Chapter 2, Problem 19P

Consider Table 2-1 in the textbook, which lists the specific gravities of various substances. (a) Explain the difference between specific gravity and specific weight. Which one (if any) is dimensionless? (b) Calculate the specific weight of all the substances in Table 2-1. For the case of bones, give answers for both the low and high values given in the table. Note: Excel is recommended for this kind of problem in which there is much repetition of calculations, but you are welcome to do the calculations by hand or with any other software. If using software like Excel, do not worry about the number of significant digits, since this is not something easy to modify in Excel. (c) As discussed in the text, there is another related property called specific volume. Calculate the specific robe of a liquid with SG = 0.592.

To determine

(a)

Explanation of the difference between specific gravity and specific weight and which one is the dimensionless.

Explanation of Solution

ss

S. No	Specific gravity	Specific weight
1	It is defined as the ratio of density of the substance to the density of reference substance at specific temperature and also known as relative density.	It is defined as the weight per unit volume of a substance.
2	It does not depend on the gravitation field.	It depends on the gravitation field.
3	It is relative quantity and used for comparison between two materials.	It is absolute quantity.
4	It is dimensionless quantity.	It has dimension ie., $\text{N}/{\text{m}}^3$ .

To determine

(b)

To calculate:

The specific weight of all the substances given in Table 2-1.

Explanation of Solution

Write the expression for specific gravity of the substance.

  $\begin{array}{l}\text{SG=}\frac{\rho }{{\rho }_{{\text{H}}_{\text{2}}\text{O}}}\\ \rho =\text{SG}\cdot {\rho }_{{\text{H}}_{\text{2}}\text{O}}\end{array}$.......(I)

Here, specific gravity of the substance is $\text{SG}$, density of substance is $\rho$, and density of water is ${\rho }_{{\text{H}}_{\text{2}}\text{O}}$.

Write the expression for specific weight of the substance.

  $\gamma =\rho g$.......(II)

Here, the specific weight of the substance is $\gamma$ and gravitation acceleration is $g$.

From Equation (I) and (II).

  $\gamma =\text{SG}\cdot {\rho }_{{\text{H}}_{\text{2}}\text{O}}\cdot g$......... (III)

Calculation:

Substitute $1$ for $\text{SG}$, ${10}^3\text{kg}/{\text{m}}^{\text{3}}$ for ${\rho }_{{\text{H}}_{\text{2}}\text{O}}$ and $9.81\text{m}/{\text{s}}^{\text{2}}$ in Equation (III) to to calculate specific weight of Water.

  $\begin{array}{c}\gamma =1\cdot \left({10}^3\text{kg}/{\text{m}}^{\text{3}}\right)\left(9.81\text{m}/{\text{s}}^{\text{2}}\right)\\ =9.81\times {10}^3\left(\text{kg}\cdot \frac{\text{m}}{{\text{s}}^{\text{2}}}\right)\left(\frac{\text{1}}{{\text{m}}^3}\right)\\ =9.81\times {10}^3\text{N}/{\text{m}}^{\text{3}}\end{array}$

Substitute $1.06$ for $\text{SG}$, ${10}^3\text{kg}/{\text{m}}^{\text{3}}$ for ${\rho }_{{\text{H}}_{\text{2}}\text{O}}$ and $9.81\text{m}/{\text{s}}^{\text{2}}$ in Equation (III) to calculate specific weight of Blood.

  $\begin{array}{c}\gamma =\left(1.06\right)\left({10}^3\text{kg}/{\text{m}}^{\text{3}}\right)\left(9.81\text{m}/{\text{s}}^{\text{2}}\right)\\ =10.40\times {10}^3\left(\text{kg}\cdot \frac{\text{m}}{{\text{s}}^{\text{2}}}\right)\left(\frac{\text{1}}{{\text{m}}^3}\right)\\ =10.40\times {10}^3\text{N}/{\text{m}}^{\text{3}}\end{array}$

Substitute $1.025$ for $\text{SG}$, ${10}^3\text{kg}/{\text{m}}^{\text{3}}$ for ${\rho }_{{\text{H}}_{\text{2}}\text{O}}$ and $9.81\text{m}/{\text{s}}^{\text{2}}$ in Equation (III) to calculate specific weight of Seawater.

