)$	$P\left(\text{bar}\right)$	$\widehat{U}\left(\text{Source}1\right),\frac{\text{J}}{\text{g}}$	$\begin{array}{l}\widehat{U}\left(\text{Source}2\right),\\ \frac{\text{ft}\cdot {\text{lb}}_{\text{f}}}{{\text{lb}}_{\text{m}}}\end{array}$	$\widehat{U}\left(\text{Source}3\right),\frac{\text{J}}{\text{g}}$
A	Solid	17	1	-79.2	0	-307.8
B	Liquid	17	1	-18.0	20,470	-246.6
C	Liquid	25	1	0	26,492	-228.6
D	Liquid	82	1	228.6	102,970	0
E	Vapor	82	1	295.6	125,390	67
F	Vapor	100	1.33	364.6	148,470	136
G	Liquid	100	1.33	265.6	115,350	37