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Adding ice to water?

  1. Sep 20, 2007 #1
    1. The problem statement, all variables and given/known data
    An insulated beaker with negligible mass contains liquid water with a mass of 0.350 kg and a temperature of 69.7degrees C.

    How much ice at a temperature of -17.5degrees C must be dropped into the water so that the final temperature of the system will be 33.0 degrees C?

    2. Relevant equations
    Take the specific heat of liquid water to be 4190 J/kgkelvin, the specific heat of ice to be 2100J/kgkelvin , and the heat of fusion for water to be 334 kJ/kg.

    3. The attempt at a solution
    i think i am supposed to calculate the heat it takes for the 69.7 degree water to reach 33 degrees. and set that equal to the mcdeltaT equation for the ice and solve for mass of the ice. but i keep getting the answer wrong and it isnt my units. also how do i take into account the phase change from ice to water, do i just add it to the mcdeltaT equation for ice? any advice is appreciated
    Last edited: Sep 20, 2007
  2. jcsd
  3. Sep 20, 2007 #2


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    You are almost there.
    Work out how many Joules it takes the water to go from 69.7->33.

    Then you can write an equation for how much energy you lose to the ice for the three stages, ice heating to 0deg + ice melting + water(from melted ice) heating from 0->33deg.

    The energy f fusion is just added in - but with the correct sign - Is energy given out when ice melts or does it take energy?

    Note. Your units are wrong for the heat of fusion, since no temperature change is involved there are no kelvin.
  4. Sep 20, 2007 #3
    so joules for water=(.350kg)(4190J/kg)(306-342.7)=-53820.55
    ice heating to 0= (unknown m)(2100J/KgK)(273-255.5)=36750
    ice melting= (unknown m)(what should go here?)
    water from ice= (unknown m)(334kJ/Kg)(1000J/1kJ)(306-273)=11022000m

    are these the right equations? and if so all i have to do is add them up and set that equal to what to solve for m?
  5. Sep 20, 2007 #4


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    All you have to do is set them equal to the energy lost by the hot water and solve for m

    Remember that it is 2100J PER kg so you have the equations the wrong way round
    Check the units on both sides of the equation, they must be the same.
    And since you are only adding temperatures there is no need to convert to kelvin, a difference in deg c = a difference in kelvin.

    Energy lost by hot water = (69.7-33) * 4190 / 0.350 = 439KJ

    Ice heating = 17.5*2100/m
    Ice melting = 334000/m ( melting ice takes heat so increases the cooling effect)
    Water heating = 33*4190/m

    This equals the heat lost by the hot water so:
    17.5*2100/m + 334000/m + 33*4190/m = 439KJ

    rearanged (17.5*2100 + 334000 + 33*4190)/m = 439KJ
    and so m = (17.5*2100 + 334000 + 33*4190)/439K

    (remember to check the numbers!)
    Last edited: Sep 20, 2007
  6. Sep 20, 2007 #5
    i'm still not getting the right answer, i get that i need to change 439 to joules, but it isnt working out what might be wrong, should 17.5 be negative?
  7. Sep 20, 2007 #6


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    Remember that it is 439K ie 439000J other than that just check the arithmetic
  8. Sep 20, 2007 #7
    i used 439000 but i still got the wrong answer and i dont see anything wrong with the arithmetic....i dont know why i am not gettin gthis i have done similar problems and gotten them right, something here just isnt adding up....
    Last edited: Sep 20, 2007
  9. Sep 20, 2007 #8
    i just tried it with all the temps in kelvin and stilli cant figure out what i am doing wrong someone please help me with this.
  10. Sep 20, 2007 #9


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    Oops, a typo in my equations.

    Ice heating = 17.5*2100 J / kg
    Ice melting = 334000 J / kg
    Water heating = 33*4190 J/kg

    so 439KJ = (17.5*2100 + 334000 + 33*4190)J/Kg * m kg
    ie m = 439000 / (17.5*2100 + 334000 + 33*4190) = 0.86Kg

    Remember to make sure the units balance! This graphics explains it quite well
  11. Sep 20, 2007 #10
    .86kg is also incorrect, i have only one more try at, this any other ideas?
  12. Sep 20, 2007 #11
    i got an answer of .122kg earlier and the program told me to " Remember that the ice turns into liquid water above the melting point." i tried to incoporate this but was unsuccessful, what does it mean in terms of an equation?
  13. Sep 20, 2007 #12


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    It means that you have to add in the energy needed to raise M kg of ice from 0 to 33deg at 4190J/kgK which we have done.
  14. Sep 20, 2007 #13
    so it should be greater or less than .122, and why is .86 incorrect if we took that into account?
  15. Sep 20, 2007 #14


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    I don't know other than rounding/arithmatic error - I never know how smart these online sites are, do they have a margin of error?
    I checked the values of specific/latent heat capacity and they are pretty close.
  16. Sep 20, 2007 #15
    if u are off by a little they usually tell u and accept the answer, .86 was just wrong, if u are close they give u some advice, and tell u to check your rounding, this was not the case.
  17. Sep 21, 2007 #16


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    This bit is fine.
    How about the latent heat of fusion?

    The part I highlighted in red is not what you should be using. Remember the ice has turned into water now so you should be using values for water.
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