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Thread: unit conversion help

  1. #1
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    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 ?
    Last edited by Tweety; Oct 19th 2010 at 01:21 PM.
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  2. #2
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    Quote Originally Posted by Tweety View Post
    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 ... $\displaystyle 3000 \, L = 3000 \cdot 10^{-3} \, m^3 = 3 \, m^3$

    $\displaystyle m = \rho \cdot V$

    $\displaystyle m = 990 \, \frac{kg}{m^3} \cdot 3 \, m^3 = 2970 \, kg$
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  3. #3
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    Quote Originally Posted by skeeter View Post
    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
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  4. #4
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    Then it is not volume, it is a rate of flow.
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  5. #5
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    Quote Originally Posted by HallsofIvy View Post
    Then it is not volume, it is a rate of flow.
    But I can still use it to work out the mass?
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  6. #6
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    Quote Originally Posted by Tweety View Post
    But I can still use it to work out the mass?
    ... you get kg/hr
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