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Heating the mixture

  1. Oct 22, 2014 #1
    In a thermally insulated container was a mixture of ice and water. Then we have put a immersion heater ( P=400W). We started measure time and at the end of every minute we writed down the heat od mixture. First two minutes the heat wasn't changing, at the end of third it raised about 2°C, at the end of fourth it raised about 8°C, t the end of fifth raised about 8°C, and so on, each by an additional 8°C.

    Task:
    a) How much ice and how much water was in a container in the beginning? (my answer: mass of water: 0,515kg, mass of ice: 0,199kg)
    b) How long did it takes to ice, untill it melted? (my answer: 2:45min)
    c) How long did it takes untill the water started to boil? And how long did it takes untill the half of water did evaporate? (my answer: 14:25min and 48:2min)


    Thanks in advance!!
     
  2. jcsd
  3. Oct 22, 2014 #2
    (Sorry i forgot to write the header: )

    Can you please check me my work?
    And sorry for my english:)
     
  4. Oct 22, 2014 #3

    berkeman

    User Avatar

    Staff: Mentor

    Welcome to the PF.

    It helps us out a lot if you show your detailed work for how you arrived at your answers. That way we can just check your work, and do not have to work it all out for ourselves. And if you have an error in your calculations, it is much easier for us to spot if we see your detailed work. :-)
     
  5. Oct 23, 2014 #4
    a) I suppose that at first the heat was absorbed to ice to melt, so that's why the temeperature of the mixture didn't raise first two minutes. And i also suppose, that at first the ice and water had same degree ( 0°C).
    To know what was the mass of each(because the mass of ice after melting was same as before), I counted it from the heat needed to raise the temperature of the mixture about 8°C , when the ice was comply melted:

    (P=400W, t=8°, c1=4200kJ/kg)
    Q=P*t=c1*t*(m1+m2)
    ---> (m1+m2)=24000/(42000*8)=5/7

    Then I get the second equation. I know that in the third minute temp raised about 2°C, so I substract the heat needed to ice to melt from the heat, which was given the mixture after three minutes.

    Q'3=Q3-Lt=P*3*60 - lt*m2

    This heat heated the mixture above 2°C: (t2=2°, lt=33200)
    Q3-Lt=c1*t2*(m1+m2)

    m2= Q3/lt - ((5/7)*c1*t2) / lt
    m2= ((P*3*60) / (33200)) - ((4200*2*5)/(332000*7))= 0,199kg

    m1... m1+m2=5/7...m1=5/7-0.199=0,515kg
     
  6. Oct 27, 2014 #5
    Is it right?
     
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