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

Answer:

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

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

$Density=\dfrac{mass}{Volume}$   ---1

Specific volume

$specific\ volume=\dfrac{Volume}{mass}$    ---2

From equation 1 and 2

We can say that

$specific\ volume=\dfrac{1}{density}\ m^3/kg$

$specific\ volume=\dfrac{1}{1025}\ m^3/kg$

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