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!

Water and Ice mixture, Find original temperature of water

  1. Apr 15, 2016 #1
    1. The problem statement, all variables and given/known data
    An 8 cm3 ice cube (temperature = 0.00 °C) is dropped into a glass with 3 dL juice which results in the ice cube being melted. By doing this, the juice drops to a temperature of 3.00 °C. What was the temperature of the juice before the ice cube was added? (Treat juice as water in this assignment. The density for ice is 920
    kg/m3
    . Assume no energy is lost to the surroundings).

    ice
    m1=(920)(8E-6) = 7.36E-3
    T1=0 C

    juice=water
    m2=(1000)(0.3) = 0.3kg
    T0water=?

    mixture
    Tfinal=3°C


    2. Relevant equations

    Q=mcΔT
    H=mLf


    3. The attempt at a solution
    Energy gained by ice = Energy lost by water

    (Ice → Water0C) + (Water0C → Water3C) = Waterinitial → Water3C

    m1Lf+m1c(T3-T0) = m2c(Tinitial-T3)
    m1[Lf+c(T3-T0)]/m2c=Tinitial-T3

    I am not sure whether in the case of water ΔT is (Tfinal-Tinitial) or vice versa.
    I get either a negative temperature, which is bs, or after swapping them around Tinitial= 5°C which I do not quite believe, because who would want to cool down water by 2°C?!

    Please help, I am on the verge of questioning my sanity after spending more than an hour on this...
    (One year ago it would've been no problem for me, but after a year of little to no physics my brain went ..dumb :D)
     
  2. jcsd
  3. Apr 15, 2016 #2

    BvU

    User Avatar
    Science Advisor
    Homework Helper
    Gold Member

    There are a c and a Lf in this calculation; what are the numerical values you use ?
    And yes, one cube for 3 dl isn't much...
     
  4. Apr 15, 2016 #3

    CWatters

    User Avatar
    Science Advisor
    Homework Helper

    Method looks ok. Show your working.
     
  5. Apr 15, 2016 #4
    values from google or specific tables
    cwater= 4180 J/KgK
    Lfice= 334E3 J/Kg

    Working? Like substituting in the numbers? I mentioned the results I got after having substituted. One was negative, close to -1 and the other was 5 after swapping the final and initial in
    Can you please focus on the question whether in the case of water/juice ΔT is (Tfinal-Tinitial) or (Tinitial-Tfinal)

    http://www4c.wolframalpha.com/Calculate/MSP/MSP28701i007f2a3h8b607b000022ab0bhbac85297h?MSPStoreType=image/gif&s=30 [Broken] =5.03
    http://www4c.wolframalpha.com/Calculate/MSP/MSP7451i161d717h0b1gbe00004g3egf9d7604ce93?MSPStoreType=image/gif&s=61 [Broken] =-0.97
    editable: http://goo.gl/dxxaum
     
    Last edited by a moderator: May 7, 2017
  6. Apr 15, 2016 #5

    haruspex

    User Avatar
    Science Advisor
    Homework Helper
    Gold Member
    2016 Award

    As BvU observes, that is not much ice for 300mL of drink.
    Make a decision, which makes sense to you? Remember ΔQwater=-ΔQice.
     
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook

Have something to add?
Draft saved Draft deleted