Question

Determine the mass density, specific volume and specific weight of a liquid whose specific gravity 0.85.

Answer 1

The specific gravity of the liquid is given as 0.85.

The formula to find specific gravity

$\[ = \frac{{Density\,of\,Liquid\,given\,(\rho )}}{{Density\,of\,Water}}\]$

$% MathType!Translator!2!1!AMS LaTeX.tdl!AMSLaTeX!\[S.G = \frac{{Density\,of\,Liquid\,given\,(\rho )}}{{Density\,of\,Water}}\]% MathType!End!2!1!$The formula for the specific volume of liquid$% MathType!Translator!2!1!AMS LaTeX.tdl!AMSLaTeX!=\[\frac{{Fluid\,volume}}{{Fluid\,mass}}\]% MathType!End!2!1!$$\[ = \,\frac{{Fluid\,volume}}{{Fluid\,mass}}\]$

The formula for the  specific weight of liquid=$% MathType!Translator!2!1!AMS LaTeX.tdl!AMSLaTeX!\[density\,of\,the\,given\,liquid\,(\rho )\, \times \,g\]% MathType!End!2!1!$$\[density\,of\,the\,given\,liquid\,(\rho )\, \times \,g\]$

Explanation: The specific gravity of the given liquid is 0.85 so,

$% MathType!Translator!2!1!AMS LaTeX.tdl!AMSLaTeX!\[S.G = \frac{{Density\,of\,Liquid\,given\,(\rho )}}{{Density\,of\,Water}}\]% MathType!End!2!1!$$\[S.G = \frac{{Density\,of\,Liquid\,given\,(\rho )}}{{Density\,of\,Water}}\]$

$% MathType!Translator!2!1!AMS LaTeX.tdl!AMSLaTeX!\[\begin{gathered} \rho = \,S.G.\, \times \,1000 \hfill \\ \rho = \,0.85\, \times \,1000 \hfill \\ \rho = \,850\,kg/{m^3} \hfill \\ \end{gathered} \]% MathType!End!2!1!$$\[\begin{gathered} \rho = \,S.G.\, \times \,1000 \hfill \\ \rho = \,0.85\, \times \,1000 \hfill \\ \rho = \,850\,kg/{m^3} \hfill \\ \end{gathered} \]$

$% MathType!Translator!2!1!AMS LaTeX.tdl!AMSLaTeX!\[Mass\,density\,of\,given\,liquid\, = \,850\,kg/{m^3}\]% MathType!End!2!1!$$\[\begin{gathered} Mass\,density\,of\,given\,liquid\, = \,850\,kg/{m^3} \hfill \\ Specific\,volume\, = \,\frac{{Fluid\,volume}}{{Fluid\,mass}} \hfill \\ Specific\,volume = \,\frac{1}{\rho } \hfill \\ = \frac{1}{{850}} = {\text{ }}1.1760{\text{ }} \times {\text{ }}{10^{-3}}{m^3}/Kg \hfill \\ \end{gathered} \]$

$% MathType!Translator!2!1!AMS LaTeX.tdl!AMSLaTeX!\[\begin{gathered} Specific\,volume\, = \,\frac{{Fluid\,volume}}{{Fluid\,mass}} \hfill \\ Specific\,volume = \,\frac{1}{\rho } \hfill \\ = \frac{1}{{850}} = {\text{ }}1.1760{\text{ }} \times {\text{ }}{10^{-3}}{m^3}/Kg \hfill \\ Specific\,volume\,of\,liquid\, = 1.1760{\text{ }} \times {\text{ }}{10^{-3}}{m^3}/Kg \hfill \\ \end{gathered} \]% MathType!End!2!1!$$\[\begin{gathered} Specific\,volume\,of\,liquid\, = 1.1760{\text{ }} \times {\text{ }}{10^{-3}}{m^3}/Kg \hfill \\ Specific\,weight = \,density\,of\,the\,given\,liquid\,(\rho )\, \times \,g \hfill \\ Specific\,weight = \rho \times g \hfill \\ Specific\,weight = 850 \times 9.81 \hfill \\ \end{gathered} \]$

$% MathType!Translator!2!1!AMS LaTeX.tdl!AMSLaTeX!\[\begin{gathered} Specific\,weight = \,density\,of\,the\,given\,liquid\,(\rho )\, \times \,g \hfill \\ Specific\,weight = \rho \times g \hfill \\ Specific\,weight = 850 \times 9.81 \hfill \\ Specific\,weight = 8338.5N/{m^3} \hfill \\ \end{gathered} \]% MathType!End!2!1!$$\[Specific\,weight = 8338.5N/{m^3}\]$

So, the specific weight of the given liquid will be:

$\[8338.5N/{m^3}\]$