1. Not finding help here? Sign up for a free 30min 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!

How many kg of ice must be dropped to make Tf= 22.7C?

  1. Apr 10, 2016 #1
    1. The problem statement, all variables and given/known data
    An insulated beaker with neglible mass contains .3kg of water at a temp of 78.6 C. How many kg of ice at -19.6 C must be dropped into the water to make Tf= 22.7C?
    Specific heat for water= 4190 J/kgK
    specific heat for ice= 2100 J/kgK
    heat of fusion for water= 334kJ/kg

    2. Relevant equations
    Q=mc(Tf-Ti)
    Q=mL

    3. The attempt at a solution
    Q1 = bring ice to 0 C
    • Q=micec(Tf-Ti)
    • Q=mice(2100 J/kgK)(0-19.6)
    • Q= 41160mice
    Q2 = melt ice (phase change)
    • Q=mL
    • Q=mice(334000 J/kg)
    Q3= cool down water
    • Q=mwaterc(Tf-Ti)
    • Q=(.3kg)(4190 J/kgK)(22.7 - 78.6)
    • Q=-70266.9 J
    Q1+Q2+Q3=0
    41160mice + 334000mice + (-70266.9)=0
    375160mice= 70266.9
    mice = .187kg

    This answer isn't correct. What did I do wrong or what did I miss? I'm thinking maybe I missed another Q but I don't know what it could be.
     
  2. jcsd
  3. Apr 10, 2016 #2

    SteamKing

    User Avatar
    Staff Emeritus
    Science Advisor
    Homework Helper

    Remember, once the ice melts, the liquid from this mixes with the liquid already in the beaker.

    The final temperature of 22.7 °C must be reached by all of the liquid in the beaker.
     
  4. Apr 10, 2016 #3
    Then would Q3 have to include both the masses?
     
  5. Apr 10, 2016 #4

    SteamKing

    User Avatar
    Staff Emeritus
    Science Advisor
    Homework Helper

    Yep. You can't separate the melt water from the warm water already in the beaker when the ice was dropped in.
     
  6. Apr 10, 2016 #5
    So then its
    Q3=(mice+mwater)c(Tf-Ti)
    =(.3kg + mice)(4190)(22.7-78.6)
    =(.3kg + mice)(-234221)
    = -70266.3 - 234221mice

    And then add this Q3 to the others and set =0?
     
  7. Apr 10, 2016 #6

    SteamKing

    User Avatar
    Staff Emeritus
    Science Advisor
    Homework Helper

    Be careful here.

    The warm water in the beaker is being cooled from 78.6 °C to 22.7 °C.
    The melt water is being warmed from 0 °C to 22.7 °C.

    Your equations should reflect this.
     
  8. Apr 10, 2016 #7
    So what I really need is 2 extra equations then?
     
  9. Apr 10, 2016 #8
    No, nevermind. Just one. For the ice water. Q4= micecice(22.7-0)
     
  10. Apr 10, 2016 #9

    SteamKing

    User Avatar
    Staff Emeritus
    Science Advisor
    Homework Helper

    Remember, the ice has already melted. cice should be cwater.
     
  11. Apr 10, 2016 #10
    Oh of course. Ok I got it, thanks!
     
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook

Have something to add?
Draft saved Draft deleted



Similar Discussions: How many kg of ice must be dropped to make Tf= 22.7C?
Loading...