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!

Grade 11 Heat Capacity question

  1. Dec 11, 2011 #1
    1. The problem statement, all variables and given/known data

    a 0.20kg lead shot is heated to 90 degrees celsius and dropped into an ideal calorimeter containing 0.50kg of water initially at 20 degrees celsius. What is the final equilibrium temperature of the lead shot? The specific heat capacity of lead is 128 Joules/(kg degrees celsius) and the specific heat capacity of water is 4186Joules/(kg degrees Celsius)

    Ml = 0.20
    Cl = 128
    Til = 90

    Mw = 0.50
    Cw = 4186
    Tiw = 20



    2. Relevant equations

    Q= mass x heat capacity x difference in temperature

    Qabsorbed + Qreleased = 0

    3. The attempt at a solution

    0.20 x 128 x (T2 - 90) + 0.50 x 4186 x(T2 - 20)

    25.6 x (t2 -90) + 2093 x (T2-20)

    I am completely stuck after this. I don't know what to do with the difference in temperature part. I've looked at example equations and it shows the temperature given getting transformed into joules? So in this equation the 90 and 20 would be replaced by different numbers with the unit Joules after. I don't understand why this is happening. Thank you for any help and I appreciate you taking the time to read this.

    This may not be very clear, so if I need to explain more please let me know
     
    Last edited: Dec 11, 2011
  2. jcsd
  3. Dec 11, 2011 #2

    gneill

    User Avatar

    Staff: Mentor

    Make an equation out of it as in your Relevant Equations section. Solve for T2.

    When units (Joules, °C, etc.) get confusing, take your equation and substitute each item with its corresponding units. Check how the units cancel and combine. They should all 'harmonize' to produce the expected units that are on the other side of the equation.
     
  4. Dec 11, 2011 #3
    I tried rearranging it to solve for T2, but I'm definitely doing something wrong.

    0.20 x 128 x (t2 - 90)

    -T2 = 0.20 x 128 - 90

    -t2 = 25.6 - 90

    -t2 = -64.4

    T2 = 64.4

    This is one of those multiple choice answer things, and this isnt an option sadly. I understand the whole unit thing, I'm just pressed for time and trying to shorten it. I really appreciate the help thank you

    EDIT: I realized i should factor this, so I did it and I got the correct answer of 20.8 degrees celsius. I know this is kind of random and im not explaining myself, but you really helped and I got the right answer. Thank you very much, you have no idea how much i appreciate it
     
    Last edited: Dec 11, 2011
  5. Dec 12, 2011 #4

    Delphi51

    User Avatar
    Homework Helper

    May I offer a little clarification at the beginning?
    Heat lost by lead = heat gained by water
    mC*ΔT = MC*ΔT
    0.2*128*(90-T2) = .5*4186*(T2-20)
    This differs from your solution in that there is an equals sign in the middle
    and your (T2-90) is replaced by (90-T2). (T2-90) would be a negative number, leading to all sorts of trouble!

    Next step is to expand both sides out so you only T2 terms and numerical terms.
    Collect like terms and solve for T2.
    Keep the = sign in every line!
     
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook




Similar Discussions: Grade 11 Heat Capacity question
  1. Grade 11 lenses question (Replies: 15)

Loading...