1. Limited time only! Sign up for a free 30min personal tutor trial with Chegg Tutors
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Calculating the final temperature of a mixture when ice & water

  1. Jul 19, 2013 #1

    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!

    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))

    any help would be great appreciated

    Additional Details
    This is taking place in a completely insulated environment.
  2. jcsd
  3. Jul 19, 2013 #2
    Try doing this in stages. How much heat is required to bring the ice to the point of melting? Does the water have that much heat in it? If it does, then how much heat is left? Is it enough to melt all the ice? And so on.
  4. Jul 20, 2013 #3
    Hello Voko,

    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))

    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<63,117 complete melting of ice occurs

    the remaining energy is 602 joules
    the total remaining water mass is 176+206=382grams
    620joules=382grams * 4.18 * (x-0)
    602= 1,597 * (x-0)

    This does not seem correct though
    Last edited: Jul 20, 2013
  5. Jul 20, 2013 #4
    I get 620 J for the residual heat.
  6. Jul 20, 2013 #5
    Sorry 620 is the number that I come to. "602" was a typo. Is this the correct calculation?
  7. Jul 20, 2013 #6
    I do not see any error.
  8. Jul 20, 2013 #7
    Actually, I do.

    This: "176+206=326" can't be right.
  9. Jul 20, 2013 #8
    good catch I have corrected the 602 error and 326 error in my post. Other than those two Items, you believe my method is the exact way to calculate this problem?
  10. Jul 20, 2013 #9
    Everything else looks good.
Share this great discussion with others via Reddit, Google+, Twitter, or Facebook