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# Water Density Calculator

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This online calculator helps to determine density of water at a particular temperature. Water density slightly varies according to the temperature. So if we know the temperature, one can calculate the water density using the formula,
$\rho$ = 1000$\left \{ 1-\left [ \frac{T+288.9414}{508929(T+68.129630)} \right ](T-3.9863)^{2} \right \}$

## Steps

Step 1 : Note down the value of temperature given in the question.

Step 2 : Substitute in the formula and find out the density of water

$\rho$ = 1000$\left \{ 1-\left [ \frac{T+288.9414}{508929(T+68.129630)} \right ](T-3.9863)^{2} \right \}$.

## Problems

Let us discuss the problems related to water density.

### Solved Examples

Question 1: Calculate the density of water at 30$^{\circ}$.
Solution:

Given that,
T = 30$^{\circ}$

Density of water at 30$^{\circ}$ is,

$\rho$ = 1000$\left \{ 1-[\frac{T+288.9414}{508929(T+68.129630)}](T-3.9863)^{2} \right \}$

$\rho$ = 1000$\left \{ 1-[\frac{30+288.9414}{508929(30+68.129630)}](30-3.9863)^{2} \right \}$

$\rho$ = 1000(1 - 0.0043217) = 1000$\times$0.995678 = 995.678kg/m3

Question 2: Calculate the water density at 45$^{\circ}$.
Solution:

Given that,
T = 45$^{\circ}$

Density of water at 45$^{\circ}$ is,

$\rho$ = 1000$\left \{ 1-[\frac{T+288.9414}{508929(T+68.129630)}](T-3.9863)^{2} \right \}$

$\rho$ = 1000$\left \{ 1-[\frac{45+288.9414}{508929(45+68.129630)}](45-3.9863)^{2} \right \}$

$\rho$ = 1000(1 - 0.0097565) = 1000$\times$0.9902435 = 990.2435kg/m3