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Homework Help: Heat transfer and latent heat

  1. Dec 8, 2009 #1
    1. The problem statement, all variables and given/known data
    How much energy is required to change a 46.0 g ice cube from ice at -10.0°C to steam at 116°C?


    2. Relevant equations
    specific heat
    c = Q/(m(Tf-Ti))
    Q = cm(Tf-Ti)

    latent heat
    Q = mL

    3. The attempt at a solution
    2.05*10*46+4.1813*100*46+2.080*16*46 + 2260*.046+334*.046
    = 21827.184J

    (please only speak in metrics or you'll lose me)

    can anyone help me figure this out?
    It's very crucial that I understand this process.
    thanks
     
  2. jcsd
  3. Dec 8, 2009 #2

    mgb_phys

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    Looks like the right technique, you should put some units in to make sure that you didn't lose anything.
    It's also normal to work in base units eg kg,

    heat ice 10deg = 2.05 J/gk *10k *46g = 943 J
    melt ice = 333.5 J/g * 46g = 15340J
    heat water 100deg = 4.1813 j/gK * 100k*46g = 19300J
    boil water = 2260 kJ/kg * 46g = 2260 J/g * 46g =104000J
    heat steam 16deg = 2.080 J/gk *16k*46g = 1530J

    Total 140K Joules
     
    Last edited: Dec 8, 2009
  4. Dec 8, 2009 #3
    apparently the table on wikipedia was using kJ/kg which threw me off

    2.05 J/(gK) *10 K*46g+4.1813 J/(gK) *100K*46g+2.080 J/(gK) *16K*46g + 2260J/g*46g+334J/g*46g
    = 141031.86

    I still don't know if this is right though

    I also need to know if I was supposed to be adding for the ice to melt of subtracting the latent heat energy (it may be releasing energy when it melts as opposed to absorbing it)
     
  5. Dec 8, 2009 #4

    mgb_phys

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    That's why I suggested writing it out as above.
    Then it's more obvious if a number is wildly wrong

    eg,
    warming ice 10 deg small amount of energy
    melting ice = a lot of energy (it takes along time for freezer to defrost)
    heating water 100deg = a lot of energy (think how long it takes to make a coffee)
    boiling water = a lot of energy, thats why steam is so powerful
    heating the steam = small energy

    The energy flow is all in the same direction (same sign) at each step you are putting energy in
     
  6. Dec 8, 2009 #5
    nothing you're saying is incredibly unreasonable... but I never tried to convert kj to j

    ????

    (I did however have numbers from wiki. that said x kJ/kg
    which is inherently equal to J/g)
     
  7. Dec 8, 2009 #6
    *bows head in shame and bumps thread*
     
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