To solve this problem we will apply the theoretical concepts and definitions given for the specific weight, density and specific gravity. Consider also that a barrel contains 159 liters or [tex]0.159 m^3[/tex]
Consider the net weight of the oil which would be
[tex]W_{oil}= 1500-110[/tex]
[tex]W_{oil}= 1390 N[/tex]
The specific weight is defined as the proportion of weight by volume thereof, therefore
[tex]\gamma = \frac{1390}{0.159}[/tex]
[tex]\gamma = 8742.14 N/m^3[/tex]
Through the information given we could find mass and density through the following relations:
Mass
[tex]W = mg \rightarrow m = \frac{W}{g}[/tex]
[tex]m= \frac{1390}{9.8}[/tex]
[tex]m = 141.84 kg[/tex]
Density
[tex]\rho = \frac{m}{V} \rightarrow \text{Here m is the mass and V the Volume}[/tex]
[tex]\rho = \frac{141.84}{0.159}[/tex]
[tex]\ rho = 892.05 kg/m^3[/tex]
Specific gravity,
[tex]S= \frac{\rho_{oil}}{\rho_{water}} \text{Her the specific gravity is the ratio between the density of the oil and the water}[/tex]
[tex]S= \frac{892.05}{1000}[/tex]
[tex]S= 0.892[/tex]
