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4 years ago

6

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)?

1 answer:

Svetradugi [14.3K]4 years ago

6 0

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

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

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

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

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3 years ago

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3 years ago