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Finding mass of ice in latent heat of fusion problem

  1. Mar 1, 2009 #1
    1. The problem statement, all variables and given/known data

    The question is: a 6 pack of cola cans contains m grams of liquid. you want to put ice into a perfectly insulated box to cool it down from Ts to Tf. The ice initially has temp Ti. So (a)how much ice would you need as a minimum. (ignore presence of cans and final state of ice will be water). (b) calculate the mass of ice initially at -5 deg celcius needed to cool 2.1kg of canned cola from 23 deg celcius to 4 deg celcius.

    2. Relevant equations

    specific heat of water = 4.18J
    Lf of water = 333.55Jkg-1
    mass(cola) x C (water) (Tf - Ts) = mass(ice) x C(ice) (0 - Ti) + mass(ice) x Lf + mass(ice) x C(water) (Tf - 0)


    3. The attempt at a solution

    I've worked out that the amount of energy needed to melt 'm'kg of ice = m(ice) x Lf(water) or Q = mL. also that heat lost by cola = heat gained by ice

    I realise there are 3 stages involved:
    (1) heat needed to raise ice to O deg celcius
    (2) heat needed to melt ice
    (3) heat needed to raise temp of water (from melted ice) to Tf from O deg celcius.

    Then I get lost. How do I find the mass of the ice to begin with? Do I rearrange the equation? If so, I have no idea how to start rearranging a long complex equation like this.

    Thank you for any guidance on this.
     
  2. jcsd
  3. Mar 1, 2009 #2

    tiny-tim

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    Welcome to PF!

    Hi aussie-girl ! Welcome to PF! :smile:
    call the mass of ice M, and write out the equations …

    then we'll see how to rearrange them :wink:
     
  4. Mar 1, 2009 #3
    Re: Welcome to PF!

    Hi tiny-tim, nice to meet you!

    So done as suggested and here is the result:

    part (a):

    m(cola) x C(water) (Tf - Ts) = M(ice) x C(ice) (0 - Ti) + M(ice) x Lf + M(ice) x C(water) (Tf - 0)

    part (b):

    2.1kg x 4.186 x (4 - 23) = M(ice) x 2.093 x (0 - -5) + M(ice) x 333.55 + 4.186 (4 - 0)


    I think I've substituted in the right figures. Please advise how I now proceed?

    Thanks, appreciate your time and help!
     
  5. Mar 1, 2009 #4

    tiny-tim

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    Hi aussie-girl! :smile:

    (i haven't checked your figures, but …)

    put M(ice) outside a bracket on the right-hand side …

    M(ice) x [2.093 x (0 - -5) + 333.55 + 4.186 (4 - 0) ] :wink:
     
  6. Mar 1, 2009 #5
    Hi again

    I'm confused by what you said. I can see that you've taken the second M(ice) out of the equation and assume you mean to put that outside a bracket on the right-hand side? but that doesn't make sense to me. are you showing me what you mean by this:

    M(ice) x [2.093 x (0 - -5) + 333.55 + 4.186 (4 - 0) ] but that can't be right because the second M(ice) is missing.

    I've been thinking about this and I can't figure it out. maybe I'm just being thick but it is 3.30am over here! can you please explain more? I'm not expecting you to solve it for me but I need more explanation.

    Thanks again...
     
  7. Mar 1, 2009 #6

    tiny-tim

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    ah yes … i should have noticed that …

    the clue's in the name, isn't it? :biggrin:

    Ma + Mb = M(a + b) …

    (oh, and I should have written M(ice) x [2.093 x (0 - -5) + 333.55] + 4.186 (4 - 0) :redface:)

    that may not mean anything to you right now, but …

    get some sleep! :zzz:

    … and it'll all be clear in the morning. :smile:
     
  8. Mar 1, 2009 #7
    Hi again

    Just before I go to bed (I need to know this stuff for a paper thats due tomorrow!) can I see if this is what you mean:

    2.1kg x 4.186 x (4 - 23) = M(ice) x [2.093 x (0 - -5) + 333.55] + 4.186 (4 - 0)

    -167.021 = M(ice) x 362.2

    so to get M(ice) on its own via rearrangement, you get:

    M(ice) = -167.021 / 362.2 = -0.46

    I think I'm going crazy :'( this doesn't seem right! the answer is a negative. God I don't know! I think I'll go to bed now...

    thanks again tiny-tim.
     
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