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Calculating the final temperature of a mixture of Ice and Water

  1. Jul 19, 2013 #1
    1. The problem statement, all variables and given/known data
    How do I calculate the following. The final temperature of a mixture of of ice and water. Where 176grams of ice at -10celsius mixes with 206 grams of water at 73.3celsius. I have tried this equation in multiple fashions and cannot seems to come to a consistent answer! Any help would be appreciated!


    2. Relevant equations



    3. The attempt at a solution

    Why does the below not work?
    water specific heat=4.184
    ice specific heat= 2.11
    heat of fusion=334

    ((specific Heat ice)X(Mass ice)x(∆temperature))+((mass ice)*(heat of fusion))+((specific Heat ice)X(Mass ice)x(x-0))=((specific Heat water)X(Mass water)x(∆temperature))
     
  2. jcsd
  3. Jul 19, 2013 #2

    ehild

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    What do you mean on ∆temperature?

    After the ice reached the melting temperature and melted, it became water at 0 degree, and its specific heat is the same as that of water.

    ehild
     
  4. Jul 20, 2013 #3
    Hello ehlid,

    I have tried this in stages but continue to do something incorrectly.

    The Amount of Energy in Water
    ((specific Heat water)X(Mass water)x(∆temperature))
    4.18 X 73.3C X 206gram=63,117Joules

    The Amount of Energy to get ice to melting point
    ((specific Heat ice)X(Mass ice)x(∆temperature))
    2.11 * 10c * 176grams=3,713Joules

    Energy for Melting
    ((mass ice)*(heat of fusion))
    334*176=58,784

    Total Energy of ice melt and bringing to melting point
    ((specific Heat ice)X(Mass ice)x(∆temperature))+((mass ice)*(heat of fusion))
    62,497

    62,497<63,117 complete melting of ice occurs

    the remaining energy is 602 joules
    the total remaining water mass is 176+206=326grams
    602joules=326grams * 4.18 * (x-0)
    602= 1,362 * (x-0)
    0.44=(x-0)

    is this correct?
     
  5. Jul 20, 2013 #4

    lurflurf

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    ^mostly
    remaining energy=
    -62,497+63,117=620 not 602
    typo?
     
  6. Jul 20, 2013 #5
    yes, 620 is correct. apologies. Is this the correct answer?
     
  7. Jul 20, 2013 #6

    ehild

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    Why did you ignore the last digit of the specific heat? 4.184 X 73.3C X 206gram=63,522Joules

    It is 1029 J

    No. Recalculate with the correct remaining energy. But the method is correct.

    ehild
     
  8. Jul 21, 2013 #7

    lurflurf

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    I wonder if the desired answer should be 0°C.
    Can you quote the question exactly?
    It seems we are to use significant 3 digits for each value.
    Then we should write -10.0°C.
    When we subtract our 3 digits will be reduced to 1.
    This leads to an answer of about 0.4°C.
    As this value is uncertain the mixture might contain some unmelted ice or be slightly warmer than 0°C. Not that it would make much difference.
     
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