  $\begin{array}{c}\gamma =\left(1.025\right)\left({10}^3\text{kg}/{\text{m}}^{\text{3}}\right)\left(9.81\text{m}/{\text{s}}^{\text{2}}\right)\\ =10.06\times {10}^3\left(\text{kg}\cdot \frac{\text{m}}{{\text{s}}^{\text{2}}}\right)\left(\frac{\text{1}}{{\text{m}}^3}\right)\\ =10.06\times {10}^3\text{N}/{\text{m}}^{\text{3}}\end{array}$

Substitute $0.68$ for $\text{SG}$, ${10}^3\text{kg}/{\text{m}}^{\text{3}}$ for ${\rho }_{{\text{H}}_{\text{2}}\text{O}}$ and $9.81\text{m}/{\text{s}}^{\text{2}}$ in Equation (III) to calculate specific weight of Gasoline.

  $\begin{array}{c}\gamma =\left(0.68\right)\left({10}^3\text{kg}/{\text{m}}^{\text{3}}\right)\left(9.81\text{m}/{\text{s}}^{\text{2}}\right)\\ =6.67\times {10}^3\left(\text{kg}\cdot \frac{\text{m}}{{\text{s}}^{\text{2}}}\right)\left(\frac{\text{1}}{{\text{m}}^3}\right)\\ =6.67\times {10}^3\text{N}/{\text{m}}^{\text{3}}\end{array}$

Substitute $0.79$ for $\text{SG}$, ${10}^3\text{kg}/{\text{m}}^{\text{3}}$ for ${\rho }_{{\text{H}}_{\text{2}}\text{O}}$ and $9.81\text{m}/{\text{s}}^{\text{2}}$ in Equation (III) to calculate specific weight of Ethyl alcohol.

  $\begin{array}{c}\gamma =\left(0.79\right)\left({10}^3\text{kg}/{\text{m}}^{\text{3}}\right)\left(9.81\text{m}/{\text{s}}^{\text{2}}\right)\\ =7.75\times {10}^3\left(\text{kg}\cdot \frac{\text{m}}{{\text{s}}^{\text{2}}}\right)\left(\frac{\text{1}}{{\text{m}}^3}\right)\\ =7.75\times {10}^3\text{N}/{\text{m}}^{\text{3}}\end{array}$

Substitute $13.6$ for $\text{SG}$, ${10}^3\text{kg}/{\text{m}}^{\text{3}}$ for ${\rho }_{{\text{H}}_{\text{2}}\text{O}}$ and $9.81\text{m}/{\text{s}}^{\text{2}}$ in Equation (III) to calculate specific weight of Mercury.

  $\begin{array}{c}\gamma =\left(13.6\right)\left({10}^3\text{kg}/{\text{m}}^{\text{3}}\right)\left(9.81\text{m}/{\text{s}}^{\text{2}}\right)\\ =133.42\times {10}^3\left(\text{kg}\cdot \frac{\text{m}}{{\text{s}}^{\text{2}}}\right)\left(\frac{\text{1}}{{\text{m}}^3}\right)\\ =133.42\times {10}^3\text{N}/{\text{m}}^{\text{3}}\end{array}$

Substitute $0.17$ for $\text{SG}$, ${10}^3\text{kg}/{\text{m}}^{\text{3}}$ for ${\rho }_{{\text{H}}_{\text{2}}\text{O}}$ and $9.81\text{m}/{\text{s}}^{\text{2}}$ in Equation (III) to calculate specific weight of Balsa wood.

