1. ## unit conversion help

I have been given the density of milk = $\displaystyle 990 kg m^{-3}$

and a volume of $\displaystyle 3000 Litres h^{-1}$ I need to work out the
mass.

I know $\displaystyle \rho\\\\\\\\\\\\( kg m^{-3} ) = \frac{m( kg)}{ v( m^{3})}$

so I need to convert $\displaystyle 3000 Litres h^{-1}$ to meters,

can some please check my working.

1L = 1000 ml

1 ml = 1cm

1m = 100cm

$\displaystyle \frac{3000 L}{1 h } \times \frac{1000ml}{1L} = \frac{3000000 ml}{h} = \frac{3000000 cm}{h}$

$\displaystyle \frac{3000000 cm}{h} \times \frac{1m}{100cm} = \frac{30000 m}{h}$

to get m^{3} = $\displaystyle (\frac{30000}{h})^{3} = 2.7 \times 10^{13} m^{3} h^{-1}$

so $\displaystyle m kg = 990 kg m^{-3} \times 2.7 \times 10^{13} m^{3} = 2.673 \times 10^{16} kg$

is this correct ?

2. Originally Posted by Tweety
I have been given the density of milk = $\displaystyle 990 kg m^{-3}$

and a volume of $\displaystyle 3000 Litres h^{-1}$ I need to work out the
mass.

I know $\displaystyle \rho\\\\\\\\\\\\( kg m^{-3} ) = \frac{m( kg)}{ v( m^{3})}$

so I need to convert $\displaystyle 3000 Litres h^{-1}$ to meters,

can some please check my working.

1L = 1000 ml

1 ml = 1cm

1m = 100cm

$\displaystyle \frac{3000 L}{1 h } \times \frac{1000ml}{1L} = \frac{3000000 ml}{h} = \frac{3000000 cm}{h}$

$\displaystyle \frac{3000000 cm}{h} \times \frac{1m}{100cm} = \frac{30000 m}{h}$

to get m^{3} = $\displaystyle (\frac{30000}{h})^{3} = 2.7 \times 10^{13} m^{3} h^{-1}$

so $\displaystyle m kg = 990 kg m^{-3} \times 2.7 \times 10^{13} m^{3} = 2.673 \times 10^{16} kg$

is this correct ?
what do you mean by $\displaystyle Litres h^{-1}$ ?

litres by itself is volume ... $\displaystyle 1 \, L = 10^3 \, cm^3 = 10^{-3} \, m^3$

so ... $\displaystyle 3000 \, L = 3000 \cdot 10^{-3} \, m^3 = 3 \, m^3$

$\displaystyle m = \rho \cdot V$

$\displaystyle m = 990 \, \frac{kg}{m^3} \cdot 3 \, m^3 = 2970 \, kg$

3. Originally Posted by skeeter
what do you mean by $\displaystyle Litres h^{-1}$ ?

litres by itself is volume ... $\displaystyle 1 \, L = 10^3 \, cm^3 = 10^{-3} \, m^3$

so ... $\displaystyle 3000 \, L = 3000 \cdot 10^{-3} \, m^3 = 3 \, m^3$

$\displaystyle m = \rho \cdot V$

$\displaystyle m = 990 \, \frac{kg}{m^3} \cdot 3 \, m^3 = 2970 \, kg$

Litres per hour

$\displaystyle \frac{3000 L}{h}$

thanks

4. Then it is not volume, it is a rate of flow.

5. Originally Posted by HallsofIvy
Then it is not volume, it is a rate of flow.
But I can still use it to work out the mass?

6. Originally Posted by Tweety
But I can still use it to work out the mass?
... you get kg/hr