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Problem with calculating final temperature of water

  1. Oct 20, 2012 #1
    A piece of ice with a mass of 60 g is transferred directly from the freezer at a
    temperature of –7°C to a polystyrene cup containing 200 g of water at 75°C. Calculate
    the final temperature of the water. Ignore any heat loss to the cup and its surroundings.

    Qheat ice + Qmelt ice + Qheat cold water + Qcool hot water
    mciceΔtice + mLf + mcwΔt + mhwcwΔthw = 0
    (0.060 kg) x (2100 J/(kg.C)) x (0°C - (-7°C)) + ( 0.060 kg) x (3.3 x 10^5) + (0.060 kg) x (4200 J /(kg.C)) x (t2 – 0°C) + (0.2) x (4200 J) x (t2 – 75°C) = 0
    882 + 19800 + 252t2 + 840t2 – 63000 = 0
    840t2 = 42318
    t2 = 50°C

    The above problem is from my book, but I am having difficulty with the equation. Where did
    the 63000 come from?
     
  2. jcsd
  3. Oct 20, 2012 #2

    TSny

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    Gold Member

    Hi, V3ra. Multiply the three red numbers.
     
  4. Oct 20, 2012 #3
    Hmm, but how would I have known to do that? I already multiplied the 0.2 and 4200 J to get 840. Where in the equation is it implied that I would use these two numbers again along with the 75°C to get 63000?
     
  5. Oct 20, 2012 #4
    It's called the distributive property.
    a*(b+c)=a*b+a*c
    You have a*(b+c)=a*b
     
  6. Oct 21, 2012 #5
    Can you please elaborate? How would someone who is approaching the above equation for the first time know to apply such a rule? The first part of the equation contains three numbers that are being multiplied, would all three be represented by "a" in the "a*(b+c)=a*b" or is this rule only applying to one part of the equation?
     
  7. Oct 21, 2012 #6

    TSny

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    v3ra, I'm not sure of your question. You are trying to simplify (0.2) x (4200) x (t2 – 75). You noted earlier that you can multiply the 0.2 and the 4200 to get 840. So now you have 840 x (t2-75).

    To reduce this, take a simple example: 3 x (6 - 2). This means you have the quantity (6-2) three times.
    So, 3 x (6 - 2) = (6-2) + (6-2) + (6-2). So, you get "6" three times and you also get (-2) three times.

    So, 3 x (6-2) = 3x(6) + 3x(-2) = 3x6 - 3x2. Thus 3 x (6-2) = 3x6 - 3x2

    Likewise, 840 x (t2-75) means that you have 840 of the quantity (t2-75). That's 840 t2 's and also 840 of the quantity (-75).

    So, 840 x (t2-75) = 840 x t2 + 840 x (-75) = 840 x t2 - 840x75 = 840t2 - 63000.
     
    Last edited: Oct 21, 2012
  8. Oct 21, 2012 #7
    I understand now. I was, for some reason, completely avoiding the "t2".
     
  9. Oct 21, 2012 #8
    No, you had t2, it was the -75 that you left out. That's the term that became the -63000.
     
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