  $\begin{array}{c}\gamma =\left(0.17\right)\left({10}^3\text{kg}/{\text{m}}^{\text{3}}\right)\left(9.81\text{m}/{\text{s}}^{\text{2}}\right)\\ =1.67\times {10}^3\left(\text{kg}\cdot \frac{\text{m}}{{\text{s}}^{\text{2}}}\right)\left(\frac{\text{1}}{{\text{m}}^3}\right)\\ =1.67\times {10}^3\text{N}/{\text{m}}^{\text{3}}\end{array}$

Substitute $0.93$ for $\text{SG}$, ${10}^3\text{kg}/{\text{m}}^{\text{3}}$ for ${\rho }_{{\text{H}}_{\text{2}}\text{O}}$ and $9.81\text{m}/{\text{s}}^{\text{2}}$ in Equation (III) to calculate specific weight of Dense oak wood.

  $\begin{array}{c}\gamma =\left(0.93\right)\left({10}^3\text{kg}/{\text{m}}^{\text{3}}\right)\left(9.81\text{m}/{\text{s}}^{\text{2}}\right)\\ =1.67\times {10}^3\left(\text{kg}\cdot \frac{\text{m}}{{\text{s}}^{\text{2}}}\right)\left(\frac{\text{1}}{{\text{m}}^3}\right)\\ =9.12\times {10}^3\text{N}/{\text{m}}^{\text{3}}\end{array}$

Substitute $19.3$ for $\text{SG}$, ${10}^3\text{kg}/{\text{m}}^{\text{3}}$ for ${\rho }_{{\text{H}}_{\text{2}}\text{O}}$ and $9.81\text{m}/{\text{s}}^{\text{2}}$ in Equation (III) to calculate specific weight of Gold.

  $\begin{array}{c}\gamma =\left(19.3\right)\left({10}^3\text{kg}/{\text{m}}^{\text{3}}\right)\left(9.81\text{m}/{\text{s}}^{\text{2}}\right)\\ =189.33\times {10}^3\left(\text{kg}\cdot \frac{\text{m}}{{\text{s}}^{\text{2}}}\right)\left(\frac{\text{1}}{{\text{m}}^3}\right)\\ =189.33\times {10}^3\text{N}/{\text{m}}^{\text{3}}\end{array}$

Substitute $1.7$ for $\text{SG}$, ${10}^3\text{kg}/{\text{m}}^{\text{3}}$ for ${\rho }_{{\text{H}}_{\text{2}}\text{O}}$ and $9.81\text{m}/{\text{s}}^{\text{2}}$ in Equation (III) to calculate specific weight of Bones.

  $\begin{array}{c}\gamma =\left(1.7\right)\left({10}^3\text{kg}/{\text{m}}^{\text{3}}\right)\left(9.81\text{m}/{\text{s}}^{\text{2}}\right)\\ =16.68\times {10}^3\left(\text{kg}\cdot \frac{\text{m}}{{\text{s}}^{\text{2}}}\right)\left(\frac{\text{1}}{{\text{m}}^3}\right)\\ =16.68\times {10}^3\text{N}/{\text{m}}^{\text{3}}\end{array}$

Substitute $2$ for $\text{SG}$, ${10}^3\text{kg}/{\text{m}}^{\text{3}}$ for ${\rho }_{{\text{H}}_{\text{2}}\text{O}}$ and $9.81\text{m}/{\text{s}}^{\text{2}}$ in Equation (III) to calculate specific weight of Bones.

  $\begin{array}{c}\gamma =\left(2\right)\left({10}^3\text{kg}/{\text{m}}^{\text{3}}\right)\left(9.81\text{m}/{\text{s}}^{\text{2}}\right)\\ =19.62\times {10}^3\left(\text{kg}\cdot \frac{\text{m}}{{\text{s}}^{\text{2}}}\right)\left(\frac{\text{1}}{{\text{m}}^3}\right)\\ =19.62\times {10}^3\text{N}/{\text{m}}^{\text{3}}\end{array}$

Substitute $0.916$ for $\text{SG}$, ${10}^3\text{kg}/{\text{m}}^{\text{3}}$ for ${\rho }_{{\text{H}}_{\text{2}}\text{O}}$ and $9.81\text{m}/{\text{s}}^{\text{2}}$ in Equation (III) to calculate specific weight of Ice (at $0°\text{C}$ ).

  $\begin{array}{c}\gamma =\left(0.916\right)\left({10}^3\text{kg}/{\text{m}}^{\text{3}}\right)\left(9.81\text{m}/{\text{s}}^{\text{2}}\right)\\ =8.99\times {10}^3\left(\text{kg}\cdot \frac{\text{m}}{{\text{s}}^{\text{2}}}\right)\left(\frac{\text{1}}{{\text{m}}^3}\right)\\ =8.99\times {10}^3\text{N}/{\text{m}}^{\text{3}}\end{array}$

Substitute $0.001204$ for $\text{SG}$, ${10}^3\text{kg}/{\text{m}}^{\text{3}}$ for ${\rho }_{{\text{H}}_{\text{2}}\text{O}}$ and $9.81\text{m}/{\text{s}}^{\text{2}}$ in Equation (III) to calculate specific weight of Air.

  $\begin{array}{c}\gamma =\left(0.001204\right)\left({10}^3\text{kg}/{\text{m}}^{\text{3}}\right)\left(9.81\text{m}/{\text{s}}^{\text{2}}\right)\\ =0.012\times {10}^3\left(\text{kg}\cdot \frac{\text{m}}{{\text{s}}^{\text{2}}}\right)\left(\frac{\text{1}}{{\text{m}}^3}\right)\\ =0.012\times {10}^3\text{N}/{\text{m}}^{\text{3}}\end{array}$

Conclusion:

The specific gravity of Water is $9.81\times {10}^3\text{N}/{\text{m}}^{\text{3}}$, specific gravity of Blood is $10.40\times {10}^3\text{N}/{\text{m}}^{\text{3}}$, specific gravity of Seawater is $10.06\times {10}^3\text{N}/{\text{m}}^{\text{3}}$, specific gravity of Gasoline is $6.67\times {10}^3\text{N}/{\text{m}}^{\text{3}}$, specific gravity of Ethyl alcohol is $7.75\times {10}^3\text{N}/{\text{m}}^{\text{3}}$, specific gravity of Mercury is $133.42\times {10}^3\text{N}/{\text{m}}^{\text{3}}$, specific gravity of Balsa wood is $1.67\times {10}^3\text{N}/{\text{m}}^{\text{3}}$, specific gravity of Dense oak wood is $9.12\times {10}^3\text{N}/{\text{m}}^{\text{3}}$, specific gravity of Gold is $189.33\times {10}^3\text{N}/{\text{m}}^{\text{3}}$, specific gravity of Bones is $16.68\times {10}^3\text{N}/{\text{m}}^{\text{3}}$, specific gravity of Bones is $19.62\times {10}^3\text{N}/{\text{m}}^{\text{3}}$, specific gravity of Ice is $8.99\times {10}^3\text{N}/{\text{m}}^{\text{3}}$, and specific gravity of Air is $0.012\times {10}^3\text{N}/{\text{m}}^{\text{3}}$.

To determine

(c)

The specific volume of liquid with specific gravity $0.592$.

Explanation of Solution

Write the expression for specific volume.

  $v=\frac{1}{\rho }$.......(IV)

Here, specific volume is $v$.

From equation (I) and (IV).

  $v=\frac{1}{\text{SG}\cdot {\rho }_{{\text{H}}_{\text{2}}\text{O}}}$.......(V)

Calculation:

Substitute the $0.592$ for $\text{SG}$ and ${10}^3\text{kg}/{\text{m}}^{\text{3}}$ for ${\rho }_{{\text{H}}_{\text{2}}\text{O}}$ in Equation (V).

  $\begin{array}{c}v=\frac{1}{\left(0.859\right)\left({10}^3\text{kg}/{\text{m}}^{\text{3}}\right)}\\ =\frac{1}{859\text{kg}/{\text{m}}^{\text{3}}}\\ =1.16\times {10}^{-3}{\text{m}}^{\text{3}}/\text{kg}\end{array}$

Conclusion:

The specific volume of liquid is $1.16\times {10}^{-3}{\text{m}}^{\text{3}}/\text{kg}$.