Question

Seawater has a specific density of 1.025. What is its specific volume in m^3/kg (to 3 significant figures of accuracy, tolerance +/- 0.000005 m^3/kg)?

Answer 1

`specific\ volume=0.00097\ m^3/kg`

Step-by-step explanation:

Given that

Specific gravity of sea water = 1.025

So density of sea water = 1.025 x 1000
`kg/m^3`

Density of sea water = 1025
`kg/m^3`

We know that

`Density=(mass)/(Volume)` ---1

Specific volume

`specific\ volume=(Volume)/(mass)` ---2

From equation 1 and 2

We can say that

`specific\ volume=(1)/(density)\ m^3/kg`

`specific\ volume=(1)/(1025)\ m^3/kg`

`specific\ volume=0.00097\ m^3/kg`