Anonymous
Anonymous asked in Education & ReferenceHomework Help · 1 month ago

chem help!!!?

I need some fancy water for a party I'm having, so I buy a hunk of a glacier. If I have 72.7 mol of glacial H2O at a temperature of -28.7 C, how much energy (in kJ) do I need to put in to serve it at a temperature of 12.3 C?

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  • 1 month ago

    Molar mass

    H₂  = 2 * 1.008 = 2.016

    O = 15.999

    Total = 18.015g/mol

    ----------------------------

    Mass of water

    m = 72.7mol * 18.015g/mol

    m = 1,309.6905 g

    -------------------------

    Energy 

    Q = mc∆T

    c

    Specific heat of water = 4.186 J/(gCº)

    Specific heat of ice = 2.04 J/(gCº)

    Energy for ice

    ∆T from -28.7ºC to 0ºC = 28.7ºC

    Q = 1,309.6905 g * 2.04 J/(gCº) * 28.7ºC

    Q ≈ 76,680 J

    Energy for water

    ∆T from to 0ºC to 12.3ºC = 12.3ºC

    Q = 1,309.6905 g * 4.186 J/(gCº) * 12.3ºC

    Q ≈ 67,433 J

    Total energy

    Q = 76,680 J + 67,433 J

    Q = 144,113 J

    Q = 144.1 kJ <––––––

     